Fundamental of Heat Transfer Chapter 2 Problem 9 Answer Key

PROBLEM 2.9 KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures. FIND: Heat flux, ′′ q x , and temperature gradient, dT/dx, for the three different coordinate systems shown. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internal generation, (4) Constant properties. ANALYSIS: The rate equation for conduction heat transfer is ′′ = − q k dT dx x , (1) where the temperature gradient is constant throughout the wall and of the form ( ) () TL T0 dT . dx L − = (2) Substituting numerical values, find the temperature gradients, (a) ( ) 2 1 600 400 K dT T T 2000 K/m dx L 0.100m − − = = = < (b) ( ) 1 2 400 600 K dT T T 2000 K/m dx L 0.100m − − = = =− < (c) ( ) 2 1 600 400 K dT T T 2000 K/m. dx L 0.100m − − = = = < The heat rates, using Eq. (1) with k = 100 W/m⋅K, are (a) 2 x W q 100 2000 K/m=-200 kW/m mK ′′ =− × ⋅ < (b) 2 x W q 100 ( 2000 K/m)=+200 kW/m mK ′′ =− − ⋅ < (c) ′′ = − ⋅ × q W mK K / m = -200 kW / m x 2 100 2000 <

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Transcript of Fundamental of Heat Transfer Chapter 2 Problem 9 Answer Key

PROBLEM 2.9 KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures. FIND: Heat flux, qx , and temperature gradient, dT/dx, for the three different coordinate systems shown. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internal generation, (4) Constant properties. ANALYSIS: The rate equation for conduction heat transfer is

= q k dTdxx

, (1) where the temperature gradient is constant throughout the wall and of the form

( ) ( )T L T 0dT .dx L

= (2) Substituting numerical values, find the temperature gradients,

(a) ( )2 1 600 400 KdT T T 2000 K/mdx L 0.100m

= = = <

(b) ( )1 2 400 600 KdT T T 2000 K/mdx L 0.100m

= = = <

(c) ( )2 1 600 400 KdT T T 2000 K/m.dx L 0.100m

= = = < The heat rates, using Eq. (1) with k = 100 W/mK, are (a) 2x

Wq 100 2000 K/m=-200 kW/mm K

= < (b) 2x

Wq 100 ( 2000 K/m)=+200 kW/mm K

= < (c) = q

Wm K

K / m = -200 kW / mx2100 2000 <

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