Unit Operations of Chemical Engineering（化工单元操作）problems&answers

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