Problems and Solutions to Chemical
Engineering Principles
化工原理教研组编
Chapter 1 Fluid Mechanics
1.The flame gas from burning the heavy oil is constituted of 8.5%CO2，7.5O2，76%N2，8%H2O(in volume).When the temperature and pressure are 500℃and 1atm, respectively, calculate the density of the mixed gas .
Solution: The molecular weight of the gaseous mixture Mn is
Mn=My+ My+ My+ My
=44×0.085+32×0.075+28×0.76+18×0.08
=28.86kg/kmol
Under 500℃,1atm,the density of the gaseous mixture is
ρ==×=0.455kg/m3
2.The reading of vacuum gauge in the equipment is 100mmHg,try to calculate the absolute pressure and the gauge pressure, respectively. Given that the atmospheric pressure in this area is 740mmHg.
Solution: The absolute pressure in the equipment is equal to that atmosphere pressure minus vacuum
P(absolute)=740―100
=640mmHg
=640×=8.53×10N/m2
The gauge pressure=-vacuum =-100mmHg
=-（100×）=―1.33×10N/m2
or the gauge pressure=-（100×1.33×10）=―1.33×10N/m2
3.As shown in the figure, the reservoir holds the oil whose density is 960kg/m3. The oil level is 9.6m higher than the bottom of the reservoir. The pressure above the oil level is atmospheric pressure. There is a round hole(Φ760mm )at the lower half of the sidewall, the center of which is 800mm from the bottom of the reservoir.
The hand-hole door is fixed by steel bolts (14mm). If the working stress of the bolts is 400kgf/cm2, how many bolts should be needed ?
Solution: Suppose the static pressure of the liquid on 0-0 level plane is p, then p is the average pressure of the liquid acting on the cover.
According to the basic hydrostatics equation
p=p+ρg h
The atmosphere pressure which acts on the outer flank of the cover is pa, then pressure difference between the inner flank and outer flank is
Δp=p―p= p+ρgh―ρgh
Δp =960×9.81(9.6 ―0.8)=8.29×10N/m2
The static pressure which acts on the cover is
Δp×=8.29×10N/m2
The pressure on every screw is
the Number of screw=3.76×10=6.23
4. There are two differential pressure meter fixed on the fluid bed reactor, as shown in the figure. It is measured that the reading are R1=400mm，R2=500mm, respectively. Try to calculate the pressures of points A and B.
Solution: There is a gaseous mixture in the U-differential pressure meter. Suppose are the densities of gas, water and mercury, respectively, then the pressure difference could be ignored for 《. Then
According to the basic hydrostatics equation
=
=1000×9.81×0.05+13600×9.81×0.05
=7161N/m2
=7161+13600×9.81×0.4=6.05×10N/m
5. As shown in the figure, the manometric tubes are connected with the equipment A , B, C, respectively. The indicating liquid in the tubes is mercury, while the top of the tube is water, water surfaces of the three equipments are at the same level. Try to determine:
Are the pressures equal to each other at the points of 1,2,3?
Are the pressures equal to each other at the points of 4,5,6?
If h1=100mm，h2=200mm，and the equipment A is open to the atmosphere(the atmospheric pressure is 760mmHg),try to calculate the pressures above the water in the equipment B and C.
Solution: 1) The pressure is different among 1, 2 and 3. They are at the same level plane of static liquid, but they don’t connect the same liquid.
2) The pressure is the same among 4, 5 and 6. They are at the same level plane of static liquid, and they connect the same liquid.
3)
then
=101330―（13600―1000）×9.81×0.1
=88970N/m2
or =12360N/m2（vacuum）
and because
then
so
=101330―（13600―1000）×9.81×0.2
=76610N/m2
or 24720N/m2（vacuum degree）
6. As shown in the figure, measure the steam pressure above the boiler by the series “U”-shape differential pressure meter, the indicating liquid of the differential pressure meter is mercury, the connected tube between the two “U ”-shape meters is full of water. Given that the distance between the mercury levels and the referring level are h1=2.3m，h2=1.2m，h3=2.5m，h4=1.4m respectively .The distance between the level of the water in the boiler and the base level is h5=3m. The atmospheric pressure Pa is 745mmHg.Try to calculate the vapor pressure P above the boiler.(in N/m2，kgf/cm2 respectively.)
Solution: Choose 2, 3 and 4 in the U-pipe of series connection as referring level. According the basic hydrostatical principle, start from 2, we can get the equation of every basic level. Then we can get the pressure of the water vapor.
or
or
or
or
From the above equations, we can get
so
×101330+13600×9.81[（2.3―1.2）+（2.5―1.4）]
―1000×9.81[（2.5―1.2）+（3―1.4）]
=364400N/m2
or =364400/9.807×10=3.72kgf/cm2
7. Based on the reading of the differential pressure meter as shown in the figure, calculate the gauge pressure of the gas in the tube line. The indicating liquids in the differential pressure meter are oil and water, respectively. the density of which are 920kg/m3 and 998kg/m3, respectively. The distance of the water and oil interfaces in the “U”shape tube R is 300mm.The inner diameter of the reservoir is 60mm,and the inside diameter of the “U”-shape tube is 6mm.When the gas pressure in the tube line is equal to the atmospheric pressure, the liquid level is flush with each other.
Solution: If the barometric pressure in the pipe equals the atmosphere pressure, then the liquid levels in the two enlarging room are at the same level. Then the relation between the enlarging room and differential pressure meter is
When the value of differential pressure meter R=300mm, the difference of liquid level between the two enlarging rooms is
Δh=R
Suppose are the densities of water and oil respectively, according to the basic hydrostatical principle
then the pressure of gas in the pipe is
p=（998―920）×9.81×0.3+920×9.81×0.003=257N/m2
8.The tube bundle of the tubular heat exchange is constituted of 121 steel tubes(Φ25×2.5mm).The air flows in the tube bundle at 9m/s.The average temperature of the air in the tube is 50℃,the pressure is 2kgf/cm2(gauge pressure).The local atmospheric pressure is 740mmHg. Try to calculate :
The mass velocity of the air;
The volume flow rate of the air in the operating condition;
The volume flow rate of the air in the standard condition.
Solution: 1) the density of air is 1.293kg/m3
pressure in operating
N/m2
the density of air under the operating condition
=1.293×
the mass flow rate of gas is
2) the volume flow rate of gas under the operating condition is
3) he volume flow rate of air under the standard condition is
9. The gas at the average pressure of 1atm flows in the pipe (Φ76(3mm).When the average pressure changes to be 5atm,if it is required that the gas flows in the tube at the same temperature , rate and mass velocity, what’s the inside diameter of the tube?
Solution: Suppose the subscribe 1 as the state under 1atm and subscribe 2 as the state under 5atm.
In the two cases
because
so
then
10. As shown in the figure, the feed liquid whose density is 850kg/m3 is sent into the tower from the elevated tank .The liquid level of the elevated tank keeps constant. The gauge pressure in the tower is 0.1kgf/cm2,and the feed rate is 5m3/h. The connected pipe is steel pipe(Φ38(2.5mm),the energy loss in the connected pipe of the feed liquid is 0.1kgf/cm2(the energy loss in the exit is not included).What is the distance between the liquid level of the elevated tank and the feed inlet?
Solution: Suppose the liquid level of header tanker as the upper reaches, and the inner side of the connecting pipe as the lower reaches. And suppose section1-1’ as basic level. We can get the equation
in the equation
therefore we can get that
the liquid level of header tanker should be 4.37m higher than the orifice for raw stuff.
11. The liquid level of the elevated tank is 8m higher than the floor. The water flows out of the pipeline(Φ108(4mm).The exit of the pipeline is 2m higher than the floor. In the given condition, the energy loss of the water flowing through system (the energy loss of the exit is not included )can be calculated byΣhf=6.5m2,where u is water velocity(in m/s).Try to calculate :
The velocity of the water at the “A--A” cross section;
The flow rate of water(in m2/h)
Solution: 1) Suppose the liquid level of header tanker as the upper reaches, and the inner side of the pipe’s exit as the lower reaches. Suppose the ground as basic level. Then we can get the equation
in the equation
based on the above，we can get
because the pipes’ diameters are the same, and water density can be considered as constant，so the velocity at section A-A
2）volume flow rate of water
12. The water at 20℃ flows through the horizontal pipe(Φ38(2.5mm) at 2.5m/s.The pipe is connected with the another horizontal (Φ53(3mm)through a conical pipe. As shown in the figure, two vertical glass tubes are inserted in either side of the conical pipe. If the energy loss through sections A and B is 1.5J/kg,try to calculate the distance between the two liquid level of the glass tube(in mm)? Draw the relative position of the liquid level in the figure.
Solution：From section A——A’ and section B——B’，then the Bernoulli equation is
in the equation
according to the continuity equation for the incompressible fluid，then
so
the pressure difference between the two sections is
=（
then =868.5/9.798=88.6mmH2O
the liquid level’s difference between the two pipes is 88.6mm。
because
so
the pipe’s liquid level B is higher than the liquid level-A.
13. The water at 20℃ is sent to the top of the scrubber from the reservoir by the centrifugal pump. The liquid level of the reservoir keeps constant, as shown in the figure. The diameter of the pipeline is Φ76(2.5mm.In the operating condition ,the reading of the vacuum gauge in the pump entrance is 185mmHg.The energy loss of the suction tube (The entrance of the pipe is not included)and the discharge tube can be calculated according to formulaΣhf，1=2u2 and Σhf?2=10u2,where u is water velocity in the suction and discharged tubes. The gauge pressure of the joint of the discharged tube and the spray nozzle is 1kgf/cm2. Try to calculate the effective power of the pump.
Solution: Suppose liquid level of storing tank as section 1---1, the connection of vacuum meter as section 2---2. And suppose section 1---1 as basic level. Then the Bernoulli equation is
in the equation
From the above，we can get the velocity of water in the pipe
the water mass flow rate
2）pump’s effective power
Suppose the liquid level of storing tank as upper reaches section 1---1; suppose the connection point of drain pipe and blow head as lower reaches section 3---3; Suppose section 1---1 as basic level.
Then we get the equation
in the equation
from the data above, we get
the pump’s effective power is
14. As shown in the figure, the inside diameter D of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm.There is no liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m.The total energy loss of the liquid flowing through the pipe can be calculated by the formula Σhf=20u2,where u is the velocity of liquid (in m/s).Try to calculate the required time when the liquid level of the reservoir drops 1m?
Solution：The problem attributes to the instability flow，the time when the surface of the reservoir descend 1m can be calculated from material balance.
Suppose F’ is instant feed into system，D’ is instant discharge out of system，dA’ is the accumulated quantity in the dθ time，then in the dθ time, the material balance equation is
F’dθ―D’dθ=dA’
And then suppose in the dθ time，the surface of reservoir descend dh，the instant flow rate of liquid in the tube is u。
In the equation F’=0
D’=
Then the equation reduces to
or
In order to work out the relation between the height of instant surface h（take the center line of drainpipe as benchmark）and the instant velocity u in equation（a），we can use the instant equation of Bernoulli.
In the equation
substitute the above values into formula，we have
9.81h=20
Put formula b into formula a，
=―5580
integration range as flollows
θ=―5580×2
=5580×2
15． A cooling cycle system of brine is shown in the figure. The density of the brine is 1100kg/m3,and the circulation is 36m3/h.The diameter of the pipeline is the same. The energy loss from A and two heat exchangers to B is 98.1J/kg,the loss from B TO A is 49J/kg.Try to calculate:
If the efficiency of the pump is 70%, what is the shaft power of it?(in kw)
If the reading of the gauge in position A is 2.5kgf/cm2,what is the reading of the gauge in position B?
Solution：1）The efficiency of the pump
take whichever section as 1-1 and 2-2. it means liquid flow starts from section1-1 and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be
since
The equation can be simplified to
The mass flow rate is
Then, the power of the pump is
2）The reading of the gauge in position B
Set up Bernoulli equation between section A and section B，and take the center of the section A as the horizon.
In the equation
input the data into the Bernoulli equation
（gage pressure）
The reading of the gauge in position B
16.In a laboratory, the acetic acid (70%) at 20℃ is sent by the glass pipe. The inside diameter of the pipe is 1.5m, and the flow rate is 10kg/min. Calculate Reynolds number in SI system and physical unit system, respectively. What is the flow pattern?
solution：1）In SI system
From appendix 17 of the book, for the acetic acid (70%), (20℃)，
d=1.5cm=0.015m
=5657 (belong to the overfall)
2）In physical unit system
17. The liquid whose density is 850kg/m3, viscosity is 8cp, flows in the steel pipe (inside diameter is 14mm). The velocity of liquid is 1m/s. Try to find :
The Reynolds number, and the flow pattern is pointed out.
The position at which the local velocity is equal to the average one.
Suppose the pipeline is horizontal pipe, and the pressure of the upstream is 1.5kgf/cm2, How long the liquid has to flow when the pressure drop to 1.3kgf/cm2?
Solution：1）The Reynolds number
(belong to the overfall)
2）The position at which the local velocity is equal to the average one
According to the formula 1——38 and 1——39，
For the position at which the local velocity is equal to the average one，then
So r=0.707R=0.707×7=4.95mm
The position at which the local velocity is equal to the average one is 4.95mm。
3）The section of upstream is 1——1’，the section of downstream is 2——2’，because the pipeline is horizontal pipe , so
According to the formula of Hagon-Poiseuille，
then ,
18. As shown in the figure, an inverse “U”-shape differential pressure meter is fixed between the cross sections of the two horizontal pipes with different diameters in order to measure the pressure difference of the two cross sections. When the flow rate is 10800kg/h, the reading of the “U”-shape differential meter is 100mm.The diameters of the two pipe are Φ60(3.5mm and Φ42(3mm,respectively. Calculate :
The energy loss of 1kg water flowing through the two cross sections;
The pressure drop which is equal to the energy loss (in N/m2).
Solution：1）The energy loss of 1kg water flowing through the two cross sections
Set up Bernoulli equation between section 1——1’ and section 2——2’，and take the axis of pipe as horizon.
In the equation
input the data into the equation
2）
19. The liquid is flowing laminarly in the straight pipe. If the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original, how many times is the energy loss produced by resistance as much as the original?
Solution：According to the formula of Hagon-Poiseuille，
Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the viscosity of liquid and the length of the pipe have not changed.
From the problem, we know the ratio between these two diameters
Thus, if the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original，the energy loss produced by resistance has increase to 16 times as much as the original.
20. The liquid is flowing turbulently in the straight pipe. If the physical property of the liquid, the length of the pipe and the diameter are constant, but the flow rate is increased to twice of the original, how many times is the energy loss produced by resistance as much as the original? For two cases, Reynolds number is between 3(103~1(105, the friction coefficient λ can be calculate by the formula “Palasus”.
Solution: According to the frictional formula in the straight pipe,
Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the diameter of the pipe and the length of the pipe have not changed.
According to the formula，
The coefficients of friction in these cases
Because of the flow rate is 2 times as much as the original, then
The viscosity and the density of the liquid are invariability，therefore
=1/2
Upon that
21. The rectangular chimney (inside cross section is 1000(1200mm) is 30m high. The effluent gases at 400℃ is flowing from bottom to top, the average molecular weight of gases is 30kg/kmol.The bottom of the chimney keeps 5mm H2O (vacuum. The density of the air can be considered as constant in the range of the height of the chimney. The environmental temperature is 20℃, and the atmospheric pressure is 1atm .The frictional coefficient of fluid flowing through the chimney can be taken as 0.05, try to calculate the flow rate of the effluent gases (in kg/h).
