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