流体力学与传热习题参考解答(英文)

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?

U 22P 2U 12P 1

w=Z2g++h f +-(Z1g++)

2ρ2ρ

U 1=0 P1=P2 Z1=0 W=60j/kg h f =30/kg

U 2=1m/s Z2=(60-30-0.5) /g =3m

∆Z =Z 2-Z 1=4-3=1m

2. The fluid (density 1200 kg/m3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?

U PU

+W =2+Z 2g+2+h f

ρ2ρ2

200

x1.013x105=-2.67x104N /m 2 P1=0 P 2=-760

ρ=1200Kg /m 3 U 1=0 h f =120J /kg

V 20U 2===1.97m /s 2

A 3600*π/4*0. 06Z 1=0 Z 2=15

+Z 1g+

2.67x1041.972

W =-+15x9.81++120=246.88J /kg

12002

N =W ρQ =246.88x1200x20/3600=1646W

3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:

h f =6.5U2

where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m3 /h.

+Z 1g+

U PU

=2+Z 2g+2+h f 2ρ2

U 1=0 P1=P2 Z 1=6m Z 2=0

U 2

+6.5U 2 h f =6.5U 6x9.81=2

. 2x3600=82m 3/h U =2.9m /s V=UA=2.9π/4x01

4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.

U a A a =U b A b U b =2.5*(33/47) 2=1.23m/s Pa

U a 2Pb U b 2

+Z a g+=+Z b g++h f Z a =Zb ρ2ρ2Pa Pb U b 2U a 2

-=-+h f =1.232/2-2.52/2+15=12.63 ρρ22

12.63

=1.29x10-3m P a -P b =∆R ρg ∆R=3

9.8x10

5. A centrifugal pump takes brine (density 1180 kg/m3 , viscosity 1.2 cp) from the bottom of a supply tank

and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?

l U 2U 2

∑h f =h fst +h flocal =4f +∑k

d 22

ρ=1180kg /m 3 l =300m, d=0.025m

v =2m 3/h μ=1.2cp=1.2x10-3Pa.s k=0.025mm k/d=0.025/25=0.001 k c =0.4 kl =1 kr =2x0.07=0.14 k el =4x0.75=3 kre =1.5-2.2

u =v/A =2/(3600xπ/4x0.0252) =1.13m/s

u ρd Re ==2.78x104 f =0.063

∑h f =4x0.0063x300/0.025x1.132/2+ (0.4+1+2x0.07+4x0.7+1.5)x1.132/2 =197.86J/kg

6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken

as 0.61, what is the flow rate of water?

u o =c

V 0=u o s 0=0.61x π/4x0.00012

=5.8x10-8m 3/s

7. Water flows through a pipe with a diameter di 100 mm as shown in figure.

a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m3 /h?

b. If the valve opens fully, what is the pressure of section 2-2', in N/m2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m3, ρHg =13600kg/m3)

(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have

u 12P1u 22P2

gZ 1++=gZ 2+++h f1-2

2ρ2ρ

P1=0 P2=g(ρHg R -ρH 2o h) =39630N/m 2

u 1=0 Z2=0 hf1-2

l u 2u 2

=4f +kc =2.13u

d 22

We can get Z1 from the valve closed

h=1.5m R=0.6m Z1=ρHg gR/ρH 2O -h =6.66m

9.81x6.66=u 2/2+2.13u 2+39630/1000

u=3.13m/s Vh =3600x π/4x0.12x3.13=88.5m 3/h

(2) when the valve opens fully, for section 1-1’ and 3-3’, we have

u 32P3u 12P1

gZ 1++=gZ 3+++h f1-3

2ρ2ρ

Z 3=0 Z1=6.66m u1=0

l +l e u 23. +1.5

+k c ) =(4x0.00625x+0.5)=4.81u 2 h f1-3=(4fd 20.0122

9.81x6.66=u /2+4.81u u =3.51m /s

For section 1-1’ and 2-2’

u 12P1u 22P2

gZ 1++=gZ 2+++h f1-2

2ρ2ρ

P1=0 Z1=6.66 Z2=0 u1=0 u2=3.51

l u 2

=(4x0.00625x15/0.1+0.5)3.512/2=26.2J /kg h f1-2=(4f+k c )

d 2

9.81x6.66=3.152/2+26.2+2

P2=329702

8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m3 /h, what is the flow rate for pipeline B? (fA =0.0075, fB =0.0045)

