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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 entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg - K). The overall U = 300 W/m2K and the area A = 15.0 m2. Calculate the heat-transfer rate and the exit-water temperature. Instead of using 370 K for our assumption, please use 361 K as the value for Tco. Find the new: 1. Tc ave where: Tc ave = 0.5(Tco+Tci) 2. Cpc at Tc ave 3. Cmin 4. NTU 5. € 6.q 7. Tco

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 entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg - K). The overall U = 300 W/m2K and the area A = 15.0 m2. Calculate the heat-transfer rate and the exit-water temperature. Instead of using 370 K for our assumption, please use 361 K as the value for Tco. Find the new: 1. Tc ave where: Tc ave = 0.5(Tco+Tci) 2. Cpc at Tc ave 3. Cmin 4. NTU 5. € 6.q 7. Tco

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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Table A.2-5. Heat Capacity of Liquid Water at 101.325 kPa (1 Atm) Heat Capacity, Cp . cal/g C kJ/kg K Temperature °C K 0 273.15 10 283.15 1.0019 20 293.15 0.9995 25 298.15 0.9989 30 303.15 0.9987 40 313.15 0.9987 4.181 100 Source: N. S. Osborne, H. F. Stimson, and D. C. Ginnings, Bur. Standards J. Res., 23, 197 (1939). € 1.00 0.80 0.60 0.40 0.20 Heat Capacity, Cp Temperature cal/g C kJ/kg K °C K 323.15 0.9992 4.183 1.0080 4.220 50 4.195 60 333.15 1.0001 4.187 4.185 70 343.15 1.0013 4.192 4.182 80 353.15 1.0029 4.199 4.181 90 363.15 1.0050 4.208 373.15 1.0076 4.219 C /C =0 min max 0.20 0.50- -0.75 1.00- 0 1 2 3 Number of transfer units, NTU = UA/Cmin € 1.00, 0.80 0.60 0.40 0,20 0 C JC. =0 min max 0 0.20 0.50- 0.75 1.00- 1 2 3 4 Number of transfer units, NTU = UA/C min (b) Figure 16.3-2. Heat-exchanger effectiveness &: (a) counterflow exchanger, (b) parallel flow exchanger. 5
Transcribed Image Text:Table A.2-5. Heat Capacity of Liquid Water at 101.325 kPa (1 Atm) Heat Capacity, Cp . cal/g C kJ/kg K Temperature °C K 0 273.15 10 283.15 1.0019 20 293.15 0.9995 25 298.15 0.9989 30 303.15 0.9987 40 313.15 0.9987 4.181 100 Source: N. S. Osborne, H. F. Stimson, and D. C. Ginnings, Bur. Standards J. Res., 23, 197 (1939). € 1.00 0.80 0.60 0.40 0.20 Heat Capacity, Cp Temperature cal/g C kJ/kg K °C K 323.15 0.9992 4.183 1.0080 4.220 50 4.195 60 333.15 1.0001 4.187 4.185 70 343.15 1.0013 4.192 4.182 80 353.15 1.0029 4.199 4.181 90 363.15 1.0050 4.208 373.15 1.0076 4.219 C /C =0 min max 0.20 0.50- -0.75 1.00- 0 1 2 3 Number of transfer units, NTU = UA/Cmin € 1.00, 0.80 0.60 0.40 0,20 0 C JC. =0 min max 0 0.20 0.50- 0.75 1.00- 1 2 3 4 Number of transfer units, NTU = UA/C min (b) Figure 16.3-2. Heat-exchanger effectiveness &: (a) counterflow exchanger, (b) parallel flow exchanger. 5
Question: 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 entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg K). The overall U = 300 W/m2 K and the area A = 15.0 m2. Calculate the heat-transfer rate and the exit-water temperature. Instead of using 370 K for our assumption, please use 361 K as the value for Tco. Find the new: 1. Tc ave where: Tc ave = 0.5(Tco+Tci) 2. Cpc at Tc ave 3. Cmin 4. NTU 5. € 6. q 7. Tco
Transcribed Image Text:Question: 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 entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg K). The overall U = 300 W/m2 K and the area A = 15.0 m2. Calculate the heat-transfer rate and the exit-water temperature. Instead of using 370 K for our assumption, please use 361 K as the value for Tco. Find the new: 1. Tc ave where: Tc ave = 0.5(Tco+Tci) 2. Cpc at Tc ave 3. Cmin 4. NTU 5. € 6. q 7. Tco
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