Solution：Suppose the bottom of the chimney is the upstream section 1——1’，and the inside cross section of the top is 2——2’, and take section 1——1’ as horizon， then set up Bernoulli equation between two sections.
In this equation
the sections of the chimney are the same，and the pressure of gas in the chimney does not change a lot）
Because the pressure of gas in the chimney doesn’t change a lot, the temperature is 400℃, and the atmospheric pressure is 1atm.
Because the pressure on the top inside of chimney is equal to the pressure of air in the same height, so we have
The density of the air in the standard status is 1.293kg/m3，so when the environmental temperature is 20℃, and the atmospheric pressure is 1atm，the density of the air
upon that
then input the data into the equation，
therefore
The velocity of the flue gas is
The mass flow rate of the flue gas is
22. The liquid is sent to the elevated tank from the reactor by pump at 2(104kg/h (as shown in the figure). The pressure keeps 20mmHg (vacuum) above the surface of the reactor, and the pressure above the surface of the elevated tank is atmosphere. The pipeline is steel pipe(Φ76(4mm), and its length is 50m.There are two wide open brake valves, a orifice meter (the resistance coefficient is 4), five standard elbows in the pipeline. The distance from the surface of the reactor to the exit of the pipeline is 1.5m. If the efficiency of the pump is 0.7, try to calculate the shaft power of the pump. It is given that the density of the liquid is 1073kg/m3, the viscosity is 0.63cP, and the absolute roughness of pipe wallε is 0.3mm.
Solution：Set up Bernoulli equation between the surface of reactor1——1’ and the inner surface of pipe 2——2’, and take section 1——1’as referring section
In this equation
input the data into the equation and we will get
According to the value of Re and ，check the fig.1——24 and find the coefficient of the friction ，and from the fig.1——26, we can find that the equivalent lengths of the pipe and the valve are
Brake valve（wide open） 0.43×2=0.86m
Standard elbow 2.2×5=11m
So
The power of the pump
23. There are a few solvents in the exhaust gas discharged in the equipment. Before vented to the atmosphere, the gas is sent through a washing tower in order to recover the solvents and avoid the environmental pollution. The flow rate of gas is 3600m3/h (under the operating condition), and the physical property of gas is almost the same as that of air at 50℃. As shown in the figure, a “U”-shape differential pressure meter whose indicating liquid is water is fixed in the pipeline before the gas entering the blower. The reading of the meter is 30mm.The inside diameters of both the delivering pipe and the exiting pipe are 250mm.The equivalent length of the pipe line, the fitting, and the valve is 50m (The resistance of entering and leaving the tower are not included), the vertical distance between the air exit and the entrance of the blower is 20m. Suppose the pressure drop of the gas through the filler layer in the tower is 200mm H2O, the roughness of the pipe wall ε can be taken as 0.15m and the atmospheric pressure can be considered as 1atm, Try to calculate the effective power of the blower.
Solution： choose the joint of the differential manometer at the entrance of the air blast as section 1——1’and medial wall of the anti-aircraft pipe as section 2——2’, the Bernoulli equation between the two sections .
In this equation
Because the pressure of the air does not change a lot inside the system, the density of air can be calculated under the conditions of 1atm and 50℃，
input the data into the equation and we will get
=
at 1atm and 50℃, the viscosity of air μ=1.96×
check the fig.1——24 and find λ=0.019
The effective work of the blower:
The mass flow rate of the liquid
The effective power of the blower
24. As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe(inside diameter is 100mm) is connected to the bottom of the reservoir.There is a gate valve fixed in the pipiline, and there is a “U”-shape differential meter whose indicating liquid is water fixed in some position 15m away from the entrance of the pipeline, one part of the meter is connected to the pipe, and the other is connected to air. The connected tube of the differential pressure meter is full of water. The length of the straight pipe between the pressure measurement point and the exit of the pipeline is 20m.
When the gate valve is closed, R=600mm，h=1500mm; when the gate valve is opened partially, R=400mm，h=1400mm.The friction coefficient λ is 0.025, and the local resistance coefficient of the pipeline entrance is 0.5. What is the flow rate (in m3/h)?
When the gate valve is wide open, what is the static pressure in the pressure measurement point (in gauge pressure, N/m2). It is given that le/d≈15, and the friction coefficient λ is still 0.025.
Solution：(1) When the gate valve is opened partially, the water discharge is
Set up Bernoulli equation between the surface of reservoir 1—1’ and the section of pressure point 2—2’，and take the center of section 2—2’ as the referring plane, then
（a）
In the equation (the gauge pressure)
We can get the value of Z2 when the gate valve is wide open. When the gate valve is fully closed, the water is standstill. We can get the following based on the hydrostatic equation.
（b）
where h=1.5m
R=0.6m
Input the known data into equation d
input the above data into equation a
9.81×6.66=
the velocity is u=3.13m/s
the flow rate of water is
2） the pressure of the point where pressure is measured when the gate valve is wide-open.
Set up Bernoulli equation between the surface of reservoir 1—1’ and the section of the inner side of the pipeline 2—2’，and take the center of the pipe as the referring plane, then
（c）
since
input the above data into equation c，
9.81
the velocity is: u=3.51 m/s
Set up Bernoulli equation between the surface of reservoir 1—1’ and 2——2’, for the same situation of water level
（d）
since
input the above data into equation d，
9.81×6.66=
the pressure is:
25. The coal gas in the gas tank is sent through a 40m steel horizontal pipe to an ordinary pressure equipment. The gauge pressure in the gas tank is not lower than 62mmH2O.There is a gate valve in the pipeline (le/d≈15),the local resistance coefficients of the entrance and the exit of the pipeline are 0.5 and 1,respectively.It is given that the average density of the coal gas is 0.75kg/m3 and the viscosity is 0.015cP under operating conditions. It is required that the flow rate of gas is 10000m3/h.Try to choose a steel pipe which has a suitable diameter. The roughness of the pipe wall ε is 0.2mm.
nominal size，mm
350
400
450
500
600
Outside diameter，mm
377
426
478
530
630
thickness of the wall，mm
6
6
6
6
6
8
8
8
8
8
——
——
——
10
10
Solution： Set up the Bernoulli equation between the section 1—1’ which is the surface of the entrance of the pipeline and the section of the outlet of the pipeline 2—2’，and take the center of the pipe as the referring plane, then
In the equation
(the gauge pressure)
input the above data into the equation and simplify it
Because of so
In the above equation,λis a function of u. So we have to use the error-trial method. Suppose the velocity of coal gas is 20m/s，then we can use the formula to calculate the inside diameter of the pipe
choose pipe size of φ426×6mm , the inside diameter of pipe d=426-2×6=414mm。
The practical velocity in the pipe is
According to the value of Re and ，and check fig.1-24 in the book，we can findλ=0.018，
Input the values of u and λinto equation a
it illustrations that the φ426×6mm is available.
26. Water at 20℃ passes through a steel pipe whose inside diameter is 300mm. In order to measure the flow rate of water, the arrangement as shown in the figure is adopted. There is a branch pipe(Φ60(3.5mm) which is parallel with a 2m main pipe. The total length including all local frictional loss is 10m.There is a rotary meter fixed in the branch pipe. From the reading of the meter, we know that the flow rate of water is 2.72m/h. Try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficients of the main pipe and the branch pipe are 0.018 and 0.03, respectively.
Solution：This problem is attributed to the parallel pipeline，take subscript1 as main pipe, and subscript 2 as branch pipe. The friction loss for parallel pipelines is
The energy loss in the branch pipe is
In the equation
input the data into equation c
The energy loss in the main pipe is
So
The water discharge of main pipe is
Total water discharge is
27.Water supply system is shown in the figure. The water surface of the elevated tank keeps constant. Water is discharged form branch pipes BC and BD, respectively. The distances between the water surface of tank and the exits of two branch pipes are both 11m.The inside diameters of pipeline AB, BC and BD are 38mm, 32mm and 26mm, respectively. .And the length of pipeline AB, BC and BD are 58m, 12.5m and 14m, respectively. The frictional coefficient λof pipeline AB and BC are both 0.03. Try to calculate :
When the valve of branch pipe BD is closed, what’s the max water discharge of branch pipe BC(in m3/h)?
When all the valve are wide open, what is the water discharge of the two branch pipes (in m3/h)? The roughness of branch pipe BD can be considered as 0.15m.The density of water is 1000kg/h, and the viscosity of it is 1Cp.
Solution： (1)When the valve of branch pipe BD is closed, the max water discharge of branch pipe BC will be found as follows
Set up the Bernoulli equation between the surface 1--1’ and the section C--C’, and take section C--C’ as the referring plane，so
in the equation
hence, （a）
（b）
（d）
（e）
input equation e into equation c，
（f）
input equation f,d into equation b
=2.45m/s
2)when all valves are wide open，the water discharge of the two branch pipes
According to the frictional equation of parallel pipelines，we
The exits of two branch pipes are in the same level, we can simplify the equation, Choose the outboard of the two additional pipes as the lower reaches
（a）
input the value of
（b）
the relation of the flow rate of the main duct and the additional pipe is
rearrange the above equation
（c）
Set up the Bernoulli equation between the surface 1--1’ and the section C--C’, and take section C--C’ as the referring plane，we have
since
simplify it
therefore （d）
in equation b and c ，
the error-trail method can be used to find the velocity
based on the value of Re and we get λ=0.034，input the value of λand into equation b，we have
=1.79m/s
input the value of ，into equation c，
=0.708×1.79+0.469×1.45=1.95m/s
input the value of ，into equation d
23.15×1.952+6.36×1.792=108.4
the result is consistent with the assumption （108.4≈107.9），so is excepted，and the water discharge of the two pipes are
28．A standard orifice meter is fixed in the pipe (Φ38(2.5mm).The diameter of the orifice meter is 16.4mm.The toluene at 20℃ flows in the pipe. The pressure difference between both sides of orifice is measured by a “U”-shape pressure meter, and the indicating liquid is mercury. The connected tube is full of toluene. The reading of the gauge is 600mm. Try to calculate the flow rate of the toluene (in kg/h).
Solution：The diameter of the orifice meter and the inside diameter of pipe
Suppose and check the fig.1-30 in the book, and find the C0=0.626
From appendix 17 of the book, the density of toluene at 20℃ is 866kg/m3, and the viscosity at 20℃ is 0.6×10N2s/m2。Thus, the velocity of the toluene on the orifice is
The mass flow rate of toluene is
Examine Re
The velocity is：
So the assumption is right.
Chapter 2 Fluid Transportation
1.1 water is used in the experiment to measure the property of pump ,as the flow rate is 26m3/h,the reading of piezometer at the exit of the pump is 152kpa and the reading of vacuum gauge at the entry of the pump is 24.7kpa,shaft power is 2.45kw, rotational speed is 2900/min，respectively. If the vertical distance between the vacuum meter and piezometer 0.4m, the pipeline diameters of entry and exit of pump are the same. The frictional resistance between two gauges is negligible. Calculate the efficiency of pump and performance of the pump under this efficiency.
Solution: set up the Bernouli equation between the surface of the vacuum meter 1-----1’ and piezometer 2-------2’
since
total heads of pump is
efficiency of pump is
since
N=2.45kW
Therefore we can get
The performance of pump is
Flow rate ，m3/h 26
Total heads，m 18.41
Shaft power ，kW 2.45
Efficiency ，% 53.1
1.2 Hot water at 60oC in reservoir is sent by a centrifugal pump at the flow rate of 40m3/h to the top of cooling water tower , then is sprayed by a sprayer and drops into a cooling reservoir，so that the hot water is cooled. Given the piezometer maintains 49kpa before water comes into the sprayer, the inlet of sprayer is 6m above the level of hot water in the reservoir. The total head loss of inhalation pipeline and vent pipeline is respectively 1m and 3m, and kinetic head in pipeline can be negligible. Try to select a fitting centrifugal pump and ascertain the installation height. Suppose the local atmosphere pressure is 101.33kpa.
Solution:Set up the Bernouli equation between the level of hot water in the reservoir 1-----1’ and the nozzle of sprayer2-------2’
Since Z2─Z1 =0.5 P2－P1=49kpa=4.9104pa
Hf,1-2=4m
So we can get 0.5+4.9104981/1000 +4─H=0
total heads of pump is H=0.5+4.9+4=9.4m
since H=He=9.4m Qe=40m3∕h
Since pure water is carried ,we can choice B type water pump .
From characteristic curves of water pump series 2-27chart B type,we can choice 3B19A type water pump
From addenda 25 ,we can get N=2.88kw η=80% Hs’=5.5m
The temperature of water is 60oc, the pressure is19918.0pa
we can get Hg==[Hs’+(Ha-10)-(-0.24)]
=[5.5+(10-10)-(-0.24)]
=3.77m
1.3 Oil products are filled in a groove under the normal pressure, the density is 760kg/m3, absolute viscosity less than 20cp, under the operating condition the vapor pressure is 80kpa, now oil products would be sent by 65y—60B oil pump into an equipment in which the gauge pressure is 177kpa . The level of oil in the groove is invariability .The entry of the equipment is 5m above the level of oil in the groove. The total heads loss of inlet pipeline and outlet pipeline is respectively 1m and 4m. Try to calculate whether the pump can work.
If the oil pump is 1.2m below the level of oil in the groove, whether the pump can operate normally? Suppose the local atmosphere pressure is 101.33kpa.
Solution: In order to calculate whether the pump can be operated, we should first count total heads under transportation assignment by condition of subject. Then according to Qe, He which are compared with Q, H provided by pump, we can ascertain whether the pump can be operable.
Following the table of textbook, we can get 65y----60B oil pump capability as follows
Flow rate，m3/h 19.8
Total head，m 38
Shaft power，kW 3.75
Efficiency，% 55
Cavitation，m 2.6
set up the Bernouli equation between the level of the groove 1-----1’ and outer section of transportation pipe, we can get
since
Substitute the above into the equation and we have
because He < H=38 m
and Qe=15 m3/h < Q=19.8 m3/h
65Y—60B oil pump is operable .
In order examine whether the pump operation is normal under the given condition, we should calculate whether the maximum suction lift of pump (Hg) is fitting. Based on the Bernouli Eq. For the liquid level and pump inlet, we have
since:
by liquid density and vapor pressure, look up chart 21 in textbook, we can get =1。Therefore
so
actual suction lift is ―1.2m < ―0.74m, this means that the pump can operate normally .
1.4 An experiment of some centrifugal pump is done and water is transported from a vessel to a reservoir, and the following experimental data can be obtained:
Q,1/min
0 100 200 300 400 500
H, m
37.2 38 37 34.5 31.8 28.5
If the pipeline size is as follows: pipe diameter Ф76(4mm, its length 355m (Including the equivalent length of local friction loss). The vertical distance between the vessel and reservoir levels of liquid at atmosphere pressure is 4.8m, the frictional factor is 0.03, try to calculate the flow rate under the operation. If the vessel is filled with air at 129.5kpa (gauge pressure) and then calculate the flow rate of the pump under this condition. The property of the transported liquid is the same as that of water.
Solution: as a pump is operating, the flow rate is decided by working point of centrifugal pump. Draw pump characteristic curves H vs Q and pipeline characteristic curves in the same coordinate picture, the meeting point of two curves is working point.
Pipeline characteristic equation can be gotten by Bernouli Eq, namely:
（a）
since △Z=4 △ △=0
the velocity of liquid in pipeline is:
λ=0.03
Substitute the above data into equation (a), we can get the pipeline characteristic curves as follows: (b)
Based on the above formula we can figure out total heads at different flow rates.