For parallel pipe line

∑h fA =∑h fB Vtotal =V A +VB

∑(l+le)u A 22.722

∑h fA =4f A =4x0.0075x10/0.053/2()

d A 23600x π/4x0.0532

=0.333J /kg

∑(l+le)u B 2

∑h fB =4f B =4x0.0045x2/0.3/2xu B 2=0.333

d B 2

u B =2.36m /s VB =uB A B =2.36x π/4x0.232=600m 3/h

10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(mo C), backed by a 150 mm of common brick, of conductivity 0.8 w/(mo C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.

a. What is the heat loss through the wall in w per square meter.

b. To reduce the heat loss to 600 w/m2 by adding a layer of cork with k 0.2 w/(mo C) on the outside of common brick, how many meters of cork are requied?

Q ∑∆t 1400-200a. ===711N /m 2

11L ∑R +0.120.15

b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m

13. Air at the normal pressure passes through the pipe (di 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:

density 1.06 kg/m3 , viscosity 0.02 cp, conductivity 0.0289 w/(mo C), and heat capacity 1 kJ/kg-K

u ρd 10x0.02x1.06 Re===1.06x104>104 -3

μ0.02x10

μω⎫

T=T 1+T2=20+100=60℃ ⎛ ⎪=1 μ22⎝⎭

c μ

Pr=p =1000x 0.02x 0.001=0.692

k 0.0289

0.14

Nu =0. 027Re 0. 8Pr 1/3=0.027x (1.06x104)

0.8

x (0.692)

1/3

=39.66

h i d i =39.66 hi =39.66x0.0289/0.02=57.22w/(m 2.k )

14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:

k=0.5 w/(mo C), Cp =4 kJ/kg-K, viscosity 10-3 N-s/m2 , density 1000 kg/m3 Find: a. Heat transfer film coefficient hi , in w/(m2 -K).

b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the hi ?

c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate hi .

d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?

e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain? Hint: for laminar flow, Nu=1.86[Re Pr]1/3 for turbulent flow Nu=0.023Re0.8 Pr 1/3

4N 2250x4

u ρd Gd ====3.98x104>104 (1) Re =μμμ0.001

Nu =0. 023Re 0. 8Pr 1/3=0.023x (3.98x10

40.8

)

⎛4⎫

⎪⎝0.5⎭

1/3

=220.1

Nuk 220.1x0.5

==5500w /(m 2k )d 0.02

(2) w 1=2w 2 Re 2=Re 1/2=2x104>104

hi =

Nu 2⎛Re 2⎫h i20.8

= =0.5 =0.50.8 ⎪Nu 1⎝Re 1⎭h i1

0.82

h i2=5500x0.5=3159w /(m k )

(3) Re 3=

0.8

u 3ρd 3

2000x0.01

=2x104>104

0.001

Pr 1/3 hi=6347w/(m 2k )

Nu =0.023Re

0.8

(4)Q=hi A i (t-tw )=400=500x2πx0.02(t-tw )

t=40℃ t w =39.41℃

(5) there methods : increase u or hi or decrease d The first is better

15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through

the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The hi (film coefficient inside pipe) is 700 w/(m2 oC)and overall heat transfer coefficient Uo (based on the outside surface of pipe) is 300w/(m2 oC), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(mo C), find:

(1) heat transfer film coefficient outside the pipe ho ? (2) the pipe length required for counter flow, in m?

(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?

(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?

