(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?

1/3

Hint: for laminar flow, Nu=1.86[Re Pr] 0.81/3

for turbulent flow Nu=0.023Re Pr

o

solution: (1)u?4m4?2250??1.99m/s ??d23600?3.14?1000?0.022Re?du??cp??0.02?1.99?1000?39800

10?34?103?1?10?3Pr???8

k0.51kNuk0.50.80.83?39800??8?3?0.575?4785.3?2.0h??0.023RePr?0.023 dd0.02?5503.1W/m2K1(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

h′=5503.1×0.574=3158.8W/m2K (3)

kNukh??0.023Re0.8Pr3

ddh??d????h?d??0.21?2?????1?0.2?1.15

h′=5503.1×1.15=6328.6W/m2K (4)

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

tw?tf? oq400?40??38.8C ?dlh3.14?0.02?5503.1(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 92oC. 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:

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)

?k?g??0.68?958?9.81?2256.7?10??0.943??h?0.943???100?92??2.5?0.282?10?3??ToL?f????323f2f143????14?0.9431.1327?10m?hA(ts?tw)?1154??5471W/(m2?oC)??5471???0.038?2.5?100?92??0.005785kg/s?20.8kg/h 32256.7?10Check the flow pattern:

??m0.005785??0.04848kg/(ms) ?do3.14?0.038Re?4???4?0.04848?688(laminar flow)

0.282?10?3If the tube is placed horizontally, then from equation (4.5-14)

?k?g??h?0.729???Todo?f?3f2f?? ??14then

??2.5???0.773???0.773?2.848?2.2 ?0.038???h′=2.2h=2.2×5471=12036.2W/m2℃ hence, m?h?? mhm′=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

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

o

o

h?0.729?L??h0.943??do1414o

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:

(1)From equation (4.3-37)

o

Uo?11??300

do2310.002231bdo1????4521hodihikdmho1970010.001729?4.868?10?5ho=642.8 W/m℃ (2)

write energy balance equation

2

1?ho?300

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=90oC ?tm=70 oC

t1=20oC

calculate the length of double tube from equation 1

L?(3)

mcp(t2?t1)Uo?do?tm?500?1000(80?20)?5.5m

3600?300??0.023?70

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

?tm?t1=20oC

t1?t2 ?t1ln?t2L??mcp(t2?t1)Uo?do?tmmcp(t2?t1)mcp?t?500?100060??ln1?ln?4.4mt1?t2Uo?do?t23600?300??0.023120Uo?do?tln1?t2(4) the scale depositing on the surfaces of the tube to give additional thermal resistances