1. It is required to cool 250 kg/h of hot liquid with inlet temperature of 393.15 K (120 °C) using a parallel flow arrangement. 1000kg/h of cooling water is available for cooling purpose at a temperature of 283.15 K (10 °C). Overall heat transfer coefficient is 1160 W/m2° C. Heat transfer surface of the exchanger is 0.25 m. Calculate the outlet temperature of liquid and water. Determine the effectiveness of the heat exchanger. Data: Specific heat of water 4187 J/kg K; Specific heat of liquid 3350 J/kg K. %3D Note: For mCph < m.CPc T1 – T2 = 3 For m.Cpc < mCPh t2- t1 T1 - t, Where T and T2 represents the inlet and outlet temperatures of hot fluid. ty and t2 represents the inlet and outlet temperatures of cold fluid. Capacity ratio =r-Cmin/Cmax Number of Transfer Units NTU UA Cmin
1. It is required to cool 250 kg/h of hot liquid with inlet temperature of 393.15 K (120 °C) using a parallel flow arrangement. 1000kg/h of cooling water is available for cooling purpose at a temperature of 283.15 K (10 °C). Overall heat transfer coefficient is 1160 W/m2° C. Heat transfer surface of the exchanger is 0.25 m. Calculate the outlet temperature of liquid and water. Determine the effectiveness of the heat exchanger. Data: Specific heat of water 4187 J/kg K; Specific heat of liquid 3350 J/kg K. %3D Note: For mCph < m.CPc T1 – T2 = 3 For m.Cpc < mCPh t2- t1 T1 - t, Where T and T2 represents the inlet and outlet temperatures of hot fluid. ty and t2 represents the inlet and outlet temperatures of cold fluid. Capacity ratio =r-Cmin/Cmax Number of Transfer Units NTU UA Cmin
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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
Transcribed Image Text:1. It is required to cool 250 kg/h of hot liquid with inlet temperature of 393.15 K (120 °C)
using a parallel flow arrangement. 1000kg/h of cooling water is available for cooling purpose
at a temperature of 283.15 K (10 °C). Overall heat transfer coefficient is 1160 W/m2° C. Heat
transfer surface of the exchanger is 0.25 m. Calculate the outlet temperature of liquid and
water. Determine the effectiveness of the heat exchanger.
Data: Specific heat of water 4187 J/kg K; Specific heat of liquid 3350 J/kg K.
Note: For mCph < mcCpc
T, – T2
=D 3
T1- t
For m.Cpc < ṁ„CPh
T1 – t,
Where T and T2 represents the inlet and outlet temperatures of hot fluid.
t, and t2 represents the inlet and outlet temperatures of cold fluid.
Capacity ratio r Cmin/Cmax
Number of Transfer Units NTU UA/Cmin
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