1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B is methane（甲烷）, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and 6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrument In meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when the change in the levels in the reservoirs is taken into account? What is the percent error in the answer to the part (a)? Solution：

pa=1000kg/m3 pc=815kg/m3 pb=0.77kg/m3 D/d=8 R=0.145m

When the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubes ??22Dx?dR (1)

44so

?d?x???R (2)

?D?2and hydrostatic equilibrium gives following relationship

p1?R?cg?p2?x?cg?R?Ag (3)

so

p1?p2?x?cg?R(?A??c)g (4)

substituting the equation (2) for x into equation (4) gives ?d?p1?p2???R?cg?R(?A??c)g (5)

?D?2（a）when the change in the level in the reservoirs is neglected, ?d?p1?p2???R?cg?R(?A??c)g?R(?A??c)g?0.145?1000?815??9.81?263Pa?D?2

（b）when the change in the levels in the reservoirs is taken into account ?d?p1?p2???R?cg?R(?A??c)g?D??d????R?cg?R(?A??c)g?D??6.5?????0.145?815?9.81?0.145?1000?815??9.81?281.8Pa51??222

error=

281.8?263281.8＝6.7％

1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R1=400mm，R2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R3=50mm. Try to calculate the pressure at point A and B.

Figure for problem 1.4

Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by ?g,?H2O,?Hg, respectively. The pressure at point A is given by hydrostatic equilibrium

pA??H2OR3g??HgR2g??g(R2?R3)g

?gis small and negligible in comparison with?Hgand ρH2O , equation above can be simplified

pA?pc=?H2OgR3??HggR2

=1000×9.81×0.05+13600×9.81×0.05

=7161N/m2

pB?pD?pA??HggR1=7161+13600×9.81×0.4=60527N/m

1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit of drainpipe are all open to air.

pa H D d h pa Figure for problem 1.5 A

Solution:

Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: p1?gz1?u122??p2??gz2?u222

Where p1=0, p2=0, and u1=0, simplification of the equation 2u2 Hg ? 1

2

The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation: ?u2??u?o??d????? ???D?22uo?D??u2?? 2

?d?Bernoulli equation is written between the throat and the station 2-2

? ? 3 ?22

Combining equation 1,2,and 3 gives

p0u02u2

1h?g12?1000?9.812?9.81 Hg??＝＝＝442?10002.44?1?1.25??1?D?

???1 ?d?Solving for H

u2H=1.39m

1.6 A liquid with a constant density ρ kg/m3 is flowing at an unknown velocity V1 m/s through a horizontal pipe of cross-sectional area A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the area is reduced gradually to A2 m2 and the pressure is p2. Assuming no friction losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured.

Solution:

In Fig1.6, the flow diagram is shown with pressure taps to measure p1 and p2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,

V1A1A2V2?

For the items in the Bernoulli equation , for a horizontal pipe, z1=z2=0

V1A1A2Then Bernoulli equation becomes, after substitutingV2? for V2,

0?V122?p1V1?0?2A1A1222??p2?

Rearranging,

A1A1222?V1(p1?p2?2?1)

V1＝p1?p22??A??1????A??1?????2??2?

Performing the same derivation but in terms of V2,

p1?p22??1????A2????A??1?2V2＝?????

1.7 A liquid whose coefficient of viscosity is μ flows below the critical velocity for laminar flow

in a circular pipe of diameter d and with mean velocity V. Show that the pressure loss in a length

?p32?Vof pipe is . 2LdOil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.

Solution:

The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area RR1 1 V ? u 2 ? 1 udA?rdr2?A0?R? 0From velocity profile equation for laminar flow 2? u ? 0 L R 2 ? 1 r ? ? 2 ? ???4?L?R?? ??p?p??substituting equation 2 for u into equation 1 and integrating p0?pL2V?D 3

32?L

rearranging equation 3 gives

2Ld 32?VL32?0.05?0.6?120?p???11520Pa22 d0.1

1.8. In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m below A and when the flow rate down the pipe is 0.02 m3/s, the pressure at B is 14715 N/m greater than that at A.

Assuming the losses in the pipe between A and B can be expressed as kV2?p?32?V2

2gwhere V is the velocity at A, find the

Figure for problem 1.8

value of k.

If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. Solution:

dA=0.15m; dB=0.075m zA-zB=l=2.5m Q=0.02 m3/s, pB-pA=14715 N/m2 Q?VA??4dAVAQ?220.020.785?0.152?4?1.132m/s

dAQ?VB??4dBVBQ?2B20.020.785?0.0752?4?4.529m/s

dWhen the fluid flows down, writing mechanical balance equation pA?zAg?VA22??pB??zBg?VB22?kVA22

2.5?9.81?1.1322?147151000?4.5322?k1.1322

24.525?0.638?14.715?10.260?0.638k

k?0.295

making the static equilibrium

pB??x?g?R?g?pA?l?g??x?g?R?HggR??pB???pA??l?gHg??g??14715?2.5?1000?9.8112600?9.81??79mm

1.9．The liquid vertically flows down through the tube from the station a to the station b, then horizontally through the tube from the station c to the station d, as shown in figure. Two segments of the tube, both ab and cd，have the same length, the diameter and roughness. Find:

（1）the expressions of

?pab?g, hfab,

?pcd?g and hfcd, respectively.

Figure for problem 1.9

（2）the relationship between readings R1and R2 in the U tube.

Solution:

(1) From Fanning equation 2lV hfab??d2

and

2lV hfcd??d2so

hfab?hfcdFluid flows from station a to station b, mechanical energy conservation gives papb ? lg ? ? h fab ?? hence

pa?pb ? lg ? h fab 2

?

from station c to station d pp c?d?hfcd??

hence

p?pcd? h fcd 3 ?

From static equation pa-pb=R1（ρˊ-ρ）g -lρg 4 pc-pd=R2（ρˊ-ρ）g 5 Substituting equation 4 in equation 2 ，then R（????）g?l?g1?lg?hfab

?

therefore

???? h fab ? R g 6 1?