Qe，l/min
0 100 200 300 400 500
He，m
4.8 6.48 11.52 19.92 31.68 46.8
The experimental data are pictured with pipeline characteristic curves and point M of pipeline and centrifugal pump characteristics curves is working point. From the figure, we can get flow rate therefore flow rate is 400i/min, or is 24m3/h.
As the pressures of the vessel and reservoir are different,, the pipeline characteristic curve varies, and the flow rate transported is changed. For this condition, we have
=
Because other condition is fixed, equation b changes to
According to the above equation, pipeline characteristic curves He vs Qe, is pictured. The meeting point of the two lines (pipeline and pump characteristic curves) in the figure of H vs Q is working point. Under this condition, pump flow rate is 310, namely 18.6m3/h。
1.5 Water is sent by two centrifugal pumps from a reservoir into a exalted groove, each pump characteristic equation is :
H=25-1ⅹ106Q2
And the pipeline characteristic equation can be expressed approximately as:
He=10+1ⅹ105Q2e
In the above two equations, the unit of Q is m3/s, and H is m.
Try to assembled two pumps so that we can get the most transfusion flux?(the water transportation is at steady state )
Solution: If two pumps are parallel and characteristic curves are invariable,characteristic curves are converted to:
H=25-1106()2
He1=10+1105QE2 (A)
H1=25-1106()2 (B)
From simultaneous equations(A) ,(B) and let H1=He1 Q1=Qe1
We can get H1=22m Q1=0.34610-2 m3/s
If two pumps are in series and characteristic curves are invariable, characteristic curves are converted to:
=25-1106Q22
He2=10+1105Qe22 (C)
=25-1106Q22 (D)
From equations (C), (D)，let H2=He2 Q2=Qe2
We can get H2=11.9 m Q2=4.3610-2 m3/s
Since Q2>Q1, we know two pumps are in series is better so that we can get the most transfusion flux
1.6 A liquid whose density is 1250kg/m3 in a ringent reservoir is sent by a single action reciprocating pump into a tower in which the gauge pressure is 1.28ⅹ106pa , the level of liquid in the reservoir is 10m below the entry of the tower, the total heads loss of pipeline system is 2m.Given the piston diameter is 70mm,the stroke is 225m,the reciprocating time is 200l/min,the total efficiency and the cubage efficiency of the pump is 0.9 and 0.95, respectively. Try to calculate the actual flow rate, the total heads loss and the shaft power.
Solution: 1)The reciprocating times per one minute of this reciprocating pump.
the academic amount of the discharged liquid is:
by condition of subject, the actual amount of the discharged liquid is:
so
2) The shaft power of this reciprocating pump.
Total heads of pump can be gotten from Bernouli Eq. The level of liquid in the resevoir is the upriver section 1--------1’,and the outlet of transportation pipeline is the downriver section 2----------2’, choose section 1------1’as the normal horizontal .
since
so
total heads of pipeline was provided by pump, so total heads of pump is:
H=64.2m
so the shaft power is:
1.7 Air at the temperature of 15oc is sent into a fan directly and then passes through a pipeline whose inside diameter is 800mm to the bottom of a stove in which the gauge pressure is 10.8kpa. The flow rate of air is 20000m3/h (counting at the state of inlet ), the absolute roughness degree of pipeline is 0.3mm. Now there is a centrifugal fan , its property is shown as follows:
Rotational velocity, r/min
Wind pressure, pa
Wind amount, m3/h
1450
12650
2180
Calculate whether fan can work? The local atmosphere pressure is 101.33 kpa.
Solution: set up Bernouli equation between the outer section of fan inlet 1-------1’and the outer section of pipe outlet 2---2’, we can get the wind pressure needed by pipeline system.
since
so
the viscosity of air at 106720N/m2 and 15oc is 1.79ⅹ105NS/m2.
So
From figure 1-24 in the textbook, we can get λ=0.0167
so
Wind pressure needed by pipeline is
+1.29×170.2=11000N./m2
namely
apply the following equation, substitute wind pressure HT’ of system with wind pressure HT at the standard state of fan sample,, namely
comparing the above results with the property of fan, flow rate in pipeline is Q=2000m3/h<21800m3/h, wind pressure needed by pipeline HT=1044mmH2O <1290mmH2O. So the fan can work.
1.8 Air is compressed by a three-stage compressor from 98.07ⅹ103pa to 62.8ⅹ105pa.Suppose the intermediate condenser can cool air to 20oc, which is sent to the next stage, all compression ratios are the same. Try to find:
1) Under the condition that all piston strokes and the reciprocating times are the same, calculate ratios of these cylinder diameters?
2) Academic power consumed by triple compressor (The process is adiabatic, the adiabatic compression index of air is 1.4 and counting based on 1kg fluid)?
Solution: 1)Ratios of all cylinder diameters
Assume ⅵ ⅶ ⅷ is respectively inspiration quantity of class 1,2,3, for temperatures of all inspiration are 20oc,so:
so
namely
Because the amount of academic inspiration is:
now the piston stroke and the reciprocating time in each stage are the same. so :
Hence, ratios of all cylinder diameters are:
2) Academic power consumed by triple compressor
and G=1kg
R’=8316 / 29=286.8J/(kg.k）
so W=286.8×293×
Chapter 3 Non- homogeneous phase matter series separation
1. “Fall the ball method” is used to measure the viscosity of liquid. A liquid is placed in the glass container. The steel ball of which the diameter measured as 6.35mm subsiding in the liquid 200mm need 7.32s, as known, the density of steel is 7900 kg/m3, and the density of the liquid is 1300 kg/m3, attempt to work out the viscosity of liquid.
Solution ：assume as the laminar flow ，According to the formula 3 －6
Then
It is known
Instead with the last formula
Check
2. Attempt to work out the subsiding speed of the raindrop in the 20℃ air of which the diameter is 1mm.
Solution ：
Using the friction pack numbers to count:
Check the picture 3 －3，
Then
Or solved with the formula of Alan ：
Check:
3. Global quartzose granule of 2650 kg/m3 dense freely subside in the 20℃, Attempt to count the maximal atom diameter obeying Stocks’ law and the minimal atom diameter obeying Newton’s low.
Solution: Instead the Stocks formula with Renault number in stagnating area, then can work out the maximal atom diameter of the Global granule with stagnant subsiding.
Getting the Renault number upper limit with , then
Instead formula 3 －6 with this formula
Solution
Check the addenda, 20 air density p=1.205 kg/m3
viscosity u=1.81*10^(-3)N·S/m^2
As so, instead the Newton formula with Renault number lower limit in onflow area, then can work out the minimal global atom diameter d with onflow subsiding.
Getting the Renault number lower limit as ，
Then
Instead it in formula 3 －8，
Solution
Instead with all the known value, then
4.A global granule with 2000 kg/m3 density freely subside in the water of 30 ℃,attempt to make a curve to show the relationship between the its subsiding speed and its diameter.
Solution: Find out the physical numbers of the water of 30 ℃ in addenda
P=995.7 kg/m3
U=80.12(10^(-5) N·S/m^2
According to formula (a) and (b) in the last exercise
Then can know the maximal subsiding speed in stagnating area
Minimal subsiding speed in stagnating area
Then obtain two series of opposite data, such as two spots on the plotting chart ：
According to the Stocks formula
According to Newton formula
On the grounds of the foregoing analysis, then can see, this chart ought to be protracted on the logarithm plotting paper. Take as the erect axis, the d as the horizontal axis. When d<dc , relationship ought to be a line across the point c, the slope is 2 ；When , relationship is a line across c′,the slope is 1/2 .
When the value is between, according to Alan formula to work out. For example ，Get D=1000um ，then
Get D=500um ，Piece may count as
so obtain 2 spots in transition:
point A ：d=1000um,ut=0.1162m/s
point B ：d=500um,ut=0.0527m/s
According to the foregoing account, make curve, as the curve CBAC′in the addenda
5. There is a multilayer dust clearing room used for removing the mineral dust of the furnace gas. The minimal diameter of the dust is 8um and the density is 4000 Kg/ m3. The inside of the room is 4.1 meters long, 1.8 wide and 4.2 tall. The gas temperature is 427 ℃，viscosity is 3.4(10^3N ·S/ m2, Density is 0.5 Kg/ m3. If the quantity of furnace gas is 2160 standard m3, attempt to fix the distance between clapboards and the number of layers.
Solution:
The distance between clapboards:
The number of layers:
Check ：
Renault number of dust declining:
Renault number of gas floating:
6. Attempt to infer formula 3-16a and formula 3-16b from formula 3-16.
Solution: (1) take 1m3 dregs to consider, in it the volume of solid should amount to its mass divide by its density, namely:
1.
Then
Equivalent，
Substitute formula 3 －16：
Also take 1m3 dregs to consider, in it the mass of solid is the product of its volume ratio and its density, and the mass of liquid is the product of its volume ratio and its density. So in the dreg the mass ratio of solid and liquid is:
Therefore
Equivalent,
Substitute formula 3 －16：
7. The calcareousness slurry with 236 kg/m3 solid concentration of which the viscidity will be enhanced to 700 kg/m3 in a series depositing trough. According to the intermittent deposit experiment, the relative data between the depositing speed of granule surface and the solid concentration in the slurry
C， Kg ( solid) /m^3 ( slurry)
236
358
425
525
600
714
U,m/h
0.157
0.05
0.0278
0.0127
0.00646
0.00158
As known, 3785 m3 slurry is needed to deal with everyday, attempt to fix the area of depositing trough.
Solution According to the formula 3 －16 the area of depositing trough is.
C,kg/m^3
，m/h
1/C,m^3/kg
1/C-1/Cc,
m^3/kg
*(1/C-1/Cc),m^2·h/kg
236
0.157
0.00424
0.00281
0.0179
236
0.157
0.00279
0.00136
0.0273
236
0.157
0.00235
0.000923
0.0332
236
0.157
0.00190
0.000475
0.0374
236
0.157
0.00167
0.000237
0.0366
236
0.157
0.00140
—
—
The moving in mass of stuff
The flux mass of solid
Then the area of depositing trough
Take the safe coefficient, Then
8. As known, in the gas including dust, the density of the dust is 2300 kg/m3, the flux of gas is 1000(m3)/h, the viscidity is 3.6(10^5N·S/ m2, the density is 0.694kg/ m2, adopting standard whirlwind separator to clear dust. If the diameter of separator is 400 mm, attempt to estimate it’s critical atom diameter, separate atom diameter and the drop of pressure.
Solution
D=0.4m
B=D/4=0.1m
h=D/2=0.2m
ui=Vs/hB=(1000/3600)/0.2(0.1=13.9m/s
According to the formula 3 －21：
According to the formula 3 －24：
According to the formula 3 －26：
9. As known in the last exercise, the gas round the entrance of the whirlwind separator of which the including dust concentration is 150g/ m2, the atom ratio is as the following
Atom diameter ，um
<3
3~5
5~10
10~20
20~30
30~40
40~50
50~60
>60
Mass ratio, %
3
11
17
27
12
9.5
7.5
6.4
6.6
Attempt to estimate the including dust concentration of the gas round the exit.
Solution first, cipher out the average atom diameter in the range of every levels, then using the solution of last exercise, according to the curve of the standard whirlwind separator（chart 3－11）cipher out the atom level efficiency of every atom, then can know the including dust concentration of the gas round the exit when working out the sum-efficiency according to formula 3－25.Make a list of efficiency cipher process as following:
Atom diameter ，um
Average atom diameter D2
Mass ratio,%
<3
2
3
0.349
0.1
0.3
3~5
4
11
0.698
0.35
3.85
5~10
7.5
17
1.301
0.60
10.2
10~20
15
27
2.618
0.86
23.22
20~30
25
12
4.363
0.96
11.52
30~40
35
9.5
6.108
0.98
9.31
40~50
45
7.5
7.853
0.99
7.43
50~60
55
6.4
9.599
1.0
6.4
>60
6.6
1.0
6.6
Sum-efficiency
The including dust concentration of the gas round the exit
10.The including dust concentration of the gas that is cleansed by a whirlwind separator is 0.7, the flux of the gas is 5000 standard, and the mass of collecting dust is 21.5. The mensuration of dust granularity in this cleansed gas or has been collected showed in the following table.
Attempt to work out:
The sum efficiency of clearing the dust;
Make the atomic level efficiency curve.
Solution: (1) The sum efficiency of clearing the dust
The mass flux of the dust in the gas round the entrance:
The mass flux of the dust in the gas round the exit:
Then the sum efficiency is:
Atomic level efficiency curve
Atomic level efficiency
According to the data of dust granularity distributing as known, take use of the last formula to work out the mass of the atoms in various size that is collected. The chart as following:
According to the data in the second and sixth row in the chart, protract on the common plotting paper, then can get the atomic level efficiency curve, see the addenda.
Chapter 4 heat transfer
1.As shown in the figure, measure the heat transfer coefficient of the solid by “flat plate method”。Heat up one side of the flat by the electric heater, and cool the other side of the flat by the cooler. At the same time, the thermocouple is used to measure the surface temperature of both sides of the flat. The surface area of the solid is 0.01m2, and the thickness is 20mm.If the reading of ampere meter is 0.58A, the reading of the voltmeter is 100V, and the temperatures of both sides of the flat are 180℃ and 30℃, respectively, calculate the thermal conductivity of the material.
SOLUTION: From the electrical current and voltage, the heat transfer rate can be calculated
From the heat conduction equation,
2. As shown in the figure 4-6, the heat transfer through the three plane walls is considered. When the surface temperatures, t1, t2, t3, t4 are 500oC，400 oC，200 oC and 100 oC, respectively, Calculate the ratio of the thermal resistance of three plane walls. Assuming the three plane walls contact tightly.
SOLUTION: The ratio of resistance for each layer is equal to its temperature ratio, as follows:
3. The wall of the fireplace is constituted of three materials, their thickness and heat-transfer coefficients are shown as follows:
No.
material
thickness, b mm
Thermal conductivity λ，
w/m(K）
1（inner layer）
Refractory brick
200
1.07
2
Insulation brick
100
0.14
3
steel
6
45
If the inside surface temperature t1 of the refractory brick is 1150 oC, the outside surface temperature t2 of the steel plate is 30 oC, and the heat loss of the fireplace wall is 300w/m2, respectively. Try to calculate the heat flux. If the result is not the same as that of the heat loss measured, analyze the reason and calculate the additive thermal resistance.
SOLUTION: From the heat transfer rate equation of the flat wall, we can compute the heat flux.
The result is greater than that of the heat loss measured, it is shown that the flat layers are contact poorly. Because the air layer exists, there is an additive heat resistance,
4. The combustion furnace consists of refractory brick (460 mm thick) and insulation brick (230 mm thick). If the inner surface temperature of refractory brick t1 and the outside surface temperature of t3 of the insulation brick are 1400 oC and 100 oC, respectively, calculate the heat flux and the interface temperatures of two layer. Assume that the two layers are contact excellently. Given that the thermal conductivity of the refractory brick is λ1=0.9+0.0007t; and the thermal conductivity of the insulation brick is λ2=0.3+0.0003t, where t can be considered as the average temperature of the layers.
