Page 479 - Bird R.B. Transport phenomena

§15.3 Use of the Macroscopic Balances to Solve Steady-State Problems with Flat Velocity Profiles 459

Air out at 0° F and 15 psia Fig. 15.3-1. The cooling of air in a countercurrent
T heat exchanger.

- Plane 2

Hot
liquid
out
- Plane 1

Air in at 300°F and 30 psia
<v> = 100 ft sec" 1

EXAMPLE 15.3-1 Two hundred pounds per hour of dry air enter the inner tube of the heat exchanger shown in
Fig. 15.3-1 at 300°F and 30 psia, with a velocity of 100 ft/sec. The air leaves the exchanger at
The Cooling of an 0°F and 15 psia, at 10 ft above the exchanger entrance. Calculate the rate of energy removal
Meal Gas across the tube wall. Assume turbulent flow and ideal gas behavior, and use the following ex-
pression for the heat capacity of air:
C = 6.39 + (9.8 X 10~ )T - (8.18 X 10~ )T 2 (15.3-1)
4
8
p
where C is in Btu/(lb-mole • R) and T is in degrees R.
p
SOLUTION For this system, the macroscopic energy balance, Eq. 15.1-3, becomes
(H 2 - Щ + \{v\ - v]) 4- g(h -h,) = Q (15.3-2)
2
The enthalpy difference may be obtained from Eq. 9.8-8, and the velocity may be obtained as
a function of temperature and pressure with the aid of the macroscopic mass balance p^ =
p v and the ideal gas law p = pRT/M. Hence Eq. 15.3-2 becomes
2 2
(15.3-3)

The explicit expression for C p in Eq. 15.3-1 may then be inserted into Eq. 15.3-3 and the inte-
gration performed. Next substitution of the numerical values gives the heat removal per
pound of fluid passing through the heat exchanger:
5
4
-Q = £[(6.39X300) + 5(9.8 X 10~ )(5.78 - 2.12X10 )
8
8
-4(8.18 X 10~ )(4.39 - 0.97X10 )]
h [1
(32.2X778)/
2 2 \ V(32.2)(778)J "
= 72.0 - 0.093 - 0.0128
= 71.9Btu/hr (15.3-4)
The rate of heat removal is then
-Qw = 14,380 Btu/hr (15.3-5)
Note, in Eq. 15.3-4, that the kinetic and potential energy contributions are negligible in com-
parison with the enthalpy change.