(a)

11⎛d 0l m d o ⎫1230.002x23

=- +=--⎪

h 0Vo ⎝h i d i kd m ⎭300700x1945x21

1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21

h 0=642. 9w /m k

()

LMTD=70℃=

∆t 1+∆t 2

Q=UoAo∆Tm=mcCp(Tcb-Tca) 2

300*2π*0.023*70L=500/3600*1000*(80-20)

L=5.4m

80-20

=86.5℃

140-20ln

140-80

L ∆t

1=1=70/86.5 L 2=0. 81L =1

L 2∆t 2

(d) scale is formed on the outside ,V0 is decreased

4m 4.

16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is

1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored) 111111=+ (1) =+ (2) Uo h i h o U ' o h ' i h o

111111

(1)-(2)= -=0.8-0.8=0.8-0.8

21152660u 1C u 2C 1C 1.5C 111

C=2859 =- ho=8127W/(m2K)

h o Uo h i

17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)

Q =W h C h (T 1-T 2)=W c C c (t 2-t 1)=VA ∆t m

W h C h (T )=' 1-T 2

W t )' 1t c (C c -2

150-100-4015

t 2=50=℃

150-802-t 15L T -T 2∆' t m 1150-8092. 5

2=1 ==1. 85

L 1T 1-T 2∆t m 215-010069. 8

L 2=1. 85m L 1=∆1m m = t ∆95 t 1 m 2. 2=

69. 8

18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored)

Q 1=h 1i A i ∆T m 1=m c 1C pc (T cb -T ca )

Q 2=h 12A i ∆T m 2=m c 2C pc (T ' cb -T ca ) h i2∆T m 2

h i 1∆T m 1

∆T m 1

∆T m 2=

m c 2∆T

=1. 2=1. 20. 8m 2 m c 1∆T m 1

80-20

= ln(116. 3-20) /(116. 3-80)

80-20

ln(T h -20) /(T h -80)

Th=118.5oC

19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (Cp 1.9 kJ/kgo C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m2 o C) and 1.7 kw/(m2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(mo C).

W=1.25Kg/s Cp=1.9Kj/kg℃

Q 2=W h C p (T 1-T 2)=1.25x1.9x (80-30)=119Kw

1Q 119x10330-102

V 0==472w/(m k )A i = ==13.9m 2 ∆t m =

d l d 130V 0∆t m 472x18.2+0+m 0ln

h 0h i d i kd m 10

20. A spherical particle (density 2650 kg/m3) settles freely in air at 20 o C (density of air 1.205 kg/m3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stokes’ Law?

Re ≤1 U t =

D p 2g (ρP-ρ)18μ

μ18μ23

D p = =

ρg ρP-ρρD p

1/3

⎛⎫18x10-10-5

D p = =3.85x10 ⎪ 1.205x9.81x 2650-1.205⎪

⎝⎭

21. A filter press(A=0.1 m2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?

for filter press V 2+2VV e =KA2θ 5 min 12+2V e =0.12x5K (1) 10min 1.62+2x1.6V e =0.12x10K (2) From (1) (2),we can see Ve=0.7 K=48

15 min V +2x0.7V=48x0.1x15 V=2.07m/h

22. The following data are obtained for a filter press (A=0.0093 m2) in a lab.

------------------------------------------------------------------------------------------------ pressure difference (kgf /cm2 ) filtering time (s) filtrate volume (m3 )

1.05 50 2.27⨯10-3 660 9.10⨯10-3 3.50 17.1 2.27⨯10-3 233 9.10⨯10-3 Find

1) filtering constant K, qe , t e at pressure difference 1.05 kgf /cm2 ?

2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kgf /cm2 ?

3) compressible constant of cake s?

For p=1.05Kg/cm

q 2+2qq e =K θ

0.000227⎛0.00227⎫

+2x q e =50K ⎪

0.00930.00093⎝⎭

0.00091⎛0.00091⎫

+2x q e =660K ⎪

0.000930.00093⎝⎭

We can see K=0.015 qe=0.026

For p=3.5Kg/cm

K=2k∆P1-s K'=2k∆P' 1-s

1-s

KA 2⎛d V ⎫⎛∆P' ⎫ K = ⎪=⎪

d θ2V+VK ' ⎝∆P⎭⎝⎭E e

23. A slurry is filtered by a 0.1 m2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows: (q+10)2 = 250(t+ 0.4) where q---l/m2 t----min

find (1) how much filtrate is got after 249.6 min?