SOLUTION: calculate the interfacial temperature by the trial-error method
Find the average conductivities for bricks.
the heat transfer flux through the flat wall,
=1400―452=948 oC
The interfacial temperature calculated is not the same as the given temperature, suppose the =950 oC, repeat the above calculation procedure,
5.Steel tube, Φ60(3mm, is packed by cork 30mm thick, furthermore, is packed by some insulating material outside as the thermal insulation layer. The outside surface temperatures of the steel tube and of the insulation layer are –110 oC and 10 oC respectively. The conductivity of the cork is 0.037(kcal/m﹒h﹒oC) and that of the insulating material is 0.06 (kcal/m﹒h﹒oC).Calculate the cooling loss of per meter tube.
SOLUTION：
From the heat transfer rate equation of the cylinder wall,
where，
hence,
The minus sign shows that the heat transfer is from environment to the system, that is the loss of cooling.
6. The outside of steam tube is packed by two insulation layers with different thermal conductivities but the same thickness. Assume that the diameter of the outside layer is twice as that of the inner layer, and the thermal conductivity of the outside layer is also twice as that of the inner layer. If the place of the two materials is changed, and other conditions are constant, what’s the thermal loss changes of per meter tube? Under the above conditions, which material is much appropriate when used as the inner layer?
SOLUTION: The average diameter of the outside layer is , and the average diameter of the inside layer is ,
From the heat transfer rate equation,
After changing the place of the two materials
After changing the place, the heat loss increases. It is shown on such conditions, the material with small thermal conductivity will be more suitable for the inner layer.
7.The inside radius and the outside radius of the hollow sphere are ri and ro respectively. Their temperature are ti and to, respectively, and ti is greater than t0. The thermal conductivity of the hollow sphere is λ. Try to deduce expression of the thermal resistance through the hollow sphere wall.
SOLUTION: From the Fouriers Law, the heat transfer rate at the radius r is
Integrate from
From which
8.In the following tubular exchanger, heat up the fluid whose mass flow rate is 29400kg/h from 20 oC to 50 oC. The fluid flows in the tube-side. The temperature of entering hot fluid is 100 oC, while the temperature of leaving hot fluid is 60 oC. Calculate the average temperature difference in the following conditions.
Single-pass 1-1 countercurrent flow heat exchange;
1-4 countercurrent flow heat exchange;
2-4 countercurrent flow heat exchange.
SOLUTION:1) The logarithm mean temperature difference
2)
From the chart of temperature difference correction coefficient ,
3) From R and P, look over the temperature difference correction coefficient chart of the two shell layer
9.In the parallel-current flow heat exchanger, water is used to cool the oil. The temperature of entering water and leaving water are 15 oC and 40 oC, respectively. While the temperature of entering and leaving oil are 150 oC and 100 oC, respectively. When production conditions vary, the leaving temperature of the oil must go down to 80 oC. Suppose that the mass flow, the entering temperatures and the physical properties of fluids are the same as the former. If the tube length of the former heat exchanger is 1m, how long has the tube to be increased to meet the production requirement? The heat loss of the heat exchanger can be ignored.
SOLULTION: The logarithmic mean temperature difference is
From the heat balance
When the temperature of the oil drops to 80℃, from the heat balance
or
From which
And
From the heat transfer rate equation,
The former heat exchanger:
The new heat exchanger:
10.The heavy oil and crude oil are flowing parallelly in the single pass double pipe exchanger, the temperatures of entering oils are 243 oC and 128 oC, respectively, while the temperature of the leaving oils are 167 oC and 157 oC, respectively. If the entering temperature and the mass flow are the same as the former, but the two oils change from current flow to counter flow, calculate the average temperature difference and their leaving temperatures. Suppose that under the two conditions, the total heat transfer coefficient and the physical properties keep the same, and the heat loss can be neglected.
SOLUTION: The superscript “ ’ ” express the parallel current flow
the two formulas are compared, we get
（a）
hence
and
or （b）
From the equation (a),
that is （c）
From the equation (b) and (c),
and
11.Air is heated from 20 oC to 80 oC in the steam heater. Air flows turbulently in the tube-side of the exchanger. The steam at saturated vapor pressure 1.8 kgf/cm2 is condensed outside the tube. If the flow rate of air is needed to increase 20%, while the inlet and outlet temperatures of the air are constant, how can we finish the task? Assume that the resistances of the tube wall and the scale can be neglected.
SOLUTION: From the above conditions given,
Suppose the superscript “ ‘ ” is in behalf of the condition of the flow rate increased, that is
From the appendix, we can obtain the saturated vapor temperature of 116.3 oC at the pressure of 1.8kgf/cm2.
Now, the air flow rate increases 20%. If we still operate in the heat exchanger, we must increase the temperature of the vapor (the pressure) in order to meet the requirement.
The heat transfer rate is
From the equation of the heat transfer rate, we have
for the former air flow rate：
After increasing the flow rate：
that is
or
so
From the appendix, we know that when the pressure of the vapor is increased to 2kgf /cm2, we can achieve the requirement.
12.The butyl alcohol at 90 oC is cooled to 50 oC. The heat transfer surface area of the exchanger is 6 m2, and the total heat transfer coefficient is 230w/(m2( oC).If the flow rate of butyl alcohol is 1930kg/h, the cooling medium is water at 18 oC, find:
The outlet temperature of the cooling water;
2) The consumption of the cooling water, in m3/h.
SOLUTION: 1) the outlet temperature of the cold water
At the temperature of 70 oC (), the heat capacity of the butyl alcohol is From the heat transfer rate equation, we have
where
hence
If we calculate that with the arithmetic mean temperature difference, that is
from which
because
hence ,we can calculate with the arithmetic mean temperature difference.
2）The consumption of the cooling water
From the heat balance of the exchanger, neglecting the heat loss,
from which,
that is
13.In the counter flow heat exchanger, the water at 20 oC is used to cool the liquid from 80 oC to 20 oC, and the specific heat of the liquid is 1.69kJ/（kg( oC）, its density is 850kg/m3.The pipe diameter of the exchanger is Φ25(2.5mm., and the water flows in the tube-side. The convective heat transfer coefficients of the water-side and the liquid-side are 0.85 and 1.70kw/（m2( oC）, respectively. Neglecting the scale resistance of pipe wall. If the outlet temperature of the water can not be over 50 oC, calculate the heat transfer surface area of the exchanger.
SOLUTION: From the heat transfer rate equation, we have
where ]
14.In the tubular exchanger, cool the oil with water, which flows in the pipe(Φ19(2mm).Given that the convective heat transfer coefficient of the water-side is 3490W/(m2(K), and that of the oil-side is 258 W/(m2(K).After the exchanger is used for some time, scales are formed in both sides of the tubular wall, scale resistance Rsi in the water-side is 0.00026 (㎡(K)/W, and scale resistance Rso in the oil side is 0.000176㎡*K/W. And the thermal conductivity λ of the wall is 45 W/(m2(K). Calculate 1)Total heat transfer coefficient Ko based on the outside surface of the tubular; 2) the increased percentage of heat resistance after the scale resistances are formed.
SOLUTION:1)The total heat transfer coefficient Ko
2) the increased percentage of the heat resistance after the scale arises
15. The hot oil with mass flow rate 3000kg/h is cooled by the cold oil in a counter current flow exchanger. Specific heat of both oils is 1.68 kJ/(kg(K), The hot oil is cooled from 100℃ to 25℃, while the cold oil is heated up from 20℃ to 40℃.The total heat transfer coefficient Ko are as follows based on the hot oil temperature:
The temperature of hot oil T，oC
100 80 60 40 30 25
Total heat transfer coefficient，
W/(m2(K)
355 350 340 310 230 160
Try to calculate the heat transfer area of the exchanger.
SOLUTION: calculate the heat transfer area with the graphical integration, that is
Calculate the relation of the T and t at first:
In the range of T=100~25oC, take some T, calculated results are shown in the following :
T
100
80
60
40
30
25
K
355
350
340
310
230
160
t
40
34.7
29.3
24
21.3
20
（T-t）
60
45.3
30.7
16
8.7
5
4.7
6.3
9.58
20.2
50.0
125
Diagrammatize,
16. Oil at 110℃ is cooled by water at 35℃ in a double pipe heat exchanger. The specific heat of the oil is 1.9kj/(kg(K). The two fluids are countercurrent flow. If the mass flow rates of water and oil are 0.67kg/s and 2.85 kg/s, respectively, the heat transfer surface area So is 16m2, and the total heat transfer coefficient K is 320 W/(m2(K), what are the outlet temperature of the water and the heat transfer rate of the heat exchanger?
SOLUTION:
and
“min” is in behalf of water
From the chart, we have
From the definition of ε, we have
hence,
17.Use the dimensional analysis method to derivate the Number Equation of natural convection heat transfer coefficient α between the wall and liquid . Given that α is the following variety’s function, α=?(λ,Cp，ρ，μ，βg,Δt,l),where Δt is the difference between the surface temperature of the wall and the liquid temperature, other symbols are the same as the above .
SOLUTION:
Consider
M，L，T，θ and Q are used to express each dimension of the variable, that is
b and c are used to express other variable, from the above we have
a=1-b
d=b-c
e=c/2
f=(3/2)c-1
g=c/2
or
The air with the rate of 60m3/h at 2atm and 20 oC is heated to 80 oC in the inner pipe of double pipe heat exchanger. The inner diameter of the pipe is Φ57(3.5mm and the length of that is 3m .Calculate the convective heat transfer coefficient of air in the pipe .
SOLUTION: At the average temperature 50 oC(（20+80）/2=50 oC),the physical properties of the air are the following:
the inlet velocity of the air is
Hence ,there is no need to revise α.
19. H2SO4 with concentration of 98% is flowing in the annular space of the double pipe heat exchanger at the rate of 0.7m/s. The average temperature of H2SO4 is 70 oC, and the average temperature of the outer wall of inner pipe is 60 oC. The diameter of inner pipe is Φ25(2.5mm, and that of the outer pipe is Φ51(3mm, try to calculate the heat flux and the convective heat transfer coefficient.
SOLUTION: At the average temperature, the physical properties of H2SO4 are as follows:
the equivalent diameter of the double-pipe annular space
Calculate in term of turbulent flow first, that is
20.Toluene flows through the coil pipe (Φ57(3.5mm , bending radius 0.6m)at the rate of 1500kg/h .It is cooled from 90 oC to 30 oC. Try to calculate the convective heat transfer coefficient of toluene in coil pipe.
SOLUTION: At the qualitative temperature 60 oC（1/2）（90+30）=60 oC), the physical properties of toluene are the following:
Calculate the α according to straight tube
so, the heat transfer coefficient of the syphon is
21.?At the normal pressure, methane at 12 oC is flowing in the shell-side of the tubular exchanger in the axial direction. When the methane leaves the exchanger, the temperature has changed to 30 oC. The inner diameter of exchanger shell is 190mm, and the tube bundle consists of 37 steel tubes(Φ19(2mm).Try to calculate the convective heat transfer coefficient of methane towards the wall.
SOLUTION: The average temperature of methane is (120+30)/2=75 oC. At this temperature, the physical properties of methane are the following:
For the convective heat transfer in the non-circular tube, the equivalent diameter is
Hence,
22. The vapor at saturation temperature ts 100 oC is condensed in the surface of a single vertical circular pipe with length 2m and diameter 0.04m (outside) .The outside surface temperature of the pipe tw is 94 oC. Calculate the quantity of the condensed vapor per hour. If the pipe is placed horizontally, what’s the quantity of the condensed vapor?
SOLUTION: From the appendix, we obtain the latent heat of vaporization of the saturated vapor, 3258kJ/kg at 100 oC.
The average temperature of the condensate film=
The following are the physical properties of the water at 97 oC :
Consider the condensate film flows laminarly, hence
Check the flow pattern:
(laminar flow)
If the tube is placed horizontally, then
hence,
Check the flow pattern:
(laminar flow)
23.There are two circular pipes with the same temperature and length placed horizontally in doors. The saturated vapor flows in the pipe. The two pipes are cooled by the air with natural convection. Assume there is no affection between the two pipes. Given that the diameter of one pipe is five times of the other, and the Gr(Pr of two pipes is between and , try to calculate the ratio of heat loss of the two pipes.
SOLUTION: This is a problem of free convection in the big container
If in the small pipe，the in the big pipe.
Suppose the diameters of the small and big pipes are d and D, respectively, so
24.In the experiment of measuring the total heat transfer coefficient of the tubular exchanger, water flows turbulently in the tube-side of exchanger, while the saturated vapor is condensed in the shell-side .The tube bundle consists of steel pipe (Φ25(2.5mm). When the velocity of the water is 1m/s, the total heat transfer coefficient Ko based on the outside surface of the tube is 2115 W/(m2(℃).If the other conditions are constant, but the velocity of the water changes to 1.5m/s, the Ko measured would be 2660 W/(m2(℃).Try to calculate the heat transfer film coefficient of the vapor condensation. The scale resistance is ingored.
SOLUTION: Suppose when the velocity of water is 1m/s, heat transfer film coefficient of water towards tube wall is and is overall heart transfer coefficient; when the velocity of water is 1.5m/s, its film coefficients is and overall heat transfer coefficient is
From the formula of the total heat transfer coefficient, we have
（a）
（b）
From the (a) and (b),
hence, (c)
and (d)
From the (c) and (d),
Substitution in (a) gives
hence, the heat transfer coefficient of vapor condensation
25.The two parallel flat plate are 5mm apart from each other in the air .One of the plate ‘s blackness is 0.1, and the temperature is 350K, while the other plate’s blackness is 0.05, the temperature is 300K. If the first plate is coated, its blackness would change to 0.025. What’s the change of heat transfer rate? Assume that the convection heat transfer can be neglected.
SOLUTION: The heat flux of the two plates
The heat flux by radiation
The total heat flux
when the blackness of one plate changes to 0.025, the heat flux by radiation becomes
The reducing percentage of the total heat transfer flux is
and the reducing percentage of the heat flux by radiation is
26.Hot air flows through the steel pipe( Φ426(9mm). The thermocouple is set in the pipe to measure the temperature of the air. In order to reduce the reading error, a pipe is used to shelter the thermoelectric couple. The blackness of the shelter pipe is 0.3, the area is 90 times as much as the contact point of the thermoelectric couple. The temperature of pipe wall is 110℃, and that of the thermoelectric couple is 220℃.Suppose that the convective heat transfer coefficient of air towards shelter pipe is 10 W/(m2(℃), and that of air with thermoelectric couple’s joint is 45 W/(m2(℃).The blackness of thermoelectric couple’s joint is 0.8.Try to calculate:1)The real temperature of air ;2) The temperature of the shelter pipe ti; 3) the reading error of the thermoelectric couple.
SOLUTION: Suppose the subscript 1 is in behalf of thermoelectric couple, 2 is in behalf the tube wall, i is in behalf of the heat preservation tube, a is in behalf of air.
The convection heat transfer rate of the air towards the thermoelectric couple
The convection heat transfer rate of the air towards the heat preservation tube
（a）
The heat transfer by radiation of thermoelectric couple towards the heat preservation tube is
where
（b）
The radiation heat transfer rate of the heat preservation tube towards the tube wall
where
（c）
when stabilization is reached,
hence,
（e）
and
or
From the （e）and （f）,we have
Consider=464K，then
= ―100<0
Consider=463K，then
=1770>0
The temperature of the heat preservation tube is about 464K or 191℃, hence
The relative error of the thermoelectric couple reading is
27.The pipe which is 50mm outside diameter, is packed by some adiabatic material. The thickness of the thermal insulation is 40mm ,and the heat transfer coefficient is 0.13 W/(m2(℃).The average temperature t1 of the outside surface of the pipe is 800℃.A magnesia thermal insulation layer is planned to be packed outside the adiabatic material, which has heat transfer coefficient 0.09 W/(m2(℃).And the outside surface temperature t3 of this layer is 87℃. If the outside temperature of the pipe is still 800℃,the ambient temperature tα is 20℃.Try to calculate the thickness of the magnesia thermal insulation layer .Assume that each layer is contact well.
