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Heat and Mass Transfer Question: From the answered problem below, assume 361 K as Tco or Cp = 4.184 kJ/kg K Water Flowing at a rate of 0.667 kg/s enters a countercurrent heat exchanger at 308 K and is heated by an oil stream (Cp = 1.89 (kJ/kg-K) at 383 K at a rate of 2.85 kg/s, the overall U = 300 W/m² K and A= 15.0m². Calculate: a) heat-transfer rate (q) b) exit water stream temperature (Tco) Given: Tci = 308 K mc = 0.667 kg/s Assuming: Tco1 = 370 K Cpc = 4.192 kJ/ kg K mH = 2.85 kg/s СpH = 1.89 kJ/kg K U = 300 W/m² k A = 15.0m²

Heat and Mass Transfer Question: From the answered problem below, assume 361 K as Tco or Cp = 4.184 kJ/kg K Water Flowing at a rate of 0.667 kg/s enters a countercurrent heat exchanger at 308 K and is heated by an oil stream (Cp = 1.89 (kJ/kg-K) at 383 K at a rate of 2.85 kg/s, the overall U = 300 W/m² K and A= 15.0m². Calculate: a) heat-transfer rate (q) b) exit water stream temperature (Tco) Given: Tci = 308 K mc = 0.667 kg/s Assuming: Tco1 = 370 K Cpc = 4.192 kJ/ kg K mH = 2.85 kg/s СpH = 1.89 kJ/kg K U = 300 W/m² k A = 15.0m²

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Solution: Cc = mc Cpc = (0.667 kg/s) (4.192 kJ/kg K) Cc = 2.79 kW/K CH (2.85 kg/s) (1.89 kJ/kg K) CH= 5.39 kW/k Cmin= = 2.79 kW/k b) q= Cc (Tco - Tci) q Cc Tco= 2.79 + Tci Tco 148kW kW K Tco = 361 K = +308 K NTU=UA/Cmin (3002) (15.0m²) W 2790 NTU = 1.61 Cmin Cc = -0.51 Cmax CH ε = 0.71 q==Cmin (THr – Ta) = 0.71(2.79 *r )(383 – 308)K - HI 9 = 148 kW
Transcribed Image Text:Solution: Cc = mc Cpc = (0.667 kg/s) (4.192 kJ/kg K) Cc = 2.79 kW/K CH (2.85 kg/s) (1.89 kJ/kg K) CH= 5.39 kW/k Cmin= = 2.79 kW/k b) q= Cc (Tco - Tci) q Cc Tco= 2.79 + Tci Tco 148kW kW K Tco = 361 K = +308 K NTU=UA/Cmin (3002) (15.0m²) W 2790 NTU = 1.61 Cmin Cc = -0.51 Cmax CH ε = 0.71 q==Cmin (THr – Ta) = 0.71(2.79 *r )(383 – 308)K - HI 9 = 148 kW
Heat and Mass Transfer Question: From the answered problem below, assume 361 K as Teo or Cp = 4.184 kJ/kg K Water Flowing at a rate of 0.667 kg/s enters a countercurrent heat exchanger at 308 K and is heated by an oil stream (Cp = 1.89 (kJ/kg K) at 383 K at a rate of 2.85 kg/s, the overall U = 300 W/m² K and A= 15.0m². Calculate: a) heat-transfer rate (q) b) exit water stream temperature (T.co) Given: mH = 2.85 kg/s Tci = 308 K СpH = 1.89 kJ/kg K mc = 0.667 kg/s U = 300 W/m² k Assuming: A = 15.0m² Tco1 = 370 K Cpc = 4.192 kJ/ kg K
Transcribed Image Text:Heat and Mass Transfer Question: From the answered problem below, assume 361 K as Teo or Cp = 4.184 kJ/kg K Water Flowing at a rate of 0.667 kg/s enters a countercurrent heat exchanger at 308 K and is heated by an oil stream (Cp = 1.89 (kJ/kg K) at 383 K at a rate of 2.85 kg/s, the overall U = 300 W/m² K and A= 15.0m². Calculate: a) heat-transfer rate (q) b) exit water stream temperature (T.co) Given: mH = 2.85 kg/s Tci = 308 K СpH = 1.89 kJ/kg K mc = 0.667 kg/s U = 300 W/m² k Assuming: A = 15.0m² Tco1 = 370 K Cpc = 4.192 kJ/ kg K
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