化工单元操作-华南理工大学化工学院教材（红色外皮）problems an

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 ?2?Dx?d2R (1) 44so

?d?x???R (2)

?D?and hydrostatic equilibrium gives following relationship

2p1?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?22（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?

（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?263＝6.7％

281.81.4 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

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

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

pa of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit of

drainpipe are all open to H D d air.

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:

p12u12p2u2 ?gz1???gz2??2?2Where 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?2

?D?uo?u2?? 2

?d?Bernoulli equation is written between the throat and the station 2-2 2p0u0u2 ? ? 3

22 ?Combining equation 1,2,and 3 gives

u21h?g12?1000?9.812?9.81 Hg??＝＝＝442?D??10002.44?1?1.25??1 ?1?? ?d?Solving for H H=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,

2

V2?V1A1 A2For the items in the Bernoulli equation , for a horizontal pipe, z1=z2=0

Then Bernoulli equation becomes, after substitutingV2?V1A1 for V2, A2A12V2V1p1A12p2 0???0??2?2?21Rearranging,

A12?V(2?1)A1 p1?p2?221V1＝p1?p22??A1???A???2????2???1???

Performing the same derivation but in terms of V2,

p1?p22??A2?1???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

?p32?Vthat the pressure loss in a length of pipe is .

Ld2Oil 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 RR11 V ? udA ? u 2 ? rdr 1

2??A0?R0

From velocity profile equation for laminar flow 2??p?pr??0 L 2 u ? 2 ?R1??????4?L ??R??substituting equation 2 for u into equation 1 and integrating p0?pL2V?D 32 ? L 3

rearranging equation 3 gives ?p32?V? Ld2

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/m2 greater than that at A.

Assuming the losses in the pipe between A and B

V2can be expressed as kwhere V is the velocity

2gFigure for problem 1.8

at A, find the 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??42dAVAVA?Q?4?2dA 0.02?1.132m/s0.785?0.152

Q??42dBVBVB?Q?42dB 0.02??4.529m/s0.785?0.0752When the fluid flows down, writing mechanical balance equation

pA22VApBVB2VA ?zAg???zBg??k?2?221.132147154.5321.1322.5?9.81????k

210002224.525?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?g??Hg??g??14715?2.5?1000?9.81??79mm

12600?9.81

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?pcd, hfab, and hfcd,

?g?gFigure for problem 1.9

respectively.

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

Solution:

(1) From Fanning equation

lV2 hfab??d2

and

lV2 hfcd??d2

so

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

??

hence

p?p a b ? 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?（1???）g?l?g?lg?hfab

?

therefore

???? h fab ? R 1 g 6

?

Substituting equation 5 in equation 3 ，then

???? ? 7 h fcdR 2 g?

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 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?L r Figure for problem 1.10

??0.4?0.004?1000?1600

0.001from Hagen-Poiseuille equation

32uL?32?0.4?2?0.001?P???1600 22d0.004h??p1600??0.163m ?g1000?9.812）maximum velocity occurs at the center of pipe, from equation 1.4-19

so umax=0.4×2=0.8m

3）when u=V=0.4m/s Eq. 1.4-17

uumax?r??1???r??

?w?22V?0.5umaxV?r?1???0.5 ?＝?0.004?umaxr?0.0040.5?0.004?0.71?0.00284m 4) kerosene:

Re?ud???0.4?0.004?800?427

0.003?p???p??0.003?1600?4800Pa ?0.001h??

?p?4800??0.611m ??g800?9.81

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—1’ and the section of pressure point 2—2’，and take the center of section 2—2’ as the referring plane, then

2u12p1u2pgZ1???gZ2??2??hf,1—2 （a）

2?2?In the equation p1?0(the gauge pressure)

p2??HggR??H2Ogh?13600?9.81?0.4?1000?9.81?1.4?39630N/m2

u1?0Z2?0

When the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).

?HOg(Z1?h)??HggR

2 （b）

where h=1.5m

R=0.6m

Substitute the known variables into equation b

13600?0.6?1.5?6.66m1000

lV215V2?hf,1_2?(?d?Kc)2?(0.025?0.1?0.5)2?2.13V2Z1?

Substitute the known variables equation a

V239630??2.13V2 9.81×6.66=21000the velocity is V =3.13m/s

the flow rate of water is

??Vh?3600?d2V?3600??0.12?3.13?88.5m3/h

44 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

V32p3V12p1gZ1???gZ3????hf,1—3 （c）

2?2?since Z1?6.66m

Z3?0

u1?0 p1?p3l?leV2?hf,1_3?(?d?Kc)235V2 ?[0.025(?15)?0.5]

0.12 ?4.81V2

input the above data into equation c，

V2?4.81V2 9.81?6.66?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

V12p1V22p2gZ1???gZ2????hf,1—2

2?2?（d）

since Z1?6.66m

Z2?0

u1?0u2?3.51m/s

p1?0(page pressure）

?h

f,1_2lV2153.512?(??Kc)?(0.025??0.5)?26.2J/kg

d20.12input the above data into equation d，

p3.512?2?26.2 9.81×6.66=21000the 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