(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)

（1）let θ=249.6 (q+10)=250x (249.6+0.4)

q=240 V=qA=240*0.1=24 (2) K =2k ∆P K' =2k ∆P'

∆P' =2∆P K ' =2K =500 (q'+10)=500x (249.6+0.4) q’=343.6 v=34.36

U 22P 2U 12P 1

w=Z2g++h f +-(Z1g++)

2ρ2ρ

U 1=0 P1=P2 Z1=0 W=60j/kg h f =30/kg

U 2=1m/s Z2=(60-30-0.5) /g =3m

∆Z =Z 2-Z 1=4-3=1m

U PU

+W =2+Z 2g+2+h f

ρ2ρ2

200

x1.013x105=-2.67x104N /m 2 P1=0 P 2=-760

ρ=1200Kg /m 3 U 1=0 h f =120J /kg

V 20U 2===1.97m /s 2

A 3600*π/4*0. 06Z 1=0 Z 2=15

+Z 1g+

2.67x1041.972

W =-+15x9.81++120=246.88J /kg

12002

N =W ρQ =246.88x1200x20/3600=1646W

3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:

h f =6.5U2

where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m3 /h.

+Z 1g+

U PU

=2+Z 2g+2+h f 2ρ2

U 1=0 P1=P2 Z 1=6m Z 2=0

U 2

+6.5U 2 h f =6.5U 6x9.81=2

. 2x3600=82m 3/h U =2.9m /s V=UA=2.9π/4x01

U a A a =U b A b U b =2.5*(33/47) 2=1.23m/s Pa

U a 2Pb U b 2

+Z a g+=+Z b g++h f Z a =Zb ρ2ρ2Pa Pb U b 2U a 2

-=-+h f =1.232/2-2.52/2+15=12.63 ρρ22

12.63

=1.29x10-3m P a -P b =∆R ρg ∆R=3

9.8x10

5. A centrifugal pump takes brine (density 1180 kg/m3 , viscosity 1.2 cp) from the bottom of a supply tank

l U 2U 2

∑h f =h fst +h flocal =4f +∑k

d 22

ρ=1180kg /m 3 l =300m, d=0.025m

v =2m 3/h μ=1.2cp=1.2x10-3Pa.s k=0.025mm k/d=0.025/25=0.001 k c =0.4 kl =1 kr =2x0.07=0.14 k el =4x0.75=3 kre =1.5-2.2

u =v/A =2/(3600xπ/4x0.0252) =1.13m/s

u ρd Re ==2.78x104 f =0.063

∑h f =4x0.0063x300/0.025x1.132/2+ (0.4+1+2x0.07+4x0.7+1.5)x1.132/2 =197.86J/kg

as 0.61, what is the flow rate of water?

u o =c

V 0=u o s 0=0.61x π/4x0.00012

=5.8x10-8m 3/s

7. Water flows through a pipe with a diameter di 100 mm as shown in figure.

b. If the valve opens fully, what is the pressure of section 2-2', in N/m2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m3, ρHg =13600kg/m3)

(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have

u 12P1u 22P2

gZ 1++=gZ 2+++h f1-2

2ρ2ρ

P1=0 P2=g(ρHg R -ρH 2o h) =39630N/m 2

u 1=0 Z2=0 hf1-2

l u 2u 2

=4f +kc =2.13u

d 22

We can get Z1 from the valve closed

h=1.5m R=0.6m Z1=ρHg gR/ρH 2O -h =6.66m

9.81x6.66=u 2/2+2.13u 2+39630/1000

u=3.13m/s Vh =3600x π/4x0.12x3.13=88.5m 3/h

(2) when the valve opens fully, for section 1-1’ and 3-3’, we have

u 32P3u 12P1

gZ 1++=gZ 3+++h f1-3

2ρ2ρ

Z 3=0 Z1=6.66m u1=0

l +l e u 23. +1.5

+k c ) =(4x0.00625x+0.5)=4.81u 2 h f1-3=(4fd 20.0122

9.81x6.66=u /2+4.81u u =3.51m /s

For section 1-1’ and 2-2’