Substituting equation 5 in equation 3 ，then

???? h fcd ? R 2 g 7

? Thus R1=R2

1.10 Water passes through a pipe of diameter di=0.004 m with the average velocity 0.4 m/s, as shown in Figure.

1) What is the pressure drop –?P when water flows through the pipe length L=2 m, in m H2O column?

2) Find the maximum velocity and point r at which

L it occurs.

3) Find the point r at which the average velocity equals the local velocity.

4）if kerosene flows through this pipe，how do the variables above change？

（the viscosity and density of Water are 0.001 Pas and 1000 kg/m3，respectively；and the viscosity and density of kerosene are 0.003 Pas and 800 kg/m3，respectively） solution: 1）Re?ud??0.4?0.004?10000.001?1600

Figure for problem 1.10

r ?from Hagen-Poiseuille equation

32uL?32?0.4?2?0.001?P?d2?0.0042?1600

h??p?g?16001000?9.81?0.163m

2）maximum velocity occurs at the center of pipe, from equation 1.4-19

Vumax?0.5 so umax=0.4×2=0.8m

3）when u=V=0.4m/s Eq. 1.4-17

uumax?r???1???r? ?w?22V?r?1???0.5 ?＝0.004u??maxr?0.0040.5?0.004?0.71?0.00284m

4) kerosene: Re?ud??0.4?0.004?8000.0030.0030.001?427

??p???p????1600?4800Pa

h???p???g?4800800?9.81?0.611m

1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.

a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entrance

is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m3/h)

b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m2). le/d≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.

Figure for problem 1.12

Solution：

(1) When the gate valve is opened partially, the water discharge is

’

Set up Bernoulli equation between the surface of reservoir 1—1and the section of pressure point 2—2’，and take the center of section 2—2’ as the referring plane, then

gZ1?u122?p1??gZ2?u222?p2???hf,1—2 （a）

In the equation p1?0(the gauge pressure)

p2??HggR??H2Ogh?13600?9.81?0.4?1000?9.81?1.4?39630N/m

u1?0Z2?02

When the gate valve is fully closed, the height of water level in the reservoir can be related ?Hg(Z1?h)??HggR

to h (the distance between the center of pipe and the meniscus of left arm of U tube).

2O （b）

where h=1.5m

R=0.6m

Substitute the known variables into equation b

Z1?13600?0.61000ld?1.5?6.66m?Kc)V2

?hf,1_2?(?2?(0.025?150.1?0.5)V2

?2.13V22Substitute the known variables equation a

9.81×6.66=

V22?396301000?2.13V

2the velocity is V =3.13m/s

the flow rate of water is

?2Vh?3600?4dV?3600??4?0.12?3.13?88.5m/h

3 2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 1—1’ and 3-3′，then

gZ1?V122?p1??gZ3?V322?p3???hf,1—3 （c）

since Z1?6.66m

Z3?0

u1?0p1?p3

?hf,1_3?(?l?led35?Kc)V22V2 ?[0.025( ?4.81V20.1?15)?0.5]2

input the above data into equation c，

9.81?6.66?V22?4.81V

2the velocity is: V=3.51 m/s

Write mechanical energy balance equation between thestations 1—1’ and 2——2’, for the same situation of water level

gZ1?V1222?p1??gZ2?V22?p2???hf,1—2 （d）

since Z1?6.66m

Z2?0

u1?0u2?3.51m/s

p1?0(page pressure）?hf,1_2?(?ld?Kc)V22?(0.025?150.1?0.5)3.5122?26.2J/kg

input the above data into equation d，

9.81×6.66=

3.5122?p21000?26.2

the pressure is: p2?32970

1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60?3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is 2.72m3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.

Solution： The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2.

The friction loss for parallel pipelines is

?hf1??hf2Vs?VS1?VS2

The energy loss in the branch pipe is

l2??hf2??2?ld2e2u222

In the equation ?2?0.03

l2??le2?10m

d2?0.053u2?3600?2.72

?4?0.343m/s2?0.053input the data into equation c

?hf2?0.03?100.053?0.34322?0.333J/kg

The energy loss in the main pipe is

?hf1??hf2??1l1u1d122?0.333

So u1?0.333?0.3?20.018?2?2.36m/s

The water discharge of main pipe is

?23Vh1?3600??0.3?2.36?601m/h

4Total water discharge is

Vh?601?2.72?603.7m/h

3

1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be 2.5R m/s, where R is the manometer reading in metres of mercury. Determine the loss of head

between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution:

Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meter p1?V122?2?z1g?po??Vo2?z2g?hf 1

rearranging the equation above, and set (z2-z1)=x p1?po?Vo?V1222??xg?hf 2

Figure for problem 1.16

from continuity equation

?d1??5?????V1?6.25V1 3 Vo?V1??d??2??o?22substituting equation 3 for Vo into equation 2 gives p1?po??39.06V1?V12222?xg?hf?19.03V1?hf?19.032.5R??2?xg?hf 4

?118.94R?xg?hffrom the hydrostatic equilibrium for manometer

p1?po?R(?Hg??)g?x?g 5

substituting equation 5 for pressure difference into equation 4 obtains R(?Hg??)g?x?g?118.94R?xg?hf 6

?rearranging equation 6 hf?R(?Hg??)g?118.94R?123.61R?118.94R?4.67R?2.288J/kg

?

1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,

calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.

The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m3 Solution: a)D0D1?1050?0.2

p1?p2?R(?Hg??)g?0.1(13600?1300)?9.81?

V2?

m?Co?D0??1???D??1?42?p1?p2???0.611?0.242?0.1(13600?1300)?9.811300?0.6118.56?0.61?4.31?2.63m/s?4D0V2??2?4?0.01?2.63?1300?0.268kg/s

2b) approximate pressure drop

p1?p2?R(?Hg??)g?0.1(13600?1300)?9.81?12066.3Pa

pressure difference due to increase of velocity in passing through the orifice

V2??2p1?p2??V2?V1222?Do?2??V2??D??1?24?13002.63(1?0.2)224?4488.8Pa

pressure drop caused by friction loss

?pf?12066.3?4488.8?7577.5Pa

2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow rate is 26m/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition.