SOLUTION: The heat transfer rate of unit tube length is
where=0.05m，=0.05+2(0.04=0.13m
=0.13+2b
The convection heat transfer rate of unit tube length
where,
when it achieve stabilization，
Calculate the b by the trial-error method, and we have: b=0.0175m=17.5mm.
28.There is 1000kg reactant with the specific heat of 3.8kJ/（kg(℃） in a container. And a coil heat exchanger (outside surface area 1m2 ) is installed in it. The saturated vapor at 1250℃ flows through the coil pipe, and is condensed to heat up the reactant. If the total heat transfer coefficient of the exchanger is 600 W/(m2(℃), and the heat loss can be neglected, try to find the time needed to heat the material from 20℃ to 90℃.
If the heat loss is considered, calculate again the time needed to heat the reactant to the same temperature. Given that the outside surface area of the container is 10m2,the convective heat transfer coefficient of the container outside wall with air is 8.5 W/(m2(℃), the ambient temperature is 20℃.
SOLUTION: It is an unsteady process. The equation of heat transfer rate at some moment is
Separate the variable and integrate,
The heat loss is not ignored:
that is
hence,
29.Some oil flows through the shell-side of the tubular exchanger at the mass flow rate of 30kg/s, and is used to heat the crude oil from 25℃ to 60℃. Its temperatures decrease from 150℃ to 110℃. Now, there is a tubular exchanger, whose specifications are: the inside diameter of the shell is 0.6m ; the shell-side is single pass, while the tube-side is double pass; there are 324 pipes (Φ19(2mm, length 3m)placed in square pitch arrangement, and the tube pitch is 25.4mm.In the shell-side, there is 25% arcuate baffle and the space length is 230mm.Check whether this exchanger can meet the above requirement of heat transfer. Given that the physical properties at average temperature are the following:
Liquid name
Specific heat,Cp
kJ/kg(℃
Viscosity,μ
Pa(S
Heat transfer coefficient,λ
W/(m2(℃)
Crude oil
1.986
0.0029
0.136
oil
2.2
0.0052
0.119
SOLUTION: Find the convection heat transfer coefficient in the tube:
The tube number of each pass==162
The flow area of each pass =
mass flux
hence,
The convection heat transfer coefficient outside the pipe :
The flow area=
=0.0348
hence,
According to the , we have
The total heat transfer coefficient is
The heat transfer area of the exchanger is :
where,
hence,
The area of the existing exchanger is
The heat transfer area should be greater than 77m2 according to the heat transfer requirement, but the area of the existing exchanger is only 58m2,so it is not feasible.
30. Water at the mass flow rate of 60m3/h is used to cool the benzene from 80℃ to 35℃ .The initial temperature of the water is 30℃, design the proper tubular exchanger.
SOLUTION: Because the benzene is caustic, it should flow in the tube.
Suppose the outlet temperature of the water
(Ⅰ).Calculate and choose the size of the equipment
1. The average temperatures of benzene and water are as follows:
benzene：
water：
From the average temperatures, the physical properties are obtained as follows：
temperature
℃
density ,
kg/m3
viscosity
Specific heat
kJ/(kg.℃)
Heat transfer coefficient
W/(m.℃)
benzene
57.5
879
0.41
1.842
0.137
water
33.5
995
0.745
4.186
0.622
2.Calculate the heat transfer rate Q
3.Calculate the average temperature difference
Calculate in terms of countercurrent flow first, that is
and then calculate to determine the quantity of shell pass
=0.85
So the single- pass in the shell would be enough.
Hence ,
Suppose the total heat transfer coefficient based on the outer area of exchanger is:
choose the plate heat exchanger(because)
From the appendix, we obtain the specs of the chosen exchanger:
D，mm 800
S，m2 230
Collocation of the tube triangle
he length of tube, L，m 6
The outside diameter of the tube，do，mm 25
The number of the tube，n 501
The flow area of the tube-side，m2 0.1574
The flow area of the shell-side，m2 0.0594
The number of tube-pass 6
The number of shell-pass 3
(Ⅱ)Calculate the pressure drop
1. The pressure drop in the shell-side
The benzene is combustible,
2.The pressure drop in the shell-side
Consider Fs=1.15
The mass flow of the water:
hence,
The pressure drop in the shell-side and tube-side can be considered as appropriate.
(III)Check the total heat transfer coefficient
1.Tube-side
2.Shell-side
3. Scale resistance
Assuming
4.The total heat transfer coefficient
Hence, the chosen exchanger is feasible.
Chapter 5 Eaporation
1. Work out the boil of the 40% NaOH water liquor under the press of 0.2 kgf/cm2in the way of empirical formula and Du Lin's regulation separately.
Solution: work out the boil point with formula
Check the addenda =28.4℃
Check the addenda then can know, under press of 0.2kgf/cm^2 ,the temperature of saturated steam is 59.7 ℃、Vaporize latent heat is 2356kJ/kg
℃
Count in the way of Dulin
Check the Dulin chart, then can know =87℃
The solution show out, the solution in the two way is so different, the latter arithmetic is more credible, because it use more actual data as the foundation.
２In the single effective evaporator, evaporate the water liquor, the operating pressure of the evaporating room is 1atm. The height of the liquor in the heating room is 1m, The concentration of liquor is 40.8% and the thickness is 1340kg/m^3. Attempting to request the boiling point of liquor and temperature falls short of decreasing 。
Solution : Check the addenda and know the boiling point of the 40.8% liquor under the average press is 120 ℃。
∴
Through the liquor static pressure the powerful temperature that arouses falls short of decreasing
Check the addenda and know the boiling point of the 40.8% liquor under press is 102 ℃。
∴
Then the liquor boiling point is
3. The evaporator with the 52 M^2 heat transfer area, under the average press, a 7% liquor is evaporated 2500kg per hour. The temperature of the material liquor is 95℃; the boiling point of the liquor in the evaporator is 103℃. The concentration of the complete liquor is 45%.The press of the heating steam is 2 Kgf/m^2's ( surface press ). The heating loss of the evaporator is 11000W. Attempt to estimate the sum-transferring coefficient of the evaporator. Assume the diluted heat could be ignored.
Solution: Check the addenda, then can know the temperature of the steam under the press of 3kgf/cm^2 is 132.9 ℃, the vaporize latent heat of the steam under the press of 1atm is 2258kJ/kg.
According to the transferring speed of the evaporator and heating balance formula, then can know
In it
∴
solution K=936W/ （m^2·℃）
4 condense 10000kg NaNO3 liquor from 5% to 25% per hour in the single effective evaporator. The temperature of the material liquor is 40℃.The vacuum of the evaporating room vacuum of 460mmHg，the surface press of the heating steam is 0.3kgf/cm^2. The sum-transferring coefficient of the evaporator is 1000W/（m^2·℃）. The heating loss is 5% of the evaporator transferring heat. Attempt to count the transferring heating area of the evaporator and the consumption of the heating steam. Assume that the difference in temperature bringing by the liquor static press could be ignored .The press is 760mmHg.
Solution: Heat transferring area of the evaporator
According to the heat transferring speed equation
In it
Check the addenda, Under the press P=（7600-4600）=300MmHg=0.41kgf/cm^2 ,The temperature of steam is 76 ℃、Vaporize latent heat is 2317Kj/kg 。
And check the addenda under the common press 25%NaNO3's liquor temperature loss
∴
Check the addenda, under the press of 1.3kgf/cm^2 the steam is 107.2 ℃、Vaporize latent heat is 2240Kj/kg
∴
and consumption of the steam is
5. Condense 10000kg NaNO3 liquor from 5% to 25% per hour in the single effective evaporator.
The temperature of the material liquor is 40℃. Heating steam without any press is 2.37kgf/cm^2,condensate is ejected in the saturation temperature. The vacuum of the evaporating room is 660mmHg. Evaporator sum-transferring coefficient is 1400 W/（㎡·℃）,heat loss could be ignored. The press is 760mmHg. Attempt to count:
（1）Evaporation capacity ；
（2）Heat vapors consumption amount ；
（3）Evaporator heat transfer area 。
Solution ：（1）Evaporation capacity
（2）Heat steam consumption amount
Through checks to such an extent that the saturated temperature of steam is 51.5 when P=760-660=100mmHg ℃，Vaporize latent heat is 2590kJ/kg
Du Lintu through NaOH's liquor checks to such an extent that liquor boiling point is
Thick picture of total heat through NaOH's liquor is checked
The amount of heat weighing apparatus through the evaporator is reckoned the knowledge
Q=Dr=WH′-Fh。
Through the annex check to such an extent that 2.37kg/cm^2 gets off that the saturated temperature of steam is 124.5 ℃、Vaporize latent heat is 2193kJ/kg
∴
Realize through the heat transfer rate equation
6. It is need to enrich some aqueous solution of 850kg/h from 35% to 15% continuously 。 One heat transfer area is 10 M^2 of a small-size evaporator may be used now 。That raw material fluid is living accedes to the evaporator below the boiling point ，Appraising the different temperature of liquors below that the operating condition fall short of decrease act as 1.8℃ in all 。Evaporating the degree of vacuum of room act as 0.8kgf/cm^2。In case the evaporator coefficient of heat transfer invariably is 1000W/（㎡·℃），The warmhearted loss is overlook 。Attempt to calculate heat steam intensity of pressure 。The atmospheric pressure is 1kgf/cm^2 in the locality 。
Solution ： check to such an extent that the saturated temperature of steam is 59.7 when 0.2kgf/cm^2 ℃、Vaporize latent heat is 2355.9kJ/kg through the annex
Reason
Therefore
Check to such an extent that heats steam intensity of pressure in 1.46kgf/cm^2 through the annex
7.Some aqueous solution of 1000kg10% is be evaporated in per hour in Twin effects - juxtaposition installation 。First effect fluid completing concentration is 15%，Second effect act as 30% 。The liquor boiling point is separately in two effects 108℃ and 95 ℃。Water that evaporates of course because the temperature is cut down when attempting to request the liquor since first effect to go into second effect is measured to reach to seize evaporation capacity percent invariably in second effect since the evaporation capacity 。
Solution ：Every the effect evaporation capacity is separately
Obtain through the amount of heat weighing apparatus since the amount evaporating water in second effect ，In immediate future
In it
Through the annex checks 95 ℃Second steam vaporize latent heat is 2271kJ/kg
∴
The percent is since what the evaporation capacity seized second evaporation capacity
8.In three-effects collateral flow vaporizing equipment, if in each effect, the influence of the difference expense of the temperature and the apparent heat, and the expense of the heat could be ignored, attempt to prove:
The vaporizing quantity of each effect is equal;
The heat-transferring rate of each effect is equal.
Solution:
Because the apparent heat and the difference expense of the temperature could be ignored, then
Therefore
Either
The weighing apparatus is reckoned the knowledge through amount of heat
∴
Reach
∴
Reasons by analogy
At the same time get yield
9.Want to design a three-effects collateral vaporizing system, to condense some liquor from concentration 10% to 50%. The material quantity entering is 22700kg/h, the temperature is 40℃. The temperature of heating steam is 121℃, the second steam of the last effect have a temperature of 51.7℃. The sum heat-transferring coefficient of every effect is
Every effect liquor specific heat ，Vaporize latent heat is separately all desirably in 4.186kj/ （kg·℃）2326kj/kg （In immediate future specific heat and vaporize latent heat may be looked upon in the interest of not changing into with temperature and concentration ）。All may overlook in case the different temperature fall short to decrease temperature 。Attempt to request the evaporator heat transfer area （In case the area of every effect is equal to ）、The amount heating steam reaches the different effect fluid the completing rate of flow 。
Solution ：（1）Evaporation capacity invariably
（2）Start appraises every effect liquor boiling point
In case
Once more ∵
Just
∴
（3）The amount of heat weighing apparatus is reckoned
In case completing the fluid rate of flow express in the way of L
First effect
In immediate future
Second effect
Third effect
Style is straighten out on the substitute
Associate with a，b，c.
Solution
:
（4）Heat transfer area
（5）Reallocated every the effective difference of temperature
（6）Repetition move （3）Reach （4）
Solution
Get
（7） Computational solution is
10.In three-effects collateral flow vaporizing equipment, the heat-transferring area of each effect is 140m2. Condense a liquor form concentration 5% to 4%. The temperature of the material liquor is 90℃. The temperature of heating the steam is 120℃; condensation-water is ejected in the saturation temperature. The second steam of the last effect has a temperature of 40℃. The sum heat-transferring coefficient of every effect vaporizer is
The specific heat of liquor all desirably is 4.2kJ/（kg·℃）。In case the different temperature fall short of to decrease and the temperature loss is overlook ，Attempt to calculate
Raw material fluid rate of flow ；
Heat vapors consumption amount ；
Solution the subject under discussion is the now available evaporator adjusting college to be calculated reckoning ，The means yet adopts the matter balance, amount of heat equilibrium and heat transfer rate equation: then
（1）The start appraises every effect effective temperature to fall short of reaching the boiling point
Through heats the steam temperature ，Checks
Follow the example of the steam temperature two times the tip ，Checks
In case
（2）Request evaporation capacity and the raw material fluid rate of flow
On the grounds of the above-mentioned equation set of substitute straighten out the known number
Antithetical couplet sets up （a），（b），（c）Divides
（3）The amount requesting the heat transfer difference as checks the effective temperature
∴
Against the initial assume △T divergence a bit ，According to the calibrate △t′Value ，Repeatedly above-mentioned calculation ，The result as follow：
F=42000kg/h
D=12100kg/h
W=37000kg/h
The above-mentioned result is bad after checks the effective temperature ，Two times be close to altogether ，As approximate value not more repeatedly calculation 。
11．Being living to run three effects feed in raw material evaporates in the unit ，Per hour by 5000kg12%'s ，The aqueous solution enriches 40% 。Raw material fluid to 80℃Below accedes to first effect 。The coefficient of heat transfer invariably to heat steam is 4kgf/cm^2。Inner place the condenser is getting near most 0.2 kgf/cm^2。Every the effect evaporator coefficient of heat transfer invariably is 。The specific heat of liquor all desirably is 3.5kj/（kg·℃）。In case the desideration of liquor is warmed up over look the temperature loss 。Attempting to request first effect-heating vapors measures the heat transfer area the evaporator （The area of every effect is equal to ）。
Solution ：（1）Evaporation capacity invariably
（2）Start appraises every effect concentration
In case every the effect evaporation capacity is equal to ，In immediate future
First effect is completed fluid May obtain through the solute weighing apparatus ，then
With run ，The three effects liquor concentration is passed an imperial examination to second effect Act as separately
（3）Estimating that every effect temperature difference decreases
The temperature as a result of liquor falls short of to decrease changing as few minutes with operation intensity of pressure ，Therefore calculate according to the constant pressure approximately 。Overlook the temperature that as a result of the powerful and going from place to place obstruction of liquor static pressure arouses in the calculation falls short of decreasing 。
Through the annex check down different concentration of constant pressure That the temperature of aqueous solution falls short of decreases △′In the interest of ：
∴∑△=2+3+7=12℃
（4）Invariably the effective temperature difference ∑△t
Through the annex check to such an extent that heats that the saturated temperature that the steam is living below the 4kgf/cm^2 is 143 ℃，The saturated temperature that two steam are living below the 0.2kgf/cm^2. is 60℃。
∴
（5）The effective temperature difference distribute in every effect
In case
Just
(6) Estimate every the effect liquor boiling point
Overlook static pressure and resistance of flow which arouse to fall short of the temperature difference, below the situation use first effect to heat the temperature of vapors and every effect to estimate every effect liquor boiling point gradually ，Result list as follows ：
Temperature ℃
Vaporize latent heat KJ/kg
Stuff fluid
Steam is heated to the effect
Effect liquor
Steam is heated to two effects
（In immediate future follow the example of two times steam ）
Two effects liquor
Steam is heated to three effects
（In immediate future two follow the example of steam two times ）
Three effects liquor
Steam inner place the condenser
（In immediate future three follow the example of steam two times ）
（7）The amount of heat weighing apparatus is reckoned
First effect ：
Second effect ：
Third effect ：
Associate with a，b，c
Solution:
（8）Every effect heat transfer area ：
Every effect heat transfer area is not equal to ，Ought to reallocate every effect effective temperature 。
（9）Redistribution effective difference of temperature
Through the style 5 —34Knowledge
（10）Duplicate the 6th ，7，8 calculations ，The main result is as follows ：
The amount of heat weighing apparatus is obtained through every effect ：
Solution:
Reach
S is close to very much through what above-mentioned calculation obtain ， Not more repeatedly calculation 。All even up the heat transfer area is 3.3m^2's first effect Heat steam whose consumption amount is 1500kg/h 。
Chapter 6 Distillation
6.1 A mixture of 50 mole percent benzene and 50 mole percent toluene at a total pressure of 99 kpa .Calculate the boiling temperature of liquid .Vapor –pressure data for these substances are given in table 1-1.
