FT 1.0 Oil flowing at the rate of 6.0 kg/s (Cpm-2.1 kJ/kg ·K) is cooled in a 1-2 heat exchanger from 370.1K to 350.2 K by 3.0 kg/s of water entering at 280.2 K. The overall heat- transfer coefficient Uo is 350 W/m²· K.Calculate the heat exchanger area required. (Hint: A heat balance must first be made to determine the outlet water temperature.Use the heat exchanger chart of Figure 4.9-4 4th edition or Figure 16.2-1 of the 5th edition of the class text to estimate the log mean temperature difference correction Factor, FT) ATm = FATS ΔΤ (16.2-6) (a) Thi z= - Tho Tco - Tel N z = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 0.2 0.9 0.8 0.7 0.6 0.5 0. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Y = Tco-Tei Thi T -

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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FT
1.0
Oil flowing at the rate of 6.0 kg/s (Cpm-2.1 kJ/kg ·K) is
cooled in a 1-2 heat exchanger from 370.1K to 350.2 K by
3.0 kg/s of water entering at 280.2 K. The overall heat-
transfer coefficient Uo is 350 W/m²· K.Calculate the heat
exchanger area required.
(Hint: A heat balance must first be made to determine
the outlet water temperature.Use the heat exchanger
chart of Figure 4.9-4 4th edition or Figure 16.2-1 of the
5th edition of the class text to estimate the log mean
temperature difference correction Factor, FT)
ATm = FATS
ΔΤ
(16.2-6)
(a)
Thi
z=
-
Tho
Tco - Tel
N
z = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4
0.2
0.9
0.8
0.7
0.6
0.5
0.
0.1
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1.0
Y =
Tco-Tei
Thi
T
-
Transcribed Image Text:FT 1.0 Oil flowing at the rate of 6.0 kg/s (Cpm-2.1 kJ/kg ·K) is cooled in a 1-2 heat exchanger from 370.1K to 350.2 K by 3.0 kg/s of water entering at 280.2 K. The overall heat- transfer coefficient Uo is 350 W/m²· K.Calculate the heat exchanger area required. (Hint: A heat balance must first be made to determine the outlet water temperature.Use the heat exchanger chart of Figure 4.9-4 4th edition or Figure 16.2-1 of the 5th edition of the class text to estimate the log mean temperature difference correction Factor, FT) ATm = FATS ΔΤ (16.2-6) (a) Thi z= - Tho Tco - Tel N z = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 0.2 0.9 0.8 0.7 0.6 0.5 0. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Y = Tco-Tei Thi T -
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