Rensselaer Polytechnic Institute

ENGR 2250

13-1 Hot exhaust gases are used in the reheat section of a Rankine cycle. Consider a commercial steel tube 5-cm
outside diameter, 4.5-cm inside diameter used to convey the steam. The air side heat transfer coefficient is
85 W/m2·K, and that of the steam side is 200 W/m2·K. Determine:
a. the overall heat transfer coefficient based on the inside tube area (in W/m2·K)
b. the overall heat transfer coefficient based on the outside tube area (in W/m2·K)
c. the overall heat transfer coefficient based on the inside area if air side fouling is 0.0015 m2·K/W and
steam side fouling is 0.0005 m2·K/W (in W/m2·K).
Approach:
Use Eq. 13-3 and Eq. 13-4, which give the overall heat
transfer coefficient in terms of its component parts.
Assumptions:
1. There are no fins.
2. There is no fouling.
Solution:
The overall heat transfer coefficient is calculated with Eq. 13-3 and Eq. 13-4:
, , , ,
1 1 1 1
i o
w
i i o o o i i i o i i o o o o o o o
R R
R
U A U A η η h A A η A η h A
′′ ′′
= = + + + +
No fins are used, so
, ,
1
η η o i = = o o .
For a circular tube ln( ) ln( )
2 2
o i o i
w
r r D D
R
π π kL kL
= =
From Appendix A-2, the thermal conductivity of commercial steel is k = 60.5 W/mK.
The areas are: and
Ai i = = π D L Ao o π D L
a) Incorporating the above expressions into the equation for the overall heat transfer coefficient based on the
inside area, noting that the length cancels, and ignoring fouling:
1 ln( )
2
i i i o i
i i i o o
A D L D D D L
U h A kL h D L
π π
π π
= + +
( ) ( )
( )
5
2 2
1 1 0.045m ln 0.050 0.045 1 0.045m
= + + 0.005 3.92 10 0.0106
200 W m K 2 60.5W mK 85W m K 0.050m
Ui
-
= + × +
=64.0 W m2K
Ui Answer
b) The overall heat transfer coefficient based on the outside area is:
2 2
W 0.045 W
64.0 =57.6
m K 0.050 m K
i
o i
o
A
U U
A
⎛ ⎞⎛ ⎞
= = ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ Answer
c) With fouling on both sides of the heat exchanger, we obtain
( ) ( )
( )
2 2
2 2
1 1 m K 0.045m ln 0.050 0.045 m K 0.045m 1 0.045m
= +0.0005 + + 0.0005 +
200 W m K W 2 60.5 W mK W 0.050 m 85W m K 0.050m
Ui
⎛ ⎞⎛ ⎞ ⎛ ⎞
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
=60.3W m2K
Ui Answer
Comment:
The overall heat transfer coefficient is low, so the effect of fouling is small. If the overall heat transfer coefficient
had been large, then the addition of fouling would have had a much g