3

Solution:

Write the mechanical energy balance equation between the suction connection and discharge connection Z1?where

Z2?Z1?0.4m

p1??2.47?10Pa(gaugep2?1.52?10Pa(gaugeu1?u2Hf,1_254u122g?p1?g?H?Z2?u222g?p2?g?Hf,1_2

pressure）pressure）

?0total heads of pump is H?0.4?1.52?105?0.247?1051000?9.81?18.41m

efficiency of pump is ??Ne/N since Ne?QH?g3600?26?18.41?1000?9.813600?1.3kW

N=2.45kW

Then mechanical efficiency

??1.32.45?100%?53.1%

The performance of pump is

Flow rate ，m3/h

Total heads，m Shaft power ，kW Efficiency ，%

2.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is ?57×3.5 mm , the orifice coefficient of Co and orifice diameter do are 0.62 and 25 mm, respectively. Frictional

26 18.41 2.45 53.1

210m1coefficient ? is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)

Solution:

Equation(1.6-9) V?0

?C0?d?1??0??D?0.620.937542Rg（?f??）??0.62?25?1???50??42?0.168?9.81(13600?1000)1000?6.44?4.12m/sMass flow rate

m?VoSo??4.12?3.144?0.0252?1000?2.02kg/s

2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V1=V2

H?p2?p1??z?Hf?g

?z=10m

?p?0.5?9.81?104?200760?1.013?105?75707Pa

?p/?g=7.7m

The relation between the hole velocity and velocity of pipe ?d??1?V?V0?0??4.12????1m/s ?2??D?Friction loss 22lu2001?0.025? Hf?4f

so

d2g220.052?9.81?5.1mH=7.7+10+5.1=22.8m

2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet? Solution:

From an energy balance,

p?pvHg?o?Hf?NPSH?gHg

Where

Po=760-640=120mmHg Pv=760-710=50mmHg

Use of the equation will give the minimum height Hg as

po?pv?Hf?NPSH Hg??g

?（0.12?0.05)?13600?9.81?1.5?3??3.55m1000?9.81

2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?

? Density of acid 1840kg/m3

-3

? Viscosity of acid 25×10 Pas

Solution:

Velocity of acid in the pipe:

mu?volumetricflowratecross?sectionalareaofpipe???4d2?m0.785?d2?30.785?1840?0.0252?3.32m/sReynolds number: Re?d?u?0.025?1840?3.3225?10?3??6109

from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9 hf??p?4flu22?d2?4?0.0085600.0253.322?450J/kg

?p?450?1840?827.5kPa

or friction factor is calculated from equation1.4-25 hf??p?4flu2?d2?4?0.046Re?0.2lu2d2＝4?0.046?6109?0.2600.0253.3222?426J/kg?p?426?1840?783.84kPa

if the pressure drop falls to 783.84/2=391.92kPa

????0.2?p???391920?4?0.046Re?＝4?0.046?????2d2????plu?1840??4?0.046?1840???3?25?10??0.22?0.2?ld1.2u1.82

600.0251.2u1079.891.8?`u20.0121.8so u?1.8391920?0.0121079..89?1.84.36?2.27m/s

new mass flowrate=0.785d2uρ=0.785×0.0252×2.27×1840=2.05kg/s

2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?

Density of acid 1840kg/m3

-3

Viscosity of acid 25×10 Pa Friction factorf?0.0056?Solution:

Write energy balance equation: p1?z1?u1220.500Re0.32 for hydraulically smooth pipe

?g2g?p?H?p2?g2?z2?u22g?hf

H?hf??g??lud2g

?4du??3

2u?3?4?d?2?123.14?0.0252?1840?3.32m/s

Re?3.32?0.025?184025?10?3?6115

f?0.0056?0.500Re??0.32?0.0056?20.561150.32?0.0087

3.322H?hf??plu?gd2g?4?0.0087600.0252?9.81?46.92

Δp=46.92×1840×9.81=847.0kpa

2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38?2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient Co＝0.63. the permanent loss in pressure is 3.5×104N/m2, the friction coefficient λ=0.024. find: (1) What is the pressure drop along the pipe AB?

(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W, 60?ficiency? (The density of fluid is 870kg/m3 )

solution： zAg?pA?uA22??w?zAg?2pA??uA22??hf

pA?pB?Ao??2hf??lud2??p0?

?16.4?????0.247 A33??u0?C01?0.24722gR?????????0.630.972?9.81?0.6?13600?870870??8.5m/s

∴u= (16.4/33)2×8.5=2.1m/s ∴pA?pB???hf?0.024?870(2) Ne?Wm??p?du??76855?0.785?0.03322300.0332.122?3.5?104?76855N/m2

?4?2.1?138W

so

the ratio of power obliterated in friction losses in AB to total power supplied to the fluid

138500?0.6?100％＝46％

3.2 A spherical quartzose particle（颗粒） with a density of 2650 kg/m3 settles（沉淀） freely in the 20℃ air, try to calculate the maximum diameter obeying Stocks’ law and the minimum diameter obeying Newton’s law.

