10.37 A simple heat exchanger is turn bend on two 2-in Schedule 40 steel pipes as shown in Fig. 10.32. Compute the pressure difference between the inlet and the outlet for a flow rate of 12.5 gal/min of ethylene glycol at 77°F.
10.37 A simple heat exchanger is turn bend on two 2-in Schedule 40 steel pipes as shown in Fig. 10.32. Compute the pressure difference between the inlet and the outlet for a flow rate of 12.5 gal/min of ethylene glycol at 77°F.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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can you please find solution for the question no: 10.37

Transcribed Image Text:10.20 Determine the energy loss for a sude
2 DN 125 Schedule 80 steel pipe to a DN 50 Schedule
80 pipe for a flow rate of 500 L/min.
10.21 Determine the energy loss for a gradual contraction from
a DN 125 Schedule 80 steel pipe to a DN 50 Schedule 80
pipe for a flow rate of 500 L/min. The cone angle for the
contraction is 105°.
the results from Problems 10.34-10.36.
10.37 A simple heat exchanger is made by installing a close re-
turn bend on two 2-in Schedule 40 steel pipes as shown
in Fig. 10.32. Compute the pressure difference between
the inlet and the outlet for a flow rate of 12.5 gal/min of
ethylene glycol at 77°F.
GURE 10.32 Problem 10.37.
-in Schedule 40 pipe
Flow
Inlet
Outlet
Close return
bend
4.00 ft
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