2l2??le2u2 ??2d22?hf2

In the equation ?2?0.03

l2??le2?10m

d2?0.053

u2?3600?2.72?4?0.343m/s?0.0532input the data into equation c

?hf2100.3432?0.03???0.333J/kg

0.0532The energy loss in the main pipe is

?hf1??hf2l1u12??1?0.333

d12So u1?0.333?0.3?2?2.36m/s

0.018?2The water discharge of main pipe is

?Vh1?3600??0.32?2.36?601m3/h

4Total water discharge is

Vh?601?2.72?603.7m3/h

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

p1poVo2V12??z1g???z2g?hf 1 ?2?2rearranging the equation above, and set (z2-z1)=x

Figure for problem 1.16

p1?po?Vo2?V12??xg?hf 2

222from continuity equation

?d1Vo?V1??d?o??5?????V1?6.25V1 3 ??2??substituting equation 3 for Vo into equation 2 gives

p1?po?39.06V12?V12??xg?hf?19.03V12?hf?19.032.5R2?118.94R?xg?hf??2?xg?hf 4

from 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?R(?Hg??)g?118.94R?xg?hf 6

rearranging equation 6

hf???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)

D010??0.2 D150p1?p2?R(?Hg??)g?0.1(13600?1300)?9.81? V2?

Co?D01???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

44b) approximate pressure drop m??D02V2????0.012?2.63?1300?0.268kg/s

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

22?Do???V?V22?22D1?V2?V12.632(1?0.24)??p1?p2?????1300?4488.8Pa

222 pressure drop caused by friction loss

4?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 26m3/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. Solution:

Write the mechanical energy balance equation between the suction connection and discharge connection

2u12p1u2p Z1???H?Z2??2?Hf,1_2

2g?g2g?gwhere

Z2?Z1?0.4m

p1??2.47?104Pa(gaugepressure）p2?1.52?105Pa(gaugepressure）u1?u2Hf,1_2?0

1.52?105?0.247?105?18.41m total heads of pump is H?0.4?1000?9.81efficiency of pump is ??Ne/N

QH?g26?18.41?1000?9.81??1.3kW 36003600N=2.45kW

Then mechanical efficiency

1.3?100%?53.1% ??2.45The performance of pump is

Flow rate ，m3/h 26 Total heads，m 18.41 Shaft power ，kW 2.45 Efficiency ，% 53.1

2.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in

2which the gauge

pressure is 0.5 kgf/cm2,

10mas shown in Fig. The total equivalent length

1of 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 coefficient ? 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)

2Rg（?f??）C00.622?0.168?9.81(13600?1000) V??044?1000 ?d0??25?1???1??? ?50??D?

0.62?6.44?4.12m/s ?0.9375

Mass flow rate

since Ne?

3.14?0.0252?1000?2.02kg/s 42) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V1=V2 m?VoSo??4.12?H?p2?p1??z?Hf ?g200?1.013?105?75707Pa 760?z=10m

?p?0.5?9.81?104??p/?g=7.7m

The relation between the hole velocity and velocity of pipe

22 ?d0??1?V?V0???4.12????1m/s ?2??D?Friction loss

lu220012?0.025??5.1m Hf?4fd2g0.052?9.81 so

H=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 Hg 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?pv?Hf?NPSH Hg?o?g

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 ? Viscosity of acid 25×10-3 Pas

Solution:

Velocity of acid in the pipe:

mu?volumetricflowratem3?????3.32m/s22?cross?sectionalareaofpipe0.785?d0.785?1840?0.025d24d?u0.025?1840?3.32?6109

25?10?3Reynolds number:

Re???from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9

lu2603.322hf??4f?4?0.0085?450J/kg

?d20.0252?p?p?450?1840?827.5kPa

or friction factor is calculated from equation1.4-25

2lu2603.322?0.2lu?0.2hf??4f?4?0.046Re＝4?0.046?6109?426J/kg?d2d20.0252?p?p?426?1840?783.84kPa

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

????plu2?0.2?p???391920?4?0.046Re?＝4?0.046??????2d2???1840??4?0.046?1840??3??25?10??0.21.8?0.2lu1.8?1.22d

60u1079.891.8?`u1.220.0120.025so

u?1.8391920?0.0121.8?4.36?2.27m/s

1079..89new 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 Viscosity of acid 25×10-3 Pa Friction factorf?0.0056?0.500 for hydraulically smooth pipe Re0.32Solution:

Write energy balance equation:

2p1u12p2u2?z1??H??z2??hf ?g2g?g2g?plu2 H?hf????gd2g?4d2u??3

u?3?412??3.32m/s 22?d?3.14?0.025?18403.32?0.025?1840?6115 ?325?10Re?f?0.0056?0.5000.5?0.0056??0.0087 0.320.32Re6115?plu2603.322H?hf????4?0.0087?46.92

?gd2g0.0252?9.81Δp=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：

22uApAuAzAg???w?zAg????hf

?2?2pApA?pB?lu2?p0 ??hf???d2?2Ao?16.4?????0.247 A?33?u0?C01?0.24722gR??????0.632?9.81?0.6?13600?870???8.5m/s

??0.97870∴u= (16.4/33)2×8.5=2.1m/s

302.12?3.5?104?76855N/m2 ∴pA?pB???hf?0.024?8700.0332(2)

Ne?Wm??p?2du??76855?0.785?0.0332?2.1?138W ?4so

the ratio of power obliterated in friction losses in AB to total power supplied to the fluid 138?100％＝46％

500?0.6

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

Ret?dcut???1, then

ut?

? dc?

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