Solution:

The gravity settling is followed Stocks’ law, so maximum diameter of particle settled can be calculated from Re that is set to 1

Re?dcut??1, then

t?ut??dc?

equation 3.2-16 for the terminal velocity

2?dc??dc(?S??)g18?

solving for critical diameter

dc?1.22432?(?S??)g

Check up the appendix

-32

The density of 20℃ air ρ=1.205 kg/m3 and viscosity μ=1.81×10N·s/m

dc?1.224?5.73?10?57.3?m3(1.81?10?3)2(2650?1.205)?1.205?5m

when Reynolds number ≥1000, the flow pattern follows Newton’s law and terminal velocity can be calculated by equation 3.2-19 ut?1.75gdp???p??? 1

critical Reynolds number is Ret??ut?dc?1000， 2

?rearranging the equation 2 gives

ut?1000???dc 3

combination of equation 1 with equation 3

1000???dc?1.74?(?S??)gdc?

solving for critical diameter ??32.33 dc(?S??)??2

??32.33dc(1.81?10?3?3)2(2650?1.205)?1.205m?1.512?10?1512um

3.3 It is desired to remove dust particles 50 microns in diameter from 226.5m/min of air, using a settling chamber for the purpose. The temperature and pressure are 21oC and 1 atm. The particle density is 2403kg/m3. What minimum dimensions of the chamber are consistent with these conditions? (the maximum permissible velocity of the air is 3m/s) solution:

to calculate terminal velocity from the equation 3.2-16

ut?dp??p???g23

18?

-5

2

The density of 21℃ air ρ=1.205 kg/m3 and viscosity μ=1.81×10N·s/m ut?dp??p???g218?＝(50?10?6)(2403?1.205)?9.81?5218?1.81?10?0.181m/s

Q?BLut so

Qut226.560?0.1812BL???20.86m 1

from equation3.3-4

Lu?Hut

the maximum permissible velocity of the air is 3m/s

L3?H0.181

L?16.58H 2

set B to be 3m, then from equation 1 L=7m And H=0.42m

3.4 A standard cyclone is to be used to separate the dust of density of 2300 kg/m3 from the gas.

-5

The flow rate of gas is 1000m3/h, the viscosity of the gas is 3.6?10N·s/ m2, and the density is 0.674 kg/ m3. If the diameter of cyclone is 400 mm, attempt to estimate the critical diameter. Solution: D=0.4m B=D/4=0.1m h=D/2=0.2m

ui?QhB?10003600?0.2?0.1?13.9m/s

According to the equation3.3-12 for N=5：

dc?9?B?9(3.6?10?5)(0.1)?N(?p??)ui5?(2300?0)?13.9?8?10?6m?8?m

3.6 A filter press of 0.1m2 filtering area is used for filtering a sample of the slurry. The filtration is carried out at constant pressure with a vacuum 500mmHg.The volume of filtrate collected in the first 5min was one liter and, after a further 5min, an additional 0.6 liter was collected. How much filtrate will be obtained when the filtration has been carried out for 15min on assuming the cake to be incompressible? Solution:

The equation for the constant-pressure filtration V22?2VVm?KAt

225min 1l. 1?2Vm?K0.1?5

2210min 1.6l. 1.6?2?1.6Vm?K?0.1?10

solving the equations above for Vm and K

Vm?0.7 and K=48

For t?15min VSolving for V=2.073 l

2?2?0.7?V?48?0.1?15

23.7 The following data are obtained for a filter press of 0.0093 m filtering area in the test

Pressure difference/kgf/cm filtering time/s filtrate/ m -31.05 50 2.27×10 660 9.10×10-3 3.50 17.1 2.27×10-3 233 9.10×10-3 232

Calculate:

(1) filtration constant K, Vm at the pressure difference of 1.05

(2) if the frame of the filter is filled with the cake at 660s, what is the final rate of filtration ?dV??? dt??E(3) and what is the compressible constant of cake n? solution:

①from equation 3.4-19a

q?2qqm?Kt For pressure differencep?1.05㎏/㎝

22

?2.27?10?3??0.0093??9.1?10?3??0.0093??3?2.27?10??2?qm?K?50 1 ?0.0093?22?3?9.1?10??2?qm?K?660 2 ?0.0093?solving the equations 1 and 2 gives qm?0.0379mm32K?1.56?10?3m/s

2②

KA?dV??=7.14?10?6m3/s ??2(V?Vm)?dt?E2

For pressure differencep?3.5㎏/㎝

?2.27?10?3??0.0093??9.1?10?3??0.0093?22

?3?2.27?10??K??1.71 3 ??2?qm?0.0093?2?3?9.1?10??K??233 4 ??2?qm?0.0093?solving the equations 3 and 4 gives ??0.0309qmmm1?n32K??4.37?10?3m/s

2????p??????????p?

then 4.37?101.56?10?3?3?3.5?????1.05?1?n

1?n?ln2.8ln3.33

solving for n=0.142

3.8 A slurry if filtered by a filter press of 0.1m2 filtering area at constant pressure, the equation for a constant pressure filtration is as follows

(q?10)?250(t?0.4)

2where q=filtrate volume per unit filtering area,in l/m2, t= filtering time, in min calculate:

(1) how much filtrate will be gotten after249.6min?

(2) If the pressure difference is double and both the resistances of the filtration medium and cake are constant, how much filtrate will be obtained after249.6min? solution: (1)

(q?10)?250(t?0.4)?250(249.6?0.4)?250

22q=240 l/m V=24 l

(2) the pressure difference is double K?K??p??p?2

2

so K′=500 ?q??10???q?10q??2????2 ?22(q?10)?10?1.41?250?10?342.5l/m

V=34.25 l

3.9 Filtration is carried out in a plate and frame filter press, with 20 frames 0.3 m square and 50mm thick. At a constant pressure difference of 248.7kN/m2, one-quarter of the total filtrate per cycle is obtained for the first 300s. Filtration is continued at a constant pressure for a further 1800s, after which the frames are full. The total volume of filtrate per cycle is 0.7 m and dismantling and refitting of the press takes 500s

It is decided to use a rotary drum filter, 1.5m long and 2.2m in diameter, in place of the filter press. Assuming that the resistance of the cloth is the same and that the filter cake is incompressible, calculate the speed of rotation of the drum which will result in the same overall rate of filtration as was obtained with the filter press. The filtration in the rotary filter is carried out at a constant pressure difference of 70kN/m2 and with 25% of the drum submerged

3

Solution:

Area of filtration: A=2×0.3×20=3.6mΔp=248.7kN/m2

3

t1=300s, V1=1/4×0.7=0.175m t2-t1=1800s, ΔV=0.7-0.175=0.525m3 or t2 =2100s, V2=0.7m3 0.175and0.7?2?0.7Vm?3.6?2100K22222

