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Research Papers: Forced Convection

# An Integral Solution for Heat Transfer in Accelerating Turbulent Boundary Layers

[+] Author and Article Information
James Sucec

Department of Mechanical Engineering, University of Maine, Orono, ME 04469-5711

J. Heat Transfer 131(11), 111702 (Aug 25, 2009) (8 pages) doi:10.1115/1.3154649 History: Received December 15, 2008; Revised April 21, 2009; Published August 25, 2009

## Abstract

An equilibrium thermal wake strength parameter is developed for a two-dimensional turbulent boundary layer flow and is then used in the combined thermal law of the wall and the wake to give an approximate temperature profile to insert into the integral form of the thermal energy equation. After the solution of the integral $x$ momentum equation, the integral thermal energy equation is solved for the local Stanton number as a function of position $x$ for accelerating turbulent boundary layers. A simple temperature distribution in the thermal “superlayer” is part of the present modeling. The analysis includes a dependence of the hydrodynamic and thermal wake strengths on the momentum thickness and enthalpy thickness Reynolds numbers, respectively. An approximate dependence of the turbulent Prandtl number, in the “log” region, on the strength of the favorable pressure gradient is proposed and incorporated into the solution. The resultant solution for the Stanton number distribution in accelerated turbulent flows is compared with experimental data in the literature. A comparison of the present predictions is also made to a finite difference solution, which uses the turbulent kinetic energy—turbulent dissipation model of turbulence, for a few cases of accelerating flows.

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## Figures

Figure 1

Predicted thermal wake strength πteq variation with β and Prt

Figure 2

Predictions and experimental data (•). Runs 1 and F-3 data (1). Run 091069-1 data (2): L=1.83 m (6.0 ft). Run 072968-1 data (3): L=2.29 m (7.5 ft).

Figure 3

Predictions and experimental data (•). Run 070869-1 data (2). Run F-2 data (1): L=1.83 m (6.0 ft). Run 092469-1, data (2): L=2.29 m (7.5 ft).

Figure 4

Comparison of present predictions and those of Jones and Launder (4) with data (•). Runs 10, 42, and F-3 data (1): L=1.83 m (6.0 ft).

Figure 5

Comparison of predictions with data (•) for different locations of the start of calculation. Run 25 data (1): L=1.83 m (6.0 ft).

Figure 6

Predictions and data (•). Runs 41 and F-4 data (1): L=1.83 m (6.0 ft). Run 100269-1 data (2): L=2.29 m (7.5 ft).

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