Solution: From the figure t-x-y at the operating pressure, the bubble point can be searched out .If have no figure t-x-y ,but the vapor pressures of component are known, then an iteration is needed to matched the correct value of x . Use the following equation :
The first step is to assume the value of temperature ,the second step is to find the values of vapor pressures and , substituting in the upper equation and calculating the value of x. If the result is the same as that of problem, then the temperature and the figure t-x-y are correct.
According to fig t-x-y for benzene—toluene at 1 atm pressure ,assume t=91.5C, From example 6-1,the values of and are found, gives :
So the bubble point of liquid is about 92c.
6.2 Vapor—pressure data for pentane(C5H12) and hexane (C6H14)are given in the attached table . Assume that the mixture is an ideal liquid . Calculate the equilibrium data for liquid at 13.3kpa pressure.
Solution: From the two following equations, the relationship of x and y can be calculate :
The calculating steps are omitted ,and the results are given in the following table ：
t,K
260.6
265
270
275
280
285
289
,N/m2
13.3
17.3
21.9
26.5
34.5
42.5
48.9
, N/m2
2.83
3.5
4.26
5.0
8.53
11.2
13.3
X
1.0
0.71
0.513
0.386
0.184
0.667
0
y
1.0
0.924
0.845
0.769
0.477
0.214
0
6.3 Use the date of problem 2, calculate (a) relative volatility .(b) the values of x and y corresponding to the average relative volatility ,and compare the result with that of problem 2.
Solution: calculate the value of with the following equation:
The results are given in the second rank of the attached table :
T,K
α
x
y
The results of problem 2
The calculating result by using
260.6
265
270
275
280
285
289
4.70
4.94
5.14
5.30
4.04
3.79
3.68
1.0
0.71
0.513
0.386
0.184
0.067
0
1.0
0.924
0.845
0.769
0.477
0.214
0
1.0
0.918
0.826
0.739
0.504
0.245
0
Form the equilibrium equation ：
The value of x-y calculated by the upper equation are shown in the upper table
The result of calculation indicates that because slightly changes with t ，so the x-y data calculated with the average relatively volatility are slightly different from the results of problem 3.
6.4 A two-component mixture containing 60 mole percent (more volatile component) is subjected to flash distillation and differential distillation at a pressure of 1 atm , respectively. The vaporization ratio f is 1/3.Under these conditions, calculate the compositions of the overhead product and the bottom product by molar fraction and molar flow. Assuming that equilibrium relation is given by equation y=0.46x + 0.549
6.5 A continuous fractionating column is used to separate 4000kg/h of a mixture of 30 percent CSand 70 percent CCl. Bottom product contain 5 percent CSat least, and the rate of recovery of CSin the overhead product is 88% by weight,required. Calculate (a) the moles flowof overhead product per hour .(b) the mole fractions of CSand CClin the overhead product, respectively
Solution: Form overall material balance
Known by the justice of the problem
Take the place of 3 types and enter 2 types
The unit converts：
（mole fraction）
6.6 A liquid containing 40 mole percent methanol and 60 mole percent water is to be separated in a continuous fractional column at 1 atm pressure .Calculate the value of q under the three following conditions (a) the feeding is liquid at 40C (b) the feeding is saturated liquid (c) the feeding is saturated vapor. The equilibrium data for methanol-water liquid at 1 atm pressure are given in the attached table.
Solution: 1）The feed at 40℃
Form the equilibrium data for methanol-water ,the bubble point of feeding is 75.3℃，Check so that the mass latent heats of methanol and water at 75.3℃ are 2521[kcal/kg]and 534[kcal/kg] from appendix. So the molar latent heats of feed is
The temperature of feed is 40℃,the bubble point is75.3℃，
Hence , the average value of temperature=1/2（40+75.3）=57.7℃，Form appendix, the specific heat of methanol at 57.7℃ is 0.64kcal/kg.℃，and the specific heat of water is
4.19（0.64×32×0.4+1×18×0.6）=79.6kj/kmol℃。
The enthalpy which per kmol of feed becomes saturated vapor need is
38600+79.6（75.3-40）=41410kj/kmol
2） Feed saturated liquid
From defining knowing，q=1
3） Feed saturated vapor
From defining knowing，q=0
6.7 Use the data of problem 6,if the column is fed with 100koml/h.The molar fractions of methanol in overhead product and bottom product are 0.95 and 0.04,respectively.A reflux ratio at the top of column is 2.5.Calculate (a) the mole flow of overhead product per hour (b) the mole flow of liquid in rectifying column (c) the mole flow of vapor in stripping column .Assume that the constant molar flow applies to this system .
Solution: Form overall material balance
Solve the eqution we can have：
And
These upper values are fixed under the three feed conditions。
The feed at 40℃，q=1.07
Feed saturated liquid ，q=1
Feed saturated vapor，q=0
6.8 The feeding at dew point is fed to a continuous fractionating column ,and the equilibrium relations are given by equations: y=0.723x + 0.263 ( in rectifying column); y=1.25x – 0.0187 (in stripping column) .Calculate(a) the compositions of feeding, overhead product and bottom product , respectively(b) the reflux ratio
Solution: The slope of the operating line in rectifying column is
We can get：R=2.61
The intercept of the rectifying line on y axis is
The intersection of the stripping line and the diagonal is
Y=X=Xw
so Xw=0.0748
Form the intersection of these two operating lines, gives:
Because the feed at dew point ,so the feed line is horizon ,and the composition of feed is
6.11 . The equilibrium data for the alcohol—water are given in example 1-7. A continuous fractionating column is to used to separate a feeding contains 15 mole percent alcohol and 85 mole percent water ,the feeding is saturated liquid. Overhead product containing 80 mole percent at least and a bottom product containing 2 mole percent alcohol .required. .A reflux ratio is 2. If get the product which is saturated liquid on the certain board place of rectifying column,then the mole flow equals to half that of overhead product ,calculate (a) how many ideal plates are needed(b) the positions of the feeding plate and the lateral line outlet .
Solution: Because rectification have lateral line outlet, so the rectification divided into the upper and the low section, The products operation line equation of outlet the above of lateral line is
The operating line equation in the low section should be calculated and published by the overall balance. Do the overall balance according to a dotted line range of figure of subject under discussion, have
Uniting vertical 2, the formula 3 and formula 4 have:
Unite it is vertical 1 and 5s,solve two line point of intersect coordinates is x=x,as the intersect of two operating lines of subject under discussion g shows. The operating line of stripping column do not change
Calculate the number of the ideal plates with graphic method, and the steps are omitted
The intercept of operating line in the upper section of rectification:
The intercept of operating line in the low section of rectification:
12 ideal plates are needed （besides reboiler），the lateral line outlet at the fifth ideal plate from the top of column，and the feeding is introduced on the tenth plate from the top of column 。
6.12 It is desired to produce an overhead product containing 95.7 mole percent A form a ideal mixture of 44 mole A and 50 mole percent B feeding into a continuous fractionating column .The average relative volatility equals to 2.5 ,and a minimum reflux ratio is 1.63.Explain the heat state of feeding and calculate the value of q .
Solution: ：Form equilibrium equation:
Form operating equation :
Substituting in eq.1),gives:
X=0.365 Y=0.59
Form definiens of the minimum reflux ratio，the intersection of the two upper equations is also that of the equilibrium curve and the feed line.
so the feed is a mixture of liquid and vapor.
Form the feed equation ( feed line ),gives:
Solve and have:
6.13 A ideal mixtures contains 50 mole percent of more volatile component is feeding to continuous fractionation column .The feed is saturated liquid. Two condensers are used in series.Patial condenser is used to provide the reflux flow which contains 88 mole percent more volatile component, and a final condenser is used to provide the qualified product containing 95 mole percent more volatile component .The rate of recovery equals to 96 percent .If 79 mole percent of the liquid on the first place of the top of column is measured ,caculate(a) the operate and minimum reflux ratio ,respectively.(b) if the mole flow of overhead product is 100kmol/h,what is the mole flow of feeding .(these percentages belong to more volatile component)
Solution :
6.14 A mixture contains 3.3 mole percent ethanol and 96.7 mole percent water is to be separated in a plate column at 1 atm pressure .Steam is saturated liquid which is heated directly by a source of vapor. Assuming constant mole flow applies to this system. If the rate of recovery of ethanol in the overhead product equals to 99 percent.(a) if the number of plates is infinite ,calculate the moles of vapor needed per 1000mole of feeding .(b) if the mole flow of vapor equals to twice of the minimum amount ,calculate(i) how many ideal plates are needed .(ii) the compositions of two products. Equilibrium data are given in the attached table .
Solution : The material—balance diagram is shown in the attached figure 1.constant molal overflow ,hence
Form overall material balance :
Form , substituting in the upper equation
Calculate the minimum moles of vapor per mole of feed
Form the minimum moles of vapor ,the slope of operating line is
Checked x equals to 0.033 by equilibrium data ,then
feed
2） When,then calculate the number of ideal plates and the compositions of two products
feed
Form
Calculate the number of ideal plates by the graphic method
The slope of operating line
Draw the slope is 4.13 of operating line through point c (0.00033,0),and intersect the feed line at the point d. Draw the rectangular steps between the operating line and the equilibrium curve from point d to turning over point c .These steps are shown in figure 2,and 5 ideal plates are needed.
6.15 A liquid of benzene and toluene is fed to continuous distillation in a plate column. Under the total reflux ratio condition , the compositions of liquid on the close together plates are 0.28,0.41,and 0.57,respectively.Calculate the Murphree plate efficiency of relatively two low layers. The equilibrium data for benzene—toluene liquid at the operating condition are given as:
x 0.26 0.38 0.51
y 0.45 0.60 0.72
Solution : Under the total reflux ratio condition
Form Murphree efficiency:
Checked by
6.16 A two-component ideal mixture contains 20 mole percent more volatile component is to be separated in a plate column .The feeding is saturated liquid and fed to the top of column .The stripping tower is composed of distillation kettle and a actual plate , the distillation kettle equals to an ideal plate . The rate of recovery of more volatile component in the overhead product is 80 percent . The density of overhead product is 0.28 ,and the relative volatility is 2.5.Calculate the composition of bottom product and the Murphree plate efficiency, respectively.
6.17 A continuous fractionating column is used to separate 2500kg/h of a mixture of 25 percent acetone and 75 percent water . Overhead product containing 99 percent acetone,required. These percentages are by weight ,and 80 mole percent of acetone in feeding is entering to the overhead product. The feeding at 20c ,and the operate reflux ratio is 2.7 times as much as the minimum . there are two condensers at the top of column.Partial condenser is used to control the reflux flow to be saturated liquid . The steam not condensed enter the final condenser to be condensed and cooled, regarded as the products. If the overall efficiency is 60%,then calculate the number of ideal plates.
Solution: The compositions of each flow rate
Known by the meaning of the question：
Calculate the number of actual plates
From the attached figure 1，at the point，the corresponding bubble point is 67℃，so the average of the feed temperature and the bubble point is 1/2（67+20）=43.5℃。
The specific heats of each component at 43.5℃ are ：
the latent heats of each component at 67℃ are:
the slope of the feed line=
From fig.2 , the minimum reflux ratio must computed from the slope of operating line ag that is tangent to the equilibrium curve, and measure the slope is 0.475 .
Solve and have：
the intercept of the rectifying line on y axis
the intercept of the stripping line on y axis
the operating line of the attached figure 2 is drawn according to the slope 0.281，and this operating line of the attached figure 3 is draw according to the intercept 0.709.
7 ideal plates are needed (except reboiler and partial condenser)
hence , the number of actual plates
6.18 The equilibrium data for the system CS-CClat the 1 atm pressure are given in Example 1-11.A continuous fractionating column is used to separate a feed contains 30 mole percent CSand 70 mole percent CCl.Overhead product containing 95 mole percent CS and a bottom product containing 2.5 mole percent CS,required. The feed is saturated liquid and the mass flow is 400kg/h. A reflux ratio equals to 1.7 times as the minimum. The distillation takes place at 61C and 1 atm . the superficial velocity based on empty tower is 0.8m/s and the distance between two boards is 0.4m.The overall efficiency is 50% .Calculate (a) The number of actual plates needed.(b) the mass flow of two products c) the cross-sectional diameter of tower (d) the effective height of tower.
Solution: (a) calculate the number of actual plates
Painted x-y picture for the equilibrium data by the question, as figure shows。
Since the feed is saturated liquid,
，from fig .X-Y
The intercept of operating line in rectifying column =
Calculate the number of ideal plates with The graphic method, and the steps is omitted .The result indicates, besides reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top
Actual plates
The mass flow rate of product
The average molecular weight of feed is
M=0.316 + 0.1154=130.6kg/kmol
Form overall material balance
D + W=30.6
0.95D + 0.025W=0.330.6
From the two upper equations, gives:
W=21.5mol/h
and
The cross-sectional diameter of tower
Since the feed is saturated liquid ,hence:
Among them
Assume that the rising vapor is ideal gas ，hence
The validity height of tower is
6.19 Caculate (a)the heat capability of condenser and the amount of condensation water (b) the heat capability of reboiler and amout of stream used for heating.Regardless of the column heat loss ,and the following conditions are given:
the operate temperatures in sections :62C(feed);47C(the top of column);75C(the bottom of column). The temperatures of reflux and overhead product are 43C.
the gauge pressure of the steam used for heating is 100kpa,and the condensation water is discharged at the saturated temperature.
The temperature of condensation water at outlet and inlet are 25C and 30C, respectively.
Solution: 1）Calculate the heat load of condensers and consumption of cooling water。The heat load of the condenser is tried to achieve by the following equation ，gives
and
From appendix ,at 47c, the latent heats of CSand CCLare345 kj/kg and 190kj/kg，respectively. The specific heats of CSand CCLat the same temperature are 0.95kj/（kg.K）and 0.8kj/（kg.K）, respectively.
