Please answer and give solutions the following questions of Heat and Mass transf
ID: 499390 • Letter: P
Question
Please answer and give solutions the following questions of Heat and Mass transfer ( chemical engineering major subject) 20-26
The condenser of a large power plant is a shell-and-tube exchanger with 30,000 tubes. Each tube executes two passes. The exchanger must exchange heat at a rate of 2x109 W. This will be accomplished using 20 oC water at 30,000 kg/s ( i.e, 1kg/s per tube)
Data:
Tube data: Steam data :
D= 0.05 m Condensation temperature = 50 degree celcius
Thin-walled Outside coefficient, ho = 11,000 W/m2.K
L= length/pass
Properties of water at 20 degree celcius :
Cp = 4182 J.kg.K , µ = 0.90 x 10-3 Pa.s , k = 0.611 W/m.K , density = 996.4 kg/m3
For flow inside tube at a specified conditionsl the following correlations can be used
a. What is the temperature of the cooling water leaving the condenser? ( A. 21.1 oC B. 35.9 oC C. 38.4oC D. 30oC )
b. The average velocity of cooling water i ( A. 1.48 m/s B. 2.04 m/s C. 2.66 m/s D. 3.08 m/s)
c. The value of Reynolds number for the coling water is ( A. 56,463 B. 51,995 C. 48,774 D. 81,258 )
d. The value of prandtl number for the cooling water inside the tube is ( A. 0.94 B. 2.22 C. 5.33 D. 6.16 )
e.the heat trasnfer coefficient for the cooling water insidet the tube is........... W/m2.K ( A. 6.254 B. 6.616 C. 7.607 D. 8.003 )
f.The overall heat-tranfer coeeficient is......... W/m2.K ( A. 3.556 B. 4.497 C. 5.763 D. 7.607 )
g.The required length per pass is approximately ( A. 5.5m B. 7.5m C. 8.0m D. 4.5m )
cya 1.2 + 11.8(S)?(Pr-1)Pr7 ,Wheref=0.0014 + R125 0.125 CFG- GfG 1.2 + 11.8(/)2 (Pr-1)Pr-1 T ,Where f-0.0014+ 032 Re0.32 f21-2Explanation / Answer
1.heat duty= mass of cooling water* specific heat* temperature difference
2*109 = 30000* 4182* (T-20), where T =outlet temperatue of cooling water
T= 35.94 deg.c
2. mass velocity in each tube = 2 kg/s ( since there are two passes on the tube side)
Density = 996.4 kg/m3
Volumetric flow rate= 2/996.4 m3/s =0.002 m3/s
Cross sectional area of each tube = (PI/4)*d2 = (22/28)* (0.05)2= 0.001964 m2
Velocity= Volumetric flow rate/ cross sectional area= 0.002/0.001964 m/s=1.018 m/s ( closest answer is 1.48 m/s. This is due to the consideration that the diameter is considered as O.D not I.D, I.D will give the velocity specified.)
3. Reynolds number= Diameter* velocity* density/ viscosity= 0.05*1.108*1000/0.9*10-3 = 56555 ( A is close answer. This Reynolds number is calculated based on outside diameter of tube)
4. Pr= Cp* visccosity/thermal conductivity = 4182*0.9*10-3/0.611= 6.2 ( D is correct)
5. for turbulent flow, hd/K= 0.023* (Re)0.8 * (Pr)2/3 =0.023* (56555)0.8 (6.2)2/3= 492
h= 492*0.611/0.05
=6012 W/m2.K= 6.012 Kw/m2.K