Answer:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU = ![\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )](https://tex.z-dn.net/?f=%5C%5C%20%5Cfrac%7B1%7D%7BC-1%7D%20ln%28%5Cfrac%7B%CE%B5-1%7D%7B%CE%B5c%20-1%7D%20%29)
NTU = ![\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.582-1%7D%20ln%28%5Cfrac%7B0.432%20-1%7D%7B0.432%20X%200.582%20%20%20-1%7D%20%29)
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As = ![\frac{NTU C{min} }{U}](https://tex.z-dn.net/?f=%5Cfrac%7BNTU%20C%7Bmin%7D%20%7D%7BU%7D)
As = ![\frac{0.661 x 5016 W. °c }{640 W/m²}](https://tex.z-dn.net/?f=%5Cfrac%7B0.661%20x%205016%20W.%20%C2%B0c%20%7D%7B640%20W%2Fm%C2%B2%7D)
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L = ![\frac{As}{\pi D}](https://tex.z-dn.net/?f=%5Cfrac%7BAs%7D%7B%5Cpi%20D%7D)
L = ![\frac{5.18 m² }{\pi (0.015 m)}](https://tex.z-dn.net/?f=%5Cfrac%7B5.18%20m%C2%B2%20%7D%7B%5Cpi%20%280.015%20m%29%7D)
L= 109.91 m
L ≅ 110 m = 11,000 cm