Consumption of cooling water is
2） Calculate the heat load of reboiler and consumption of the steam used for heating. The heat load of reboiler can be calculated by using the following equation，gives
From appendix ,at 75c, the latent heats of CSand CCLare300 kj/kg and 185kj/kg,respectively. .
The steam used for heating
From appendix , the latent heats of steam at 1kgf/cm pressure is 2260kj/kg
6.20 A mixture of benzene, toluene and ethylbenzene carries on partial gasification. The operate pressure is 1 atm and the temperature is 120C. Raw materials liquid contains 5 percent benzene and is an ideal liquid. The following methods are considered for this operation. For each method calculate the compositions of liquid and vapor . The Antoine equation can be used to estimate the vapor—pressures of benzene, toluene and ethylbenzene .
t ---boiling point,c
p--vapor pressure ,kpa
phase equilibrium phase constant method
relatively volatility method.
Antoine constant for each component are given in the following table
Component
A
B
C
Benzene
6.023
1206.35
220.24
Toluene
6.078
1343.94
219.58
Ethylbenzene
6.079
1421.91
212.93
Solution: 1）Calculate the equilibrium composition with the phase equilibrium constant.
Calculate the phase equilibrium constant.
Benzene ,toluene and ethylbenzene are represented by 1,2 and 3.
Calculate the equilibrium compositions
Form the two upper equations：
and
the results are correct
2）calculate the equilibrium composition with relative volatility.
Calculate the relative volatility, each component relative to ethylbenzene
Calculate the equilibrium compositions
The results are given:
component
1
0.05
4.67
0.2335
0.148
2
0.375
2.05
0.769
0.487
3
0.575
1
0.575
0.365
1.0
1.5775
1.0
The result of calculations of two kinds of methods are self-same..
6.21 A continuous fractionating column is to used to separate a mixture of A,B,C and D(the volatility of component is descend in turn). The rate of recovery of B in the overhead product is 95 percent and that of C in the bottom product is 95percent, required. Estimate the compositions of two products with Hengstebecn method. assuming the feeding is ideal system . the compositions of component are given in the attached table.
Solution: Depending on the meaning of the question, it is a light component part to define component part B, component part C is a heavy key component part。
Relative volatility ,each components relative to C
Form Eq.,the value of x can be calculated and given in the following table：
component
A
B
C
D
2.58
1.99
1
0.845
A basis of 100 kmol of feed is chosen :
The moles of B in the overhead product
The moles of B in the bottom product
The moles of C in the bottom product
=30.4kmol
The moles of C in the overhead product
For mapping convenient, it uses rectangular coordinate instead of it, and changeandinto logarithm values，gives
So component B assigns the coordinates of some b are（0.299，1.279）
So component C assigns the coordinates of some c are（0，-1.279）
The distribution relation line of the component is painted in the figure of the subject under discussion。
the moles of component A and D ,and the compositions of two products
From,checked from the figure
solve and have：
Form，checked from the figure
solve and have：
According to above-mentioned results, calculate products to form , rank the result to the following table：
component
Raw materials liquid
Overhead product
Bottom product
Moles of component
kmol
Mole fraction
Moles of component kmol
Mole fraction
Moles of component kmol
Mole fraction
A
B
C
D
6
17
32
45
100
0.06
0.17
0.32
0.45
1.0
5.966
16.15
1.6
0.59
24.28
0.246
0.665
0.0659
0.0231
1.0
0.034
0.85
30.4
44.44
75.72
0.0004
0.0112
0.4015
0.5869
1.0
6.22 Repeat problem 21 ,if the feed is saturated liquid and a reflux ratio equal to 1.5 times as the minimum .from the Fenslse-Underwood equation, calculate(a) how many ideal plates are required (b) the positions of the feeding plate.
Solution: calculate the minimum reflux ratio
Feed saturated liquid，so q=1
calculatevalue of following equation first，gives
the value forθis found after replications , θ=1.603,gives
so the correct value for θis1.603
2）calculate the number of ideal plates
form Gilliland diagram ,
Solve and have:N14
So the minimum number of ideal plates in rectifying column is
since
hence N7
So the feed is introduced on the eight ideal plate from the top.
Chapter 7 Absorption
7.1 As shown in handbook ,if 100g water contains 1g ammonia ,the equilibrium
partial pressure of ammonia over a solution is 0.987KPa , Assuming that
Henry’s law applies to this system ,what are (1) the solubility coefficient H kmol/
.(2) the slope of equilibrium line (the equilibrium constant),m.
Solution :(1)calculate H
According to Eq(7-2)
And
The corresponding density of solution C can be worked out by using
the following means: We can suppose that the density of the
solution is the same as that of the water because it is very weak (the influence of ammonia can be negligible ) , and the chosen value is 1000kg/
SO:
(2) Calculate m
From Eq (7-5)
7.2 When the temperature is 10 and the pressure is 101.3KPa , the solubility of oxygen in water can be represented by equation p=3.31*x , where
superscripts p (kPa) and x refer to the partial pressure of oxygen in the vapor phase and the mole fraction of oxygen in the liquid phase, respectively. Assume that water is fully contact with the air on that condition , calculate how much oxygen can be solute in the per cubic meter of water ?
Solution: the mole fraction of oxygen in the air is 0.21,hence:
p = P y =1x0.21=0.21amt
Because the x is very small , it can be approximately equal to molar ratio X , that is
So
7.3 A mixture of gas is constituted of 2% carbon dioxide and 98% air (in volume).the gas is at 506.6kpa and 30 c. as shown in handbook ,when the temperature is 30c ,the Henry constant of the carbon dioxide in water is 1.88*10 kpa ,calculate (1) the solubility coefficient H, in kmol/(m*kpa).(2) the slope of equilibrium line ,m.(3) how much carbon dioxide can be solute in 100g water .
Solution (1) Calculate H
From Eq(7-4) :
(2) calculate m: from Eq (7-6)
(3) when y is 0.02 ,the carbon dioxide solute in 100g water is
p=5*760*0.02=76mmHg
Because the value is very small, it can be approximately equal to X, so
So: there is 0.0138g carbon dioxide in 100 g water
7.4 O2 and CO are diffusing to each other through a pipe at 101.3kpa and 0c . At each section (points A and B, 0.2 cm away, the partial pressures of O2 are 13.33kpa and 6.67kpa, respectively. The diffusion cofficient D is 0.185cm/s. Calculate the molar flow rate of oxygen under the two following conditions(kmol/m*s)(1) CO and O2 are taking equimolar counter diffusion (2) CO is a stagnant component.
Solution: 1）
From Eq(7-16) : SO:
(2) From Eq (7-20):
7.5 There is 2mm deep water in a dish ,and it slowly evaporates into air at 20c . we assume that the evaporation need go through a air film which is 5mm thin, and the partial pressure of the vapor outside the film is zero, The diffusion cofficient is 2.60*10, How long does it takes when all the water vapor into the air at 101.3kap?
Solution: this circumstance belongs to diffusion which component A diffuses through stagnant component B. and the diffusive distance (the thickness of air) Z is
Z=5mm=0.005m
The partial pressure of the vapor over the surface of water (the vapor pressure of water at 20c)
the partial pressure of the vapor outside the film of the air
and
From Eq(7-20),the molal flux of water
so:
so we can get the result of the down speed of surface of the solution
Where M= the molecular weight of water ,18kg/kmol
= the density of water at 20c
So
So the time
7.6 According to the Maxwell-Gilliland equation, estimate the diffusion coefficients D for NHand HCL in the air, respectively. The temperature is 0C and the pressure is 101.3kpa,then you should compare the result with the data given in fig 2-2.
Solution：1）calculate the diffusivity D for ammonia in the air
As shown in fig7-4 : The molar volume of air
The molar volume of ammonia
And The average molecular weight of air
The molecular weight of ammonia
Using the Maxwell—Gilliland equation (7-23), calculate the diffusivity D for ammonia in the air at 0 c and 1 atm
2） calculate the diffusivity D for HCI in the air
As shown in fig 7-4:
The molar volume of HCL
And the molecular weight of HCL
Using Eq (7-23)，calculate the diffusivity D for HCL in the air at 0Cand 1 atm
7.7 The vapor of methanol mixed with air is absorbed in water, the temperature is 27c and the pressure is 101.3 kpa .The molar density of methanol in bulk of liquid and gas phase are very weak. Henry’s law applies to this system, H=1.995kmol/m*kpa,
,
Calculate (1) (2) The percent of gas resistance in the whole resistance Where H=solubility coefficient =Individual mass-transfer coefficient for liquid phase =Individual mass-transfer coefficient for gas phase =Overall mass-transfer coefficient based on gas phase
Solution : 1)
2)
7.8 The vapor of methanol mixed with air is absorbed in water , using a packed tower ,the operating temperature is 27C, and the pressure is 101.3 kpa. In the steady state of operations, the partial pressure of methanol of gas phase is 5 kpa and the molar density in liquid phase is 2.11kmol/mon one section inside the tower. According to some data given in Problem 7, calculate the rate of absorption in this section.
Solution : According to equation (3-39),calculate the rate of absorption
From Problem 7 :
And H=1.955kmol/（m3.kN/m2）
So the equilibrium partial pressure of methanol on this section is :
7.9 A countercurrent flow tower is used to absorb HS from the air- HS steam fed to it ,using pure water as the absorbing liquid ,(Pure water is used in the countercurrent flow tower to absorb H2S from the air-H2S mixed gas),the tower is operating at 25c and 101.3kpa . The density of HS is changed (reduced)from 2% into 1% (in volume). Henry’s law applies to this system ,and Henry constant E=5.52*10kpa .if the amount of the absorbent is 1.2 times as much as the smallest amount according to theory , (1)calculate the liquid-gas ratio L/V and (outlet liquid) concentrations X1(2)Repeat calculate the liquid-gas ratio L/V and (outlet liquid) concentrations,if the operating pressure is 101.3 kpa
Solution: 1)From Eq (7-6), the slope of equilibrium line m is :
And :
Form Eq (7-54) , calculate the limiting liquid—gas ratio
So operating liquid-gas ratio is
the terminal concentration X1
2) The slope of equilibrium line
So
The terminal concentration X1
7.11 The vapor of ammonia mixed with the air is absorbed in water at 101.3kpa.The mole fraction of ammonia is 0.1,the mixture of gas is fed to the bottom of the tower at 40c, the volume of entering gas is 0.556m/s and the superficial gas velocity based on empty tower is 1.2m/s.,the amount of the absorbent is 1.1 times as much as the smallest amount according to theory ,and 95 percent of ammonia is recovered(absorbed) . The average of is 0.0556kmol/(m*s)
(where is the overall gas- phase coefficient ,based on unit packed volume) .The water is fed to the top of the tower at 20c, because it gives off heats of solution when absorbing ammonia , so the closer ammonia gets to the bottom of tower, the higher temperature of it is(will be) . According to heat effection , calculate the density of ammonia and temperature in the tower ,and the corresponding equilibrium concentration in liquid, then write in the apposing figure ,calculate the cross-section diameter of tower and the height of the packed section.
THE APPOSING MAP
The temperature
Of ammonia solution
c
The concentration of ammonia solution
Kmol(ammonia)/kmol(water)
The gas-phase equilibrium concentration of ammonia Y
Kmol(ammonia)/kmol(air)
20
0
0
23.5
0.005
0.0056
26
0.01
0.010
29
0.015
0.018
31.5
0.02
0.027
34
0.025
0.04
36.5
0.03
0.054
39.5
0.035
0.074
42
0.04
0.097
44.5
0.045
0.125
47
0.05
0.156
Solution :
Cross-section area of tower
Calculate,1) using logarithmic means driving force：
2) Schematicallyad
From fig7-18,searching out ≈10.5
Overall height of transfer unit based on gas phase
The height of the packed section
7.12 A countercurrent flow tower is used to absorb SOmixed with the air ,using pure water as the absorbing liquid , (Pure water is used in the countercurrent flow tower to absorb H2S from the air-H2S mixed gas),the volume of entering gas is 5000m(standard)/h, and 10 percent of SOin the gas .95 percent of SOis absorbed in water. In the state of operations ,the equilibrium equation is ,calculate (1) if the amount of the water is 1.5 as much as the smallest amount according to theorization ,how much water must be fed to the tower.(2)According to above condition ,calculate how many ideal plates are required for the tower by Schematicallyad.(3)If the number of the ideal plates is the same as Question (2),but 98 percent of SOis absorbed, under this condition ,how much water will be fed to the tower ?
Solution:
According to the data given in the figure , we need make a equilibrium curve OE in the X—Y right—angle coordinate (apposing map 1),and read the equilibrium concentration in liquid corresponding to Y=0.1111
So:
Cross--sectional area of tower
Cross—sectional diameter of tower
According to X1，X2，Y1，Y2，plot the operating line BT in the apposing map 1. we can read the value of Y and the corresponding overall driving force in gas phase ，and write in the following figure ：
Y
0.0056
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.1111
0
0.0015
0.006
0.009
0.013
0.019
0.027
0.035
0.046
0.057
0.072
0.090
0.0056
0.0085
0.014
0.021
0.027
0.031
0.033
0.035
0.034
0.034
0.028
0.021
180
117.5
71.5
47.7
37.0
32.3
30.3
28.6
29.4
29.4
35.7
47.5
According to the data given in the above figure ,we can draw a curve ——Y（apposing map 2），calculate the acreage surrounding by the curve, Y=0.1111，Y=0.0056and three =0 beeline
and
so
The height of the packed section
7.13 The butane mixed with the air is absorbed in a sieve-plate tower containing eight ideal plates .The absorbing liquid is a non-volatilization oil having a molecular weight of 250 and a density of 900kg/m.The absorption takes place at 101.3kpa and 15c. 5 percent of butane in the entering gas .The butane is to be recovered to the extent of 95% ,the vapor pressure of butane at 15C is 194.5kpa,and liquid butane has a density of 580kg/m.Assume that Raoult’s and Dalton’s laws apply .(1)calculate the cubic meters of fresh absorbing oil per cubic meter of butane recovered .(Caculate the amout of absorbing oil is required (in volume) when per cubic meter of butane is recovered) (2)Repeat caculate (1), on the assumption that the operating pressure is 3034.4kpa and that all other factors remain constant.(conditions reain the same)
Solution: 1）calculate the mass flow rate of water
2） Calculate ideal plates
(a) Schematicallyad
Plot the operating line BE and the equilibrium line OE (Y=26.7X) in the X-Y right-angle coordinate .Draw the rectangular steps between the operating line and the equilibrium line from the point B, and we can get N
So : ideal plates
(b) Using Kremser.A. map
The absorption coefficient is 95%，and X2=0，so the absorption coefficient of relativity，under the limiting amount of water according to theory condition ，。Looking into fig7-21 according to these data, we can get:
The absorption factor in the state of operation
Looking into fig 7-21（or calculate according to equation 7-77）,we can know when A=1.43，,the ideal plates。
So the result is the same as that of schematicallyad
3) This question should be estimate by Kremser.A. map
absorption coefficient，ideal plates
Form fig 7-21， reading
So:
7.14 A countercurrent flow tower is to absorb minimum 99 percent of the HS in the air-HS stream fed to it ,using tri-ethanol amine as the absorbing liquid .(Tri-ethanol amine is used in a countercurrent flow tower to absorb H2S in the air-H2S stream fed to it ,the absorptivity is at least 99 percent)The operating temperature is 300K and the pressure is 101.3kpa.2.91 percent of HS in the entering gas ,the equilibrium equation is Y=2X. The inlet liquid is pure absorbefacient (absorbent )and the mole fraction of HS in the outlet liquid is 0.013kmol(HS)/kmol(solvent) .The mole flow rate of the chemically inert gas (gas through the cross-section)is 0.015kmol/(m*s). Overall absorption coefficient based on gas phase ka ,is 0.000395kmol/(m*s*kpa), try to calculate the total height of the packed section.
