Unformatted text preview: of energy equation to solve for it. ( 29 ( 29 kg/s 0.175 350)(1000)(380 260)0)(330 (0.03)(250 TT c TT ) c m ( = m out H, in H, H P, in c, out C, C P H = = & & 3. min C P, C C C W/K 75 0) (0.03)(250 c m C = = = = & max H P, H H C W/K 175 00) (0.175)(10 c m C = = = = & 0.429 175 75 C C = C max min R = = 4. U = 15 W/(m 2 ⋅ K) 5. For the effectiveness, we calculate ( 29 ( 29 ( 29 ( 29 0.583 260380 260330 TT TT = TT C TT C = in C, in H, in C, out C, in C, in H, min in C, out C, C = = ε 6. Our NTU relationship is 1.028 129) 0.583)(0.4 ( 183 .5 ln 10.429 1 1C 1ln 1C 1 = NTU R R = = ε ε 7. Our surface area is then 2 min m 5.14 15 ) (75)(1.228 U NTU C = A = =...
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 Fall '07
 SOMERTON
 Energy, Heat, Heat Transfer, Mass flow rate, 75 W, 5.14 m

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