ProbSet4s - ME 416 Computer Assisted Design of Thermal...

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1 ME 416 Computer Assisted Design of Thermal Systems Practice Problem Set #4 Heat Exchangers 1. Setting up our solution table Problem Type: Rating Problem Hot Fluid (Water) Cold Fluid (Water) T H,in = 80 ° C T c,in = 20 ° C T H,out = 59 ° C T c,out = 61 ° C T H,avg = 70 ° C T c,avg = 41 ° C c P,H = c P (@80 ° C) = 4197 J/kg K c P,H = c P (@20 ° C) = 4182 J/kg K m H = 10,000 kg/hr = 2.78 kg/s m c = 5,000 kg/hr = 1.39 kg/s C H = 11,668 W/K = C max C c = 5,813 W/K = C min UA = 11,600 W/K C r = 0.50 ε = 0.69 NTU = 2.0 bold entries are calculated Using our rating problem methodology. 1. c P,H = c P (@80 ° C) = 4197 J/kg K c P,H = c P (@20 ° C) = 4182 J/kg K 2. C H = m H c P,H = (2.78)(4197) = 11,668 W/K = C max C c = m c c P,c = (1.39)(4182) = 5,813 W/K = C min C r = C min /C max = (5,813)/(11,668) = 0.50 3. UA = 11,600 W/K
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ME 416 CAD of Thermal Systems 2 4. NTU = UA/C min = (11,600)/(5,813) = 2.0 5. For a tube and shell heat exchanger with one shell pass and two tube passes the appropriate NTU- ε relation is ε = 2 1+ C r + 1+ C r 2 ( 29 1/2 1+ exp -NTU 1+ C r 2 ( 29 1/2 1- exp -NTU 1+ C r 2 ( 29 1/2 -1 Substituting gives ε = 0.69 6. q act = ε C min (T H,in - T c,in ) = (0.69)(5813)(80-20) = 240,658 W 7. T H,out = T H,in - q act /C H = 80 - (240,658)/(11,668) = 59 ° C T c,out = T c,in + q act /C c = 20 + (240,658)/(5813) = 61 ° C 8. T H,avg = (80+59)/2 = 70 ° C T c,avg
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This note was uploaded on 07/25/2008 for the course ME 416 taught by Professor Somerton during the Fall '07 term at Michigan State University.

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ProbSet4s - ME 416 Computer Assisted Design of Thermal...

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