?2?0.175Vm?KAt?3.6?300K22

Vm=0.2627m And K?0.7?1.4Vm3.6?2100223

?0.49?1.4?0.262712.96?2100?3.1517?10?5

capacity

Q?0.72100?500?2.692?10?4m/s

3

for the rotary drum filter

Area of filtration: A=πdL=3.14×2.2×1.5=10.362m2

Operating pressure:Δp=70kN/m And

the drum submerged : φ=25%

For rotary drum filter the filtration coefficient: keep Vm unchanged, K changes with changing in pressure difference K??K?p??p?3.1517?10?52

70248.7?8.871?10?6

the capacity of rotary drum filter is equal to that of press filter

2.692?10?4??nV?n???K?A2?n?V2m????Vm?n?????8.871?10?6?10.362n2?0.25?0.26272??0.2627???Solving for n from the equation above by trial and error

2.692?10?4??nV?n???2K?A???2?Vm?Vm??n???n??0.0002381n??0.069?0.2627?

??n=0.0088=0.048rpm(转/min)

3.10 A press filter of 0.093m2 filtering surface was used to separate the suspension containing calcium carbonate, operating in constant pressure. The volume of filtrate collected was 2.27×10-3

m during the 50 s, volume of filtrate collected was 3.35×10 m during the 100 s. How much will the filtrate be obtained as filtering for 200 s? solution：

2-33-33

filtering area A=0.093 m; filtrate V=2.27×10 m for t=50s and V=3.35×10 m for t=100s for constant pressure filtration

V23-33

?2VVm?KAt

2substituting variables given above into the equation gives

2.272?10?6?2?2.27?10?2?3.35?10?3Vm?0.093K?50 1 and

223.35?102?6?3Vm?0.093K?100 2

combining equations 1 and 2 Vm=3.85×10-4 m3 and

K=1.567×10-5 m2/s

Substitution of the filtering-constants Vm and K into the equation for constant-pressure equation gives

V2?7.70?10-3

?4V?1.567?10?5At

2solving for volume of filtrate gained for 200s V=4.835×10 m

3.11 A slurry with incompressible cake is filtered by a 1m2 filter press at constant pressure. If the filtering constant K is 10 m2/min, operating pressure difference ?p is 2 atm and the septum resistance can be ignored. Find

(1) the filtrate volume, in m3 when the filtering time is 10 min.

(2) if the septum resistance must be considered and Vm is 1 m3, what is the filtrate volume V, in m3 when the filtering time is 10 min ?

(3) what is the filtration rate dV/dt when the filtering time is 10 min for case (1)? Solution:

Equation for the constant-pressure filtration

V223

?2VVm?KAt

filter medium resistance set to be negligible

V2?KAt

32V?AKt?1?10?10?10m

②The filtering medium resistance is taken into account

V2?2VVm?KAt

2V2?2V?10?1?10

2V??2?4?4?10023?9.05m

③ the equation for constant-pressure filtering is taken derivative as filtering medium resistance doesn’t be taken into account dVdt?KA2V2?10?12?10?0.5

4.2. A flat furnace wall is constructed of a 114-mm layer of Sil-o-cel brick, with a thermal conductivity of 0.138 W/(m·°C) backed by a 229-mm layer of common brick, of conductivity 1.38 W/(m·°C). The temperature of the inner face of the wall is 760°C, and that of the outer face is 76.6°C. (a) What is the heat loss through the wall? (b) What is the temperature of the interface between the refractory brick and the common brick? (c) Supposing that the contact between the two brick layers is poor and that a ―contact resistance‖ of 0.088°C·m2/W is present, what would be the heat loss? Solution:

From equation4.2-11 (a) heat loss:

qA???T?Rii??TB1k1?B2k2?760?76.60.1140.138?0.2291.38?688.89W/m

2(b) the temperature at interface: qA??TB1k1?760?t0.1140.138?688.89W/m

2t=191℃

(c) contact resistance is 0.088°C·m2/W, heat loss: qA???T?Rii??TB1k1?B2k2?0.088?760?76.60.1140.138?0.2291.38?0.088?683.4W/m

2

4.4. The vapor pipe (do=426 mm) is covered by a 426-mm insulating layer (k=0.615 W/m?C). If the temperature of outer surface of pipe is 177 C and the temperature outside the insulating layer is 38 C, what are the heat loss per meter pipe and the temperature profile within the insulating layer?

Solution：

Similar to compound resistances in series through the flat wall, the total resistance across insulating layer is as follows

o

o

o

R?BkAm?0.4260.615??(1.278?0.426)/ln(1.278/0.426)L?0.2357L

Heat loss per meter tube through the wall is

qL??TR?177?380.2357?589.7W/m

and temperature distribution is qL??TR?177?t0.615?2??r?0.213?lnr0.213r0.213?589.7W/m

t?177?2277.5?r?0.213?ln

4.5. The outer diameter of a steel tube is 150mm. The tube wall is backed by two insulating layers to reduce the heat loss. The ratio of thermal conductivity of two insulating materials is k2/k1?2 and both insulating materials have the same thickness 30mm. If the temperature

difference between pipe wall and outer surface of insulating material is constant, which insulating material should be packed inside to enable less heat loss? solution：

k1（lower thermal conductivity）

the insulating material with a lower thermal conductivity placed outside, rate of heat loss Q???t?Ri??tb1k2Am1?b2k1Am2 Am2?2?L(r3?r2)lnr3r2

iand Am1?2?L(r2?r1)lnr2r1?, thickness of two layers are equal b1=b2=b, and k2/k1?2

Q??tb2k1Am1?bk1Am2k1b12Am1?t?1Am2

contrarily, place the material with lower thermal conductivity inside, heat loss Q???tb1k1Am1?b2k2Am2??tbk1Am1?b2k1Am2?k1b1Am1?t?12Am2

1

Q?Q?2?Am1Am2Am1?1 for

Am1Am2?1

1?2Am2so with less heat loss obtained as the material with lower thermal conductivity is placed inside