u 12P1u 22P2

gZ 1++=gZ 2+++h f1-2

2ρ2ρ

P1=0 Z1=6.66 Z2=0 u1=0 u2=3.51

l u 2

=(4x0.00625x15/0.1+0.5)3.512/2=26.2J /kg h f1-2=(4f+k c )

d 2

9.81x6.66=3.152/2+26.2+2

P2=329702

For parallel pipe line

∑h fA =∑h fB Vtotal =V A +VB

∑(l+le)u A 22.722

∑h fA =4f A =4x0.0075x10/0.053/2()

d A 23600x π/4x0.0532

=0.333J /kg

∑(l+le)u B 2

∑h fB =4f B =4x0.0045x2/0.3/2xu B 2=0.333

d B 2

u B =2.36m /s VB =uB A B =2.36x π/4x0.232=600m 3/h

a. What is the heat loss through the wall in w per square meter.

b. To reduce the heat loss to 600 w/m2 by adding a layer of cork with k 0.2 w/(mo C) on the outside of common brick, how many meters of cork are requied?

Q ∑∆t 1400-200a. ===711N /m 2

11L ∑R +0.120.15

b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m

density 1.06 kg/m3 , viscosity 0.02 cp, conductivity 0.0289 w/(mo C), and heat capacity 1 kJ/kg-K

u ρd 10x0.02x1.06 Re===1.06x104>104 -3

μ0.02x10

μω⎫

T=T 1+T2=20+100=60℃ ⎛ ⎪=1 μ22⎝⎭

c μ

Pr=p =1000x 0.02x 0.001=0.692

k 0.0289

0.14

Nu =0. 027Re 0. 8Pr 1/3=0.027x (1.06x104)

0.8

x (0.692)

1/3

=39.66

h i d i =39.66 hi =39.66x0.0289/0.02=57.22w/(m 2.k )

14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:

k=0.5 w/(mo C), Cp =4 kJ/kg-K, viscosity 10-3 N-s/m2 , density 1000 kg/m3 Find: a. Heat transfer film coefficient hi , in w/(m2 -K).

b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the hi ?

c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate hi .

d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?

4N 2250x4

u ρd Gd ====3.98x104>104 (1) Re =μμμ0.001

Nu =0. 023Re 0. 8Pr 1/3=0.023x (3.98x10

40.8

)

⎛4⎫

⎪⎝0.5⎭

1/3

=220.1

Nuk 220.1x0.5

==5500w /(m 2k )d 0.02

(2) w 1=2w 2 Re 2=Re 1/2=2x104>104

hi =

Nu 2⎛Re 2⎫h i20.8

= =0.5 =0.50.8 ⎪Nu 1⎝Re 1⎭h i1

0.82

h i2=5500x0.5=3159w /(m k )

(3) Re 3=

0.8

u 3ρd 3

2000x0.01

=2x104>104

0.001

Pr 1/3 hi=6347w/(m 2k )

Nu =0.023Re

0.8

(4)Q=hi A i (t-tw )=400=500x2πx0.02(t-tw )

t=40℃ t w =39.41℃

(5) there methods : increase u or hi or decrease d The first is better

15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through

(1) heat transfer film coefficient outside the pipe ho ? (2) the pipe length required for counter flow, in m?

(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?

(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?

(a)

11⎛d 0l m d o ⎫1230.002x23

=- +=--⎪

h 0Vo ⎝h i d i kd m ⎭300700x1945x21

1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21

h 0=642. 9w /m k

()

LMTD=70℃=

∆t 1+∆t 2

Q=UoAo∆Tm=mcCp(Tcb-Tca) 2

300*2π*0.023*70L=500/3600*1000*(80-20)

L=5.4m

80-20

=86.5℃

140-20ln

140-80

L ∆t

1=1=70/86.5 L 2=0. 81L =1

L 2∆t 2

(d) scale is formed on the outside ,V0 is decreased

4m 4.

