PROBLEM The heat rate, q, can be evaluated from an energy balance on the cold fluid, 225 kg/h J. 3600s/h

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1 PROBLEM KNOWN: Concentric tube heat exchanger. FIND: Length of the exchanger. SCHEMATIC: ASSUMPTIONS: (1) Negligible heat loss to surroundings, () Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A-6, Water ( T c = ( ) C/ = 338 K): c p,c = 4188 J/kg K ANALYSIS: From the rate equation, Eq , with A o = πd o L, L = q/uoπdo T l m. The heat rate, q, can be evaluated from an energy balance on the cold fluid, 5 kg/h q = m& ccc( Tc,o Tc,i ) = 4188J/kg K( 95 35) K = 15,705 W. 3600s/h In order to evaluate T l m, we need to know whether the exchanger is operating in CF or PF. From an energy balance on the hot fluid, find 5 kg/h J Th,o = Th,i q/m& hch = 10 C 15,705W/ 095 = s/h kg K Since T h,o T c,o it follows that Hxer operation must be CF. From Eq , T1 T ( 10 95) ( ) Tlm,CF = = C= 81.4 ln( T/ 1 T) ln115/55.1 Substituting numerical values, the HXer length is L = 15,705W/550W/m Kπ 0.10m 81.4K = 1.1m. COMMENTS: The ε-ntu method could also be used. It would be necessary to perform the hot fluid energy balance to determine if CF operation existed. The capacity rate ratio is C min /C max = From Eqs and 11.0 with q evaluated from an energy balance on the hot fluid, Th,i Th,o ε = = = Th,i Tc,i From Fig , find NTU 1.5 giving W W L = NTU C min /UoπDo /550 π ( 0.10m) 1.14m. K m K Note the good agreement in both methods.

2 PROBLEM 11.4 KNOWN: A very long, concentric tube heat exchanger having hot and cold water inlet temperatures, 85 C and 15 C, respectively; flow rate of hot water is twice that of the cold water. FIND: Outlet temperatures for counterflow and parallel flow operation. SCHEMATIC: ASSUMPTIONS: (1) Equivalent hot and cold water specific heats, () Negligible kinetic and potential energy changes, (3) No heat loss to surroundings. ANALYSIS: The heat rate for a concentric tube heat exchanger with very large surface area operating in the counterflow mode is q = qmax = Cmin Th,i Tc,i where C min = C c. From an energy balance on the hot fluid, q = Ch Th,i T h,o. Combining the above relations and rearranging, find Cmin C T c h,o = ( Th,i Tc,i) + Th,i = ( Th,i Tc,i) + T h,i. Ch Ch Substituting numerical values, 1 Th,o = ( 85 15) C+ 85 C = 50 For parallel flow operation, the hot and cold outlet temperatures will be equal; that is, T c,o = T h,o. Hence, ( ) = ( ) Cc Tc,o Tc,i Ch Th,i T h,o. Setting T c,o = T h,o and rearranging, Cc Cc Th,o = Th,i+ T c,i / 1+ Ch Ch 1 1 Th,o = C/ = 61.7 COMMENTS: Note that while ε = 1 for CF operation, for PF operation find ε = q/q max = 0.67.

3 PROBLEM KNOWN: A shell and tube Hxer (two shells, four tube passes) heats 10,000 kg/h of pressurized water from 35 C to 10 C with 5,000 kg/h water entering at 300 FIND: Required heat transfer area, A s. SCHEMATIC: ASSUMPTIONS: (1) Negligible heat loss to surroundings, () Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A-6, Water 150 C, T h 500 K): c p = 4660 J/kg K. Tc 350K : = c p = 4195 J/kg K; Table A-6, Water (Assume T h,o ANALYSIS: The rate equation, Eq , can be written in the form As = q/u T l m and from Eq , T1 T T lm = F Tlm,CF where T lm,cf =. ln( T/ 1 T) From an energy balance on the cold fluid, the heat rate is 10,000 kg/h J 5 q = m& ccp,c Tc,o Tc,i = K= W. 3600s/h kg K From an energy balance on the hot fluid, the outlet temperature is kg J Th,o = Th,i q/m& hcp,h = 300 C W/ 4660 = s kg K From Fig , determine F from values of P and R, where P = (10 35) C/(300 35) C = 0.3, R = ( ) C/(10-35) C = 1.8, and F The log-mean temperature difference based upon a CF arrangement follows from Eq. (3); find ( ) Tlm = ( ) ( ) K/ l n = 143.3K. ( ) A 5 s = W/1500W/m K K = 4.75m COMMENTS: (1) Check T h 500 K used in property determination; T h = ( ) C/ = 497 K. () Using the NTU-ε method, determine first the capacity rate ratio, C min /C max = Then ( ) q Cmax Tc,o Tc,i C ε = = = qmax Cmin Th,i Tc,i C From Fig , find that NTU = AU/C min 1.1 giving A s = 4.7 m. (1) (,3)

4 PROBLEM 11.3 KNOWN: Single pass, cross-flow heat exchanger with hot exhaust gases (mixed) to heat water (unmixed) FIND: Required surface area. SCHEMATIC: ASSUMPTIONS: (1) Negligible heat loss to surroundings, () Negligible kinetic and potential energy changes, (3) Exhaust gas properties assumed to be those of air. PROPERTIES: Table A-6, Water ( T c = ( ) C/ = 38 K): c p = 4184 J/kg K; Table A-4, Air (1 atm, T h = ( ) C/ = 436 K): c p = 1019 J/kg K. ANALYSIS: The rate equation for the heat exchanger follows from Eqs and The area is given as A = q/u Tlm = q/uf Tl m,cf (1) where F is determined from Fig using P = = 0.6 and R = =.50 giving F From an energy balance on the cold fluid, find kg J q = m& ccc( Tc,o Tc,i ) = ( 80 30) K= 67,600 W. s kg K From Eq , the LMTD for counter-flow conditions is T1 T ( 5 80) ( ) Tlm,CF = = C= ln( T/ 1 T) ln145/70 Substituting numerical values resulting from Eqs. (-4) into Eq. (1), find the required surface area to be A 67,600W/00W/m K K 33.1m. = = COMMENTS: Note that the properties of the exhaust gases were not needed in this method of analysis. If the ε-ntu method were used, find first C h /C c = 0.40 with C min = C h = 501 W/K. From Eqs and 11.0, with C h = C min, ε = q/q max = (T h,i T h,o )/(T h,i T c,i ) = (5 100)/(5 30) = Using Fig with C min /C max = 0.4 and ε = 0.64, find NTU = UA/C min 1.4. Hence, A = NTU C min /U W/K/00W/m K = 35.m. Note agreement with above result. () (3) (4)

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