4.7. Air at the normal pressure passes through the pipe (di 20 mm) and is heated from 20C to 100

o

o

C. What is the film heat transfer coefficient hi between the air and pipe wall if the average

o

velocity of air is 10 m/s? The properties of air at 60 C are as follows:

Air: density ?=1.06 kg/m3, viscosity μ= 0.02 cp, conductivity k=0.0289 W/(m?C), and heat capacity cp =1 kJ/(kg?K). Solution: Re?du??0.02?10?1.060.02?10?3 o

?cp?k?10600

Pr??1000?0.02?100.0289Re0.8?3?0.692

hi?0.023?47.6kdiPr0.4?0.0230.02890.02?106000.8?0.6920.4?0.03324?1660.5?0.863

4.8. Methyl alcohol flowing in the inner pipe of a double-pipe exchanger is cooled with water flowing in the jacket. The inner pipe is made from 25-mm Schedule 40 steel pipe. The thermal

o

conductivity of steel is 45 W/(m·C). The individual coefficients and fouling factors are given in

the following Table. What is the overall coefficient, based on the outside area of the inner pipe?

TABLE Data for Problem 8

Coefficient/ W/(m2·°C)

Alcohol coefficient hi 1020 Water coefficient ho 1700 Inside fouling factor hdi 5680

Outside fouling factor hdo 2840 Solution:

From equation (4.3-37) Uo?1dodihdi?dodihi?bdokdm?1ho?1hdo

from appendix 3, the wall thickness is 3.25mm and outside diameter is 33.5mm for nominal diameter 25mm Schedule 40 standard steel pipe

dm?33.5?272?30.25mm

Uo?133.527?5680?33.527?1020?0.0032545?333.530.251?5?11700?12840?4?2.184?10?4?1.216?10C?7.998?10?5.882?10?3.521?10?4

?407.4W/m2o

4.9. Benzene with mass flow 1800kg?h-1flows through the annulus of double pipe exchanger. Its temperature changes from 20oC to 80 oC. The dimension of inner tube is ?19×2.5mm, and the dimension of outer tube is Φ38×3mm. What is individual heat transfer coefficient of benzene?

Solution:

Outside diameter of inner tube do =19mm, inside diameter of outer tube Di=32mm The equivalent diameter of annular space of double tube de is

?de?44(Di?do)?Di?do?32?19?13mm

22?(Di?do)from equation

k?deu????h?0.023??de???tm?80?202o0.8?cp???k?????0.4 1

?50C

-3

3

the properties of benzene: μ=0.45×10Pas，cp=1.81kJ/kg.℃，k=0.138W/m℃，ρ= 850kg/m at average temperature of 50℃

and

average velocity of benzene flowing through the annular space of double tube is calculated as

follow m??4?22(Di?do)u

u?4m??(D?d)2i2o?4?18003600???8500.032?2?0.0192??1.13m/s

Substituting the variables into equation 1 gives k?deu??h?0.023?de??＝0.023????0.8?cp???k?????0.40.138?0.013?1.13?850????30.013?0.45?10?0.80.8?1.81?10?0.45?10??0.138?23?3????0.4

?0.244?27748?5.90.4＝1780W/(m·?C)

4.10. Air at 2 atm and 20oC flows through in the pipe of tubular exchanger at 60m3?h-1, and is heated to 80C.The dimension of pipe is ?57×3.5mm and its length is 3m long. What is

individual heat transfer coefficient of air side? Solution:

Temperature of air at inlet and outlet t1=20C, t2=80C. average temperature tm?t1?t22?20?802?50C

o

o

o

o

from App.7, viscosity of air at 50 oC μ=0.019×10-3Pas, and density at 2atm, on assuming that a air follows ideal law,

??PMRT?2?290.08206?(50?273)?2.188kg/m

3

from App. 11, thermal conductivity of air

k?0.0242?0.03182?0.028W/mK

from App.13 specific heat of air cp=1KJ/kgK The velocity of air flowing through tube

u?4V?d2?4?603600?3.14?0.052?8.49m/s

Re?du??cp?k?0.05?8.49?2.1880.019?103?3?48884.5

Pr??1?10?0.019?100.028kdRe0.8?3?0.68

0.0280.05h?kNud?0.023Pr0.4?0.023?48884.5?0.8?0.68?0.4?62.3W/m2K

4.12. 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/(m ?C), cp =4 kJ/(kg?K), viscosity 10-3 N?s/m , density 1000 kg/m Find: (1) Heat transfer film coefficient hi , in W/(m2?K).

(2) If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the hi? (3) 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 .

(4) When the average temperature of fluid and quantity of heat flow per meter of tube are 40 C and 400 W/m, respectively, what is the average temperature of pipe wall for case a?

(5)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]

0.81/3

for turbulent flow Nu=0.023Re Pr

1/3

o

o

2

3

solution: (1)u?4m?4?22503600?3.14?1000?0.022??ddu?2?1.99m/s

Re??cp?k?0.02?1.99?1000103?3?39800

Pr??4?10?1?100.5?3?8

1h?kNud?0.0232kdRe0.8Pr3?0.0230.50.02?39800??8?30.81?0.575?4785.3?2.0

?5503.1W/mK(2) h?u0.8

individual coefficient h′ as the flow rate decreases to 1125 kg/h h??u?????h?u?0.8?1125????2250??0.8?0.574

2

h′=5503.1×0.574=3158.8W/mK (3)

h?kNud?0.0230.2kd1Re0.20.8Pr3

h??d?????hd???2?????1??1.15

h′=5503.1×1.15=6328.6W/m2K (4)

q?Ah(tf?tw)??dlh(tf?tw)

tw?tf?q?dlh?40?4003.14?0.02?5503.1?38.8C

o

(5)

to increase the velocity of fluid flowing through the pipe

4.13 The saturated vapor at temperature of 100oC is condensed to saturated liquid on the surface of a single vertical pipe with length 2.5m and diameter 38mm (outside).The temperature of pipe wall is 92C. What is the quantity of condensed vapor per hour? If the pipe is placed horizontally, what is the quantity of condensed vapor per hour? Solution:

o

From the appendix 6, we obtain the latent heat of the saturated vapor, 2256.7kJ/kg at 100 oC.

