An air-to-water compact heat exchanger is to be designed to serve as an intercooler for a 3.7 MW gas turbine plant. The exchanger is to meet the following heat transfer and pressure drop performance specifications

Question

Air-side Operating Conditions



The exchanger is to have a cross-flow configuration with both fluids unmixed. The heat

exchanger surface proposed for the exchanger consists of flattened tubes with continuous

aluminum fins specified as a 11.32– 0.737 – SR surface in Ref. 10. The heat exchanger is

shown schematically below.



The measured heat transfer and friction characteristic for this exchanger surface are

shown in the graph below



Geometrical details for the proposed surface are





The design should specify the core size, the air flow frontal area, and the flow length. The

water velocity inside the tubes is 1.34 m/s. See Problem 10.53 for the calculation of the

water side heat transfer coefficient.

Note: (i) the free-flow area is defined such that the mass velocity, G, is the air mass flow

rate per unit free flow area, (ii) the core pressure drop is given by ?p = fG2L/2?rh where

L is the length of the core in the air flow direction, (iii) the fin length, Lf, is defined such

that Lf = 2A/P where A is the fin cross-sectional area for heat conduction and P is the

effective fin perimeter.

GIVEN

? Air-to-Water Intercooler with the geometry and requirements specified above

? From Problem 10.53: Water side convective heat transfer coefficient (hc,H2O) = 7580 W/(m2 K)

FIND

(a) The air flow frontal area (Aair)

(b) The flow length (L)

(c) The core size

ASSUMPTIONS

? Steady state

? Entrance effects are negligible

? Flow acceleration effects are negligible

? Negligible fouling resistance

? Negligible variation in thermal resistance

? The thermal resistance of the tube wall is negligible

PROPERTIES AND CONSTANTS

From Appendix 2, Table 28, for air at the mean temperature of 77°C



From Appendix 2, Table 13, for water at 20°C, c = 4182 J/(kg K)

From Appendix 2, Table 12, the thermal conductivity of aluminum at 320 K (ka) = 238 W/(m K)

Answer 1

The outlet water temperature is given by the conservation of energy



The effectiveness required for the specified performance is given by Equation (10.22a)



The heat capacity rate ratio is



From Figure 10.21 for cross-flow heat exchangers with e = 0.9 and Cmin/Cmax = 0.122,

NTUmax = 2.75 = Uair Aair/Cmin.

The solution will require iteration. For the first iteration, let ReD = 104 = 4 rh G/?

Solving for the mass velocity



From the graphical data at Re = 104



We must calculate the fin efficiency per Chapter 2



The fin efficiency from Equation (2.72) is



The total fin efficiency can be calculated from Equation (2.73)



and the overall heat transfer coefficient from Equation (2.74) is



The core pressure drop is



Repeating this procedure until ?p/p1 = 7.6% yields the following results