Solution: According to equation 7-63,
The overall number of transfer units based on gas phase
:
From Eq (7-50a) ,we can get
And X=0, Y=mX
So
So:
（——component absorption）
According to Problem ,we can know that the operating liquid-gas ratio is1.4 as much as the limiting liquid-gas ratio
And：
So：
7.15 An absorption tower is to recover 99 percent of ammonia in an air stream ,using pure water as the absorbing liquid .(Pure water is used in an absorption tower to absorb ammonia in an air stream ,the absorptivity is 99 percent),the height of the packed section is 3m ,the absorption takes place at 101.3kpa and 20c. The mass flow rate of gas V is 580kg/(m.h),and 6 percent of ammonia in the gas in volume . The mass flow rate of water L is 770kg/( m.h). The tower is countercurrent operated under isothermal temperature, the equilibrium equation Y=0.9X, ka,but has nothing to do with L, try to make out how the height of the packed section to change in order to keep the absorption coefficient unchanged when the conditions of operation have be changed as following (1)the operating pressure is 2 times as much as the original.(2)the mass flow rate of water is one time more than the original3) the mass flow rate of gas is two times as much as the original
Solution:
The average molecular weight of the mixed gas M=29×0.94+17×0.06=28.28
1）
Form Eq(7-6),
So
Form Eq(7-62) and Eq(7-43a)
So: is changed with the operating pressure
So
So the height of the packed section reduce 1.802m against the original
2）
when the mass flow rate of gas increasing, has not remarkable effect
the height of the packed section reduce 0.605m against the original
3）
when mass flow rate of gas increasing , corresponding grow，according to Poblem:
So the height of the packed section increase 4.92m against the original
7.16 A mixture of 1 percent acetone and 99 percent air is absorbed in a sieve-plate tower. The absorption takes place at 101.3kpa and 27C. the acetone is to be recovered to the extent of 90 percent ( the absorptivity of acetone is 90 percent ,requried).the molar flow rate of entering gas is 30kmol/h,the molar flow rate of water is 90kmol/h, the equilibrium equation Y=2.53X,how many ideal plates are required for the tower?
Solution :1) Schematicallyad
Plot the operating line BT and the equilibrium line OE in the Y-X
right-angle coordinate.
The point B coordinate： X=X1=0.003
Y=Y1=0.0101
The point A coordinate： X=X2=0
Y=Y2=0.00101
The slop of operating line m=2.53
From the point B, and draw rectangular steps between the line OE and BT
Get the ideal plates。
2）Analytical method
V=29.7kmol/h，L=90 kmol/h，m=2.53
So
So
So
Chapter 10 Drying
10.1 The total pressure and humidity of the humid air are 50kPa and 60 C , respectively, and the relative humidity is 40 percent. Calculate (a) the partial pressure of aqueous vapor in the humid air (b) humidity (c) the density of humid air.
Solution: a）The partial pressure of aqueous vapor in the humid air
From aqueous vapor appendix table, the vapor pressure of aqueous vapor at 60℃ is 149.4mmHg。
b） Humidity
c） The humidity volume is：
=2.24(m3humid air)/(kg dry air )
So the density of humid air
10.2 Use Humidity chart for humid air to find the values of blank space in the attached list, and draw the solving course flow chart of question four.
THE ATTACHED LIST
Number
Dry-bulb temperature
C
Wet-bulb
Temperature
C
Humidity
Kg/kg dry air
relative
Humidity
%
Enthalpy
KJ/kg dry air
Partial pressure of vapor
Kpa
Dew point
C
1
(60)
(35)
2
(40)
(25)
3
(20)
(75)
4
(30)
(4)
Solution: the values in the bracket data of the attached list are known, and the other values of items can be found from H-I chart。
number
Dry-bulb temperature
C
Wet-bulb
Temperature
C
Humidity
Kg/kg dry air
relative
Humidity
%
Enthalpy
KJ/kg dry air
Partial pressure of vapor
kpa
Dew point
c
1
（60）
（35）
0.026
21
130
31
29
2
（40）
28
0.020
43
92
24
（25）
3
（18）
18
0.011
（75）
48
14
17
4
（30）
29
0.025
93
94
（30）
28.5
The solving course flow chart of question four.（Under right）
10.3 Humid air at dry-bulb temperature 20C and humidity 0.009(kg water)/(kg dry air) is pre-heated to 50C by pre-heater, then enters atmosphere dryer. The air leaves dryer at the relative humidity 80%. Assuming the air experienced equal enthalpy drying process. Calculate (a) The change of the enthalpy of 1 cubic meter original humid air in the course of pre-heating (b) The content of moisture that 1 cubic meter of original humid air obtains from the dryer .
Solution: (a) The change of the enthalpy of 1 cubic meter original humid air in the course of pre-heater
From humidity chart :
At t=20c and H=0.009, the enthalpy Iis found to be 45kJ/(kg dry air)
At 50c and H=0.009, the enthalpy Iis found to be 73kJ/(kg dry air)
humid air
The specific volume of original humid air
humid air/m3 dry air
Hence: The change of the enthalpy of 1 cubic meter original humid air is:
b）From humidity chart，the equal enthalpy line through the point 50C and H=0.009 intersects the 80 percent line at 0.0182kg water/kg air，this is the humidity of the air leaving the dryer.
Hence: The content of moisture that 1 cubic meter of original humid air obtains from the dryer is:
kg water/m3humid air
10.4 A dryer is used to dry material from 5 percent moisture to 0.5 percent moisture (wet basis). The production capacity of dryer is 1.5 kg dry material/s. Hot air enters at 127℃ and 0.007kg water/kg dry air and leaves at 82℃.The temperature of the material in the inlet and outlet are 21℃ and 66℃, respectively. The specific heat of wet material is 1.8KJ/（kg(℃）。If the heat loss of the dyer can be ignored, then calculate a) the consumption of the wet air. b) the temperature of the air leaving the dryer.
Solution：The moisture content of the material (dry basis) is
From the material balance, we have:
From the enthalpy balance, we have:
Where：
kg water/kg dry air （3）
From these, the ，L=6.61kg wet air/s
10.3 Carry some humid air（to=25℃,Ho=0.0204kg water/kg dry air） into the atmosphere dryer after pre-heating. Calculate： a) how much heat will be required for pre-heating humidity air to 80℃.( KJ/kg dry air)； b)the relative humidity corresponding to 120℃.
Solution：a)calculate how much heat will be required.
Form the temperature and humidity of the air, the point on the humidity chart for this air is located
At the enthalpy Io is found to be 75kj/kgbone-dry air
And
At ，the enthalpy Iis found to be 134kJ/kg bone-dry air，
Hence：The heat required for preheating humid air to 80℃ is：
2）Calculate the relative humidity of the air at 120℃.
At 120℃ and 1 atm ,the vapor pressure of air is
Hence:
10.8 Humid materiel is dried at a atmosphere dryer，and the width of network of metal is 0. 8m。The flow rate of the air across the material layer is 2m/s, at the average temperature 75℃,the air humidity is 0.018(kg water)/(kg dry air). Try to calculate the content of moisture vaporized from each meter conveying belt per hour under the constant drying conditions.
Solution：From thermal diffusivity，
L=u(3600
humid air/kg dry air
From the H-I chart, when the wet-bulb temperature of air is 34℃，the corresponding latent heat of water is 2416kj/kg.
The rate of drying under constant drying conditions can be calculated by the following equation:
So the content of moisture vaporized is：
1.92×0.8=1.536kg water/（m .h）
10.2 A humid material is dried at a dryer.55 hours are required to reduce moisture content from 35 percent to 10 percent. The critical moisture content was found to be 15 percent and the equilibrium moisture 4 percent. If under the same drying conditions, it is required that the moisture content of the material drops to 0. 05 from 0.35.Assuming that the rate of drying is proportional to the free-moisture content (X—X*), try to calculate how long will it take to dry the humid material.
Solution: The drying time at constant rate stage is
The drying time at falling rate stage is
So the required drying time under the new condition is
10.3 A atmosphere parallel-counterflow dryer is to be designed to dry a humid material from 1.0kg moisture /kg dry material to 0.1kg moisture /kg dry material using hot air. The air enters at 135℃ and 0.01kg water/kg dry air, and leaves at 60℃. Assuming the air experienced equal enthalpy drying process. According to experiment, the rate of drying under the constant-rate period can be represented by equation -dx/dτ=30ΔH kg water /(kg dry material. H);the rate of drying can be represented by equation -dx/dτ=1.2x kg water/(kg dry material*h). Try to calculate how long should it takes to dry this material?
Solution：Because the air experienced equal enthalpy drying process, the values of t1，H1，and t2 can be used to find the following values from the humidity chart:
kg water/kg dry air
kg water/kg dry air
From material balance.
Because the rates of drying in the meeting point between two stages are the same, so:
Assume that the humidity of air is Ho corresponding to the critical moisture content of material Xo.
Hence: （1）
（2）
From Eqs (1) and (2):
Xo=0.74kgwater/kg dry material
Ho=0.0184kgwater/kg dry air
The time of drying, in constant rate stage
Separate variables and integrate:
2 The time of drying in falling rate stage
10.4 Carry out the drying experiment under the constant drying conditions. The drying area is 0.2m2, and the mass of dry material is 1.5kg. The experimental data is given in the following table. Try to draw drying-rate curves, and calculate the equilibrium moisture content and the critical moisture content.
Time, h
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
The quality of humid material，kg
44.0
37.0
30.0
24.0
19.0
17.5
17
17
Solution：The results are given in the following table:
Time θ，h
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
G-Gc
29.1
22
15.0
9.0
4.0
2.5
2.0
2.0
1.94
1.47
1.0
0.6
0.27
0.16
0.13
0.13
1.705
1.235
0.8
0.435
0.215
0.145
0.13
0.47
0.47
0.4
0.33
0.11
0.03
0
177.5
175
150
125
37.5
12.5
0
According to results, drawing drying –rate curves, as shown in the figure.
From the figure, the
The first step of the falling-rate period: The second step of the falling-rate period:
10.6 A certain drying process is to be designed to dry a humid supplies, using the waste gas of circulation. The fresh gas enters at 20℃and 0.012kgwater/kg dry air humidity and leaves at 50℃.Some waste gas at 0.079 humidity enters the pre-heater after mixing, The rate of circulation (the flow of the dry air in the waste air is proportion to the flow of the dry air in the mixed gas)is 0.8.The mixture enters the parallel-counter flow dryer after raising temperature, and some of the waste gas leaving the dryer are recycled, the remain is discharged. The moisture content of humid material is reduced from 47% to 5%.Calculate：1）The flow of the fresh air ；2）how much heat will be required for the system .The flow of the humid supplies is 1.5×103kg/h。Assume that the heat loss of pre-heater can be ignored, and the operation is equal enthalpy drying process.
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Solution：1） From the material balance
The humidity of the air entering pre-heater
and
the moisture content of the supplies (dry basis):
so the mass flow of the fresh water
2）the heat will be calculated for the system
Because the operation is equal enthalpy drying process.
and
According to Ho，t2 and H2, the values and data required can be found from the H-I chart or calculated by the following equation:
The enthalpy of the air entering pre-heater is
kj/kg wet air
10.1 A crystal material is to be dried in an atmosphere rotary countercurrent dryer. The original air at 25℃ and 55% percent humidity is pre-heated to 85℃ by pre-heater, then enters the dryer and leaves it at 30℃。The temperature of the humid material rises from 24℃ to 60℃，and the moisture content is reduced from 3.7 percent to 0.2 percent(wet basis).The mass flow rate of the products is 1000kg/h。The specific heat of the dry material is 0.36KJ/（kg(℃）.and the diameter and length of the dryer are 1.3m and 7m,respectively. the convective heat transfer coefficient of air in the outside of dryer αT is 8.37kcal/m2h*℃），Calculate the mass flow rate of dry air and the consumption of the heating steam. Assuming the gauge pressure of the heating steam is0.5kgf/cm2.
Solution：1）The material balance
where :
Hence :W=998(0.0384-0.002)=36.33kg water/h
the vapor pressure of water at 25 C P=23.76mmHg
Hence: L(H-0.0109)=36.33
2）The enthalpy balance
Based on the whole system
Based on pre-heater
where：
From the Eqs (1) and (2)，the L=2375kg/h，H2=0.0262kgwater/kg wet air
So the consumption of the heating steam
The latent heat of the water at 1.533kgf/cm2 pressure is 2223kj/kg
[Other solution ]the consumption of air can be calculated by the simple method
We have: L=2373kg/h
10.2 An atmosphere parallel-counterflow dryer is used to dry a crystal material under the following conditions:( 1)The fresh air at 27℃ and 0.008kgwater/kgdry air humidity enters the dryer at t1 90℃ and leaves at t2=45℃。 2） The material enters dryer at 25℃ and 6 percent in humidity, and leaves at 40℃ and 0.3 percent in humidity（wet basis）.The density of material is ρs=1020kg/m3，and the specific heat of dry material is Cs=0.32kcal/(kg dry material*℃)， and the average diameter of the particle is dp=3mm；The inlet mass flow rate of the humid material is 2500kg/h.If 0.8h is required to dry the material and the heat loss of the dryer equals 20 percent of the heat that the air conveys to material, then calculate the length ,diameter, rotational speed and sloping rate of the dryer.
解：1）the material and enthalpy balance during the time of drying ，the moisture content of material (wet basis) is：
kg water/kg bone-dry material
kg water/kg bone-dry material
the mass flow of the bone-dry material：
kgbone-dry material/h
The evaporation capacity of the moisture is
From the temperature and humidity of the air, the point on the humidity chart for this air is located
At t1=90℃，H1=0.008kg water/kg bone-dry air, the
At kg water/kg bone-dry air, the
From the enthalpy balance equation10-32a and 10-35a :
where
hence：L=2.9kgbone-dry air/s
From the material balance:
kg water/kgbone-dry air
The diameter of dryer can be calculated by using the max mass flow rate of the air, and the mass flow rate of the air leaving dryer is :
Choose the mass flux of the air according to the diameter of the particle，so the cross-section area of dryer is
Choose the diameter and cross-section area of dryer 1.4m and 1.54m2, respectively
The length of dryer can be calculated by using the heat transfer rate equation,
The can be calculated by using the following equation
The length of preheating area Z1
Assume the temperature of the surface of the evaporation material is
The heat transfer at the preheating area is
According to the enthalpy balance, calculate the temperature at the intersection of the preheating area and the evaporation area:
We have：
The length of the evaporation area
The heat loss of air is：
From the appendix, at 32℃, the latent heat of water is 2417kj/kg。
We have：
the length of the heating equipment
The heat transfer by air in a dryer is
Hence:
because the value is in the range of 4—10,so the calculation result is right .
4) The rotational speed and sloping rate of the dryer
because the diameter of the dryer is not very large，we can use large rotational speed,
let n=6rpm
from Eq 9-74:
The solution above the equation is T=0.006
Because the value of T in the range of 0.1 ~1，the result is right。
Check and reexamine the calculation process and result
where：
Because the value of is in the range of 0.05~0.15，the size of dryer is agreement with the designing requirement.
| 课件名称： | 华南理工大学：化工原理习题及（试卷习题） |
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| 课件类型： | 试卷习题 |
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