The average temperature of the condensate film=(100+92)/2=96℃

The properties of the saturated water at 96 oC are as follows: μ=0.282×10-3Pas(from appendix 8); ρ=958kg/m3; k=0.68W/m.K(from appendix5)

from equation (4.5-12)

1?k?g?h?0.943???TL?of?3f2f1????4?0.68?958?9.81?2256.7?10?0.943???100?92??2.5?0.282?10?3?323?4????0.9431.1327?10?15?14?5471W/(m?C)2om?hA(ts?tw)??5471???0.038?2.5?100?92?2256.7?103?0.005785kg/s?20.8kg/h

Check the flow pattern:

??m?do?0.0057853.14?0.038?0.04848kg/(ms)

Re?4???4?0.048480.282?10?3?688(laminar flow)

If the tube is placed horizontally, then from equation (4.5-14)

1?k?g?h??0.729???Td?oof?3f2f?4? ??then

11h?h?0.729?L?0.943??do????4?2.5?4?0.773???0.773?2.848?2.2

?0.038?h′=2.2h=2.2×5471=12036.2W/m2℃ hence,

m?m?h?h

m′=2.2×20.8=45.8kg/h

4.14. In a double pipe exchange (Φ23?2 mm), the cold fluid (cp=1 kJ/(kg ?K), flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the annular. The inlet and outlet temperatures of cold fluid are 20 C and 80 C, and the inlet and outlet temperatures of hot fluid

oo are 150 C and 90 C , respectively. The hi (heat transfer coefficient inside a pipe) is 700 W/(m2 ?o

o

o

C)and overall heat transfer coefficient Uo (based on the outside surface of pipe) is 300W/(m2 ?

o

C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45

o

W/(m?C), find:

(1) heat transfer 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 oC) which 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 C), explain why? Solution:

o

(1)From equation (4.3-37) Uo?1dodihi?bdokdm10.001729?4.868?10?5?1ho?123119700?0.002234521?1ho?300

?1ho?300

ho=642.8 W/m2℃ (2)

write energy balance equation

mcp(t2?t1)?UoAo?tm?Uo?doL?tm 1

logarithmic mean temperature difference for countercurrent flow

T1=150C t2=80C oo?t1=150-80=70 oC ; ?t2=90-20=70 oC so

T2=90C o?tm=70C

o

t1=20oC

calculate the length of double tube from equation 1

L?mcp(t2?t1)Uo?do?tm?500?1000(80?20)3600?300??0.023?70?5.5m

(3)

T=140oC t2=80oC ?t1=140-80=60 oC ; ?t2=140-20=120 oC so

?tm?t1?t2ln?t1?t2

t1=20C

oL??mcp(t2?t1)Uo?do?tm?mcp(t2?t1)Uo?dot1?t2ln?t1?t2?mcpUo?doln?t1?t2??500?10003600?300??0.023ln60120?4.4m(4) the scale depositing on the surfaces of the tube to give additional thermal resistances

4.15. Water flows turbulently in the pipe of Φ25?2.5 mm shell tube exchanger. When the velocity of water V is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 W/(m2 ?oC). If the V becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 W/( m2? oC ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored) Solution:

From equation (4.3-37), neglecting thermal resistance of tube wall gives Uo?11hi?1ho

when the velocity of water is 1m/s, the individual coefficient of water to wall of tubes is hi , overall coefficient Uo?11hi?1ho?2115 1

when the velocity of water is 1.5m/s, the individual coefficient of water to wall of tubes is h′i , and h′i related to hi by equation(4.4-25) hi??u?????hi?u?0.8?1.50.8?1.383

and overall coefficient is Uo?11hi??1ho?111.383hi?1ho?2660 2

solving equation 1 and 2 for hi and ho

2 2

hi=2858 W/( m?oC ); ho=8135 W/( m?oC )

4.17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 C and 40 C, and those of oil are 150 C and 100 C, respectively. If the outlet temperature of oil decreases to 80 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) solution：

energy balance for oil q?m1cp1(150?100) （1） for cooling water q?m2cp2(40?15) （2）

q?m1cp1(150?100)?m2cp2(40?80)?UA?tm (3)

o

o

o

o

o

temperature different operating in parallel

?t1?150?15?135℃

?t2?100?40?60℃ ?tm?135?60ln13560?92.5℃

new condition for oil q?m1cp1(150?80) （4） for cooling water q?m2cp2(t2?15) （5）

??80)?UA?tm? (6) q??m1cp1(150?T1?)?m2cp2(t2q?q?m1cp1(150?80)m1cp1(150?100)???15)m2cp2(t2m2cp2(40?15)??UA?tmUA?tm (7)

solving the equation 7 for t′2=50℃

average temperature difference

?t1?150?15?135℃

?t2?80?50?30℃

?tm?135?30ln13530?69.8℃

from equation 7

q?q?150?80150?100?L?69.8L92.5 solving for L?1=1.86L=1.86m

4.18. Air which passes through the pipe in turbulent flow is heated from 20 C to 80 C. The saturated vapor at 116.3 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) solution： average temperature of air: t?the properties of air at 50 ℃：

3

??1.093kg/m Pr?0.698 cp?1.017kJ/kg.k k?0.028w/m.k

o

o

o

20?802=50C

O

to increase the vapor pressure of steam to meet the requirement for rising in the rate of air by 20% , the temperature of steam denoted by T

q?mcp(80?20)?UA?tm （1） ? （2） q??m?cp(t?20)?U?A?tmm??1.2m

∵ the individual coefficient with phase change is very large compared with the convective coefficient of air, the resistance of steam side is so very small in comparison with other side

resistance and neglected.