16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is

111111

(1)-(2)= -=0.8-0.8=0.8-0.8

21152660u 1C u 2C 1C 1.5C 111

C=2859 =- ho=8127W/(m2K)

h o Uo h i

Q =W h C h (T 1-T 2)=W c C c (t 2-t 1)=VA ∆t m

W h C h (T )=' 1-T 2

W t )' 1t c (C c -2

150-100-4015

t 2=50=℃

150-802-t 15L T -T 2∆' t m 1150-8092. 5

2=1 ==1. 85

L 1T 1-T 2∆t m 215-010069. 8

L 2=1. 85m L 1=∆1m m = t ∆95 t 1 m 2. 2=

69. 8

Q 1=h 1i A i ∆T m 1=m c 1C pc (T cb -T ca )

Q 2=h 12A i ∆T m 2=m c 2C pc (T ' cb -T ca ) h i2∆T m 2

h i 1∆T m 1

∆T m 1

∆T m 2=

m c 2∆T

=1. 2=1. 20. 8m 2 m c 1∆T m 1

80-20

= ln(116. 3-20) /(116. 3-80)

80-20

ln(T h -20) /(T h -80)

Th=118.5oC

W=1.25Kg/s Cp=1.9Kj/kg℃

Q 2=W h C p (T 1-T 2)=1.25x1.9x (80-30)=119Kw

1Q 119x10330-102

V 0==472w/(m k )A i = ==13.9m 2 ∆t m =

d l d 130V 0∆t m 472x18.2+0+m 0ln

h 0h i d i kd m 10

Re ≤1 U t =

D p 2g (ρP-ρ)18μ

μ18μ23

D p = =

ρg ρP-ρρD p

1/3

⎛⎫18x10-10-5

D p = =3.85x10 ⎪ 1.205x9.81x 2650-1.205⎪

⎝⎭

for filter press V 2+2VV e =KA2θ 5 min 12+2V e =0.12x5K (1) 10min 1.62+2x1.6V e =0.12x10K (2) From (1) (2),we can see Ve=0.7 K=48

15 min V +2x0.7V=48x0.1x15 V=2.07m/h

22. The following data are obtained for a filter press (A=0.0093 m2) in a lab.

------------------------------------------------------------------------------------------------ pressure difference (kgf /cm2 ) filtering time (s) filtrate volume (m3 )

1.05 50 2.27⨯10-3 660 9.10⨯10-3 3.50 17.1 2.27⨯10-3 233 9.10⨯10-3 Find

1) filtering constant K, qe , t e at pressure difference 1.05 kgf /cm2 ?

2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kgf /cm2 ?

3) compressible constant of cake s?

For p=1.05Kg/cm

q 2+2qq e =K θ

0.000227⎛0.00227⎫

+2x q e =50K ⎪

0.00930.00093⎝⎭

0.00091⎛0.00091⎫

+2x q e =660K ⎪

0.000930.00093⎝⎭

We can see K=0.015 qe=0.026

For p=3.5Kg/cm

K=2k∆P1-s K'=2k∆P' 1-s

1-s

KA 2⎛d V ⎫⎛∆P' ⎫ K = ⎪=⎪

d θ2V+VK ' ⎝∆P⎭⎝⎭E e

23. A slurry is filtered by a 0.1 m2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows: (q+10)2 = 250(t+ 0.4) where q---l/m2 t----min

find (1) how much filtrate is got after 249.6 min?

(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)

（1）let θ=249.6 (q+10)=250x (249.6+0.4)

q=240 V=qA=240*0.1=24 (2) K =2k ∆P K' =2k ∆P'

∆P' =2∆P K ' =2K =500 (q'+10)=500x (249.6+0.4) q’=343.6 v=34.36

流体力学与传热习题参考解答(英文)

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