# problem 3.3

## Mechanical Engineering 309 with Dent at University of Alabama - Tuscaloosa *

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- University of Alabama - Tuscaloosa
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- Mechanical Engineering 309
- Dent
- problem 3.3

PROBLEM 3.3 KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside air conditions. FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute and plot the electrical power requirement as a function of ,o T ∞ for the range -30 ≤ ,o T ∞ ≤ 0°C with h o of 2, 20, 65 and 100 W/m 2 ⋅K. Comment on heater operation needs for low h o . If h ~ V n , where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Uniform heater flux, h q′′ , (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for a unit surface area, ,i s,i s,i ,o h io TT TT q 1h Lk 1h ∞∞ −− ′′ += + ( ) s,i ,o ,i s,i h oi 22 15 C 10 C TT T T 25 C 15 C q 0.004 m 1 1 Lk 1h 1h 1.4 W m K 65 W m K 10 W m K ∞∞ −− −− − ′′ =−= − + + ⋅ ⋅ ⋅ oo oo () 22 h q 1370 100 W m 1270 W m ′′ =− = < (b) The heater electrical power requirement as a function of the exterior air temperature for different exterior convection coefficients is shown in the plot. When h o = 2 W/m 2 ⋅K, the heater is unecessary, since the glass is maintained at 15°C by the interior air. If h ~ V n , we conclude that, with higher vehicle speeds, the exterior convection will increase, requiring increased heat power to maintain the 15°C condition. -30 -20 -10 0 Exterior air temperature, Tinfo (C) 0 500 1000 1500 2000 2500 3000 3500 Heat er power ( W / m ^ 2 ) h = 20 W/m^2.K h = 65 W/m^2.K h = 100 W/m^2.K COMMENTS: With h q ′′ = 0, the inner surface temperature with ,o T ∞ = -10°C would be given by ,i s,i i ,i ,o i o TT 1 h 0.10 0.846, T T 1 h L k 1 h 0.118 ∞ ∞∞ − === −++ or ()s,i T 25 C 0.846 35 C 4.6 C=− =− ooo . Frank P. Incropera PROBLEM

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