U?U?h?h?1.20.8?1.157 (3)

logarithmic mean temperature difference ?tm?ln80?20116.3?20116.3?80?61.5℃ (4)

increase in temperature of steam with rising in vapor pressure of steam, then, mean temperature difference ?t?m?80?20lnT?20T?80 （5）

combining equations (1) , (2) , (4),and（5） q?q?m?cp(80?20)mcp(80?20)??U?A?tmUA?tm

q?qq?q?m?m?1.2, and (6)

??U?A?tmUA?tm??U??tmU?tm??h??tmh?tm?1.157?6061.5lnT?20T?80?1.1288lnT?20T?80 (7)

combining（6）and（7） and solving for T=118.5℃

4.19. Water flows through the pipe of a Φ25?2.5 mm shell-tube exchanger from 20 C to 50 C. The warm fluid (cp 1.9 kJ/(kg?C), flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 C to 30 C in a countercurrent flow with water. heat transfer coefficients of water and warm fluid are 0.85kW/(m?C) and 1.7 kW/(m?C). What is the overall heat transfer coefficient Uo (based on outer surface area of tube) and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45W/(m?C). solution： q?m2cp2(T1?T2)?U0A0?tm (1) logarithmic mean temperature

?tm1?(80?50)?(30?20)ln3010?18.2℃

o

2 o

2 o

o

o

o

o

o

overall coefficient

Uo?125850?20?0.0025?2545?22.5?11700?471.6W/m2.K

from equation 1 Ao?

o

4.20 Heated oil (heat capacity cp =3.35kJ/kg·C) is cooled by water (heat capacity

m2cp2(T1?T2)Uo?tm?1.25?1900(80?30)471.6?18.2?13.8 m

2

cp=4.187kJ/kg·C) and they flow countercurrently through a double pipe exchanger with inner pipe dimension ?180×10mm. Water at 15 oC of inlet temperature goes through the pipe and leaves at 55 oC. Oil passes through the annulus with mass flow rate 500kg/h and inlet and outlet temperatures of oil are 90 oC and 40 oC, respectively. The heat transfer coefficients hi and ho for

o

water and oil are 1000W/(m·C) and 299W/(m·C)，and heat resistances of pipe wall and fouling as well as heat loss can be ignored. Find (1) water flow rate, in kg/h.

(2) overall heat transfer coefficient Uo based on outer surface of pipe. (3) LMTD ?tm and pipe length of pipe, in m.

(4) if inlet temperature of water becomes 20C and oil inlet temperature keeps the same, what happens to that?

(5) in order to enhance heat transfer, what ways can be employed? Solution:

（1） enthalpy balance q?m1cp1(t2?t1)?m2cp2(T1?T2) m1?4.187?（55-15）=500?3.35?（90-40） m1=500.8kg/h

(2) overall coefficient based on outside surface of tube: Uo?1d0hidi?1h0?11801000?160?1299?224W/m2.k

o

2o2o

(3) t1?15℃???t2?55℃

T2?40℃???T1?90℃

?t1?25℃ ?t2?35℃

?tm?35?25ln3525?29.7℃

energy balance q?m1cp1(t2?t1)?UOAo?tm

500.83600?4.187?10?(55?15)?224???0.18L?29.7

3 L=4.64m (4) when t1?20℃

energy balance

500.8?4.187?（t′2-20）=500?3.35?（90-T′2）

t′2=92-0.8 T′2 (1) q?q???20)m2cp2(t2m2cp2(55?15)??UA?tmUA?tm

simplifying the equation above

q?q???20t255?15???tm?tm???tm29.7 (2)

???tm??(T2??20)90?t2ln?90?t2T2??20 (3)

combining equations 12 and 3 ln0.8T2??2T2??20?14.86?0.15T2?72?0.8T2? （4）

solving for outlet temperature of warm fluid by trial-error

T′2=50℃

Substituting this temperature into equation 1 gives t′2=52℃

4.22 There are two same shell-tube exchangers (1-1 pass) in stock. Each one consists of 64 tubes with dimension ?27?3.5 mm and heat transfer area 20 m. The saturated vapor with condensation latent heat 2054 kJ/kg at 170 oC goes through the shell and condenses to saturated liquid. Air at 30 oC passes fully turbulently through the tubes with mass flow rate 2.5 kg/s.

(1)If two exchangers are installed parallelly, mass flow rate of air (cp= 1 kJ/kg?K) is 1.25kg/s for each exchanger and heat transfer coefficient of air is 38 W/m2?K. If the flow pattern of air is fully turbulent, what are outlet temperature of air and vapor flow rate?

(2)If two exchangers work in series with air mass flow rate 2.5kg/s, what are outlet temperature of air and vapor flow rate now? The heat resistances of pipe wall and vapor and the heat loss are ignored.

(answer: 93.6oC, 0.077kg/s; 121.4oC, 0.11kg /s)

Solution :

(1)Write energy balance equation

mcp(t2?t1)?UoAo?tm?UoAot2?t1lnT?t1T?t22

so

lnT?t1T?t2?UoAomcp

set Uo=hi because the thermal resistance from saturated vapor to wall of tube and resistance of wall are very small and negligible

ln170?30170?t2??38?201.25?100038?201.25?1000?0.608

170?30170?t2?1.837

t2=93.8 oC

the rate of condensate for one condenser mh1?mcp(t2?t1)?1.25?1000?93.8?30?2054?1000?0.0388kg/s

?sum of condensate for two condensers in parallel mh=2mh1=2×0.0388=0.0776kg/s

(2) two condensers operate in series

heat transfer coefficient between the air and wall of tubes ?UoUo?hi???u??????h???u??i???0.8?20.8?1.74

U′38=66.12 W/m2?K. o=1.74Uo=1.74×

lnT?t1?T?t2???2AoUomcp

so

ln170?30?170?t2?66.12?402.5?1000?1.058

170?30?170?t2?2.88

ot′2=121.4C

mh???t1)m?cp(t2?2.5?1000?121.4?30?2054?1000?0.111kg/s

?