Can someone explain the step by step process of this solution like how it happened, including terms (ex. V, A, Q)

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter10: Heat Exchangers
Section: Chapter Questions
Problem 10.7P
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Can someone explain the step by step process of this solution like how it happened, including terms (ex. V, A, Q)

ANS,
STEP I:
Given data:
velocity v= 2m/see
16 tubes in a single Pass Cooler
Diameter of
dz 30mm =) 0.03m
the tube
density 8= 0.85 9lei
Transcribed Image Text:ANS, STEP I: Given data: velocity v= 2m/see 16 tubes in a single Pass Cooler Diameter of dz 30mm =) 0.03m the tube density 8= 0.85 9lei
16 x Yolume flow Yate of one tobe
STEP 2:
distauce
In one second fluild Covered
avea of
volume flow in
one second = Cvoss section
second
the tord
the tube X distemce oveved in one
Total volwme flow rate =
Q: (6 × AメV
= 16x Rx V
16x Co.03)x2
Q = 0.022619 m3/see
lit (sec
Q= 0. 6226l9 X1000
Q : 22.61· lit/sec
[ima/sec: lo0o lite
- 23 lit/sec
Hence
Answey
OPtion (d)
Transcribed Image Text:16 x Yolume flow Yate of one tobe STEP 2: distauce In one second fluild Covered avea of volume flow in one second = Cvoss section second the tord the tube X distemce oveved in one Total volwme flow rate = Q: (6 × AメV = 16x Rx V 16x Co.03)x2 Q = 0.022619 m3/see lit (sec Q= 0. 6226l9 X1000 Q : 22.61· lit/sec [ima/sec: lo0o lite - 23 lit/sec Hence Answey OPtion (d)
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