Scribd
Upload
1K views2 pages

Water at An Average of 70

The document describes heat transfer through a 2-inch steel pipe with 700F water flowing inside and 2200F steam condensing on the outside. It provides the internal and external heat transfer coefficients and asks to calculate the heat loss per foot of pipe using (a) resistances, (b) the overall heat transfer coefficient based on the inner pipe area, and (c) the overall heat transfer coefficient based on the outer pipe area. Calculations are shown for the heat transfer resistances and the heat loss is found to be 26,743.69 BTU/hr for all three methods.

Uploaded by

manuel
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
1K views2 pages

Water at An Average of 70

The document describes heat transfer through a 2-inch steel pipe with 700F water flowing inside and 2200F steam condensing on the outside. It provides the internal and external heat transfer coefficients and asks to calculate the heat loss per foot of pipe using (a) resistances, (b) the overall heat transfer coefficient based on the inner pipe area, and (c) the overall heat transfer coefficient based on the outer pipe area. Calculations are shown for the heat transfer resistances and the heat loss is found to be 26,743.69 BTU/hr for all three methods.

Uploaded by

manuel
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 2

Water at an average of 70F is flowing in a 2-in. steel pipe, schedule 40.

Steam at
220F is condensing on the outside of the pipe. The convective coefficient for the
water inside the pipe is h = 500 btu/hft 2F and the condensing steam coefficient on
the outside is h = 1500 btu/hft2F.
(a) Calculate the heat loss per unit length of 1 ft of pipe using resistances.
(b) Repeat, using the overall Ui based on the inside area Ai.
(c) Repeat, using Uo.

Ti = 70F
T0 = 220F
L = 1.0 ft
hi = 500 btu/hft2F (water)
ho = 1500 btu/hft2F (steam)
From Appendix 3-16
ksteel = 45.1 W/mk

ksteel = 45.1 W/mk ( 1.73073 ) = 26.06 btu/hft2F


From Appendix 5-1
For 2-in steel pipe Di = 2.067 in , Do = 2.375 in
ri = 1.0335 in , ro = 1.1875 in
Ai = 2Lri = 2(1.0ft)(1.0335 in/12) = 0.5421 ft2
Ao = 2(1.0ft)(1.1875 in/12) = 0.6218 ft2

Eq. (4.3-6)

Alm =

A o A i
Ao
ln
Ai

==

0.62180.5421
0.6218
ln
0.5421

Resistances
Ri =

1
hi A i

1
(500)(0.5421)

= 3.689x10-3

= 0.5810 ft2

RA =

r or i
kA A lm

Ro =

1
ho A o

(a) q =

(b) Ui =

1.18751.0335
26.06 (0.5810)(12)

1
(1500)(0.6218)

Ti
Ri + RA+ Ro

1
Ai R

q = UiAi( Ti

= 8.476x10-4

= 1.0722x10-3

70220
0.003689+ 0.0008476+0.0010722

1
(0.5421)(0.0056088)

= -26743.69 btu/h or

= 328.89 btu/hft2F or

= 328.89(0.5421)(70-220) = -26743.69 btu/h

q = 26743.69 btu/h

(c) Uo =

1
Ao R

q = UoAo( Ti

1
(0.6218)(0.0056088)

= 286.73 btu/hft2F

= 286.73(0.6218)(70-220) = -26743.69 btu/h

q = 26743.69 btu/h

You might also like