Page 1 PROBLEM 3.1 KNOWN: One-dimensional, plane wall ...

Hence,k T T h T T ∞LkT∞+ Ts,1T hLs,2 =.k1+ hL( s,1 − s,2 ) = ( s,2 − )PROBLEM 3.7 (Cont.)To determine the wind chill effect, we must determine the heat loss for the windy day and useit to evaluate the hypothetical ambient air temperature, T ∞ ′ , which would provide the sameheat loss on a calm day, Hence,Ts,1 T∞ Ts,1T∞ q′′ = − =− ′⎡L 1⎤ ⎡L 1⎤⎢+ +⎣k h⎥ ⎦⎢windy ⎣k h⎥⎦calmFrom these relations, we can now find the results sought:0.003 m 1+q′′ 2(a) calm 0.2 W/m⋅ K 65 W/m K 0.015 + 0.0154= ⋅ =q′′ windy 0.003 m 1+0.015 + 0.040.2 W/m ⋅K 25 W/m2⋅Kq′′calm = 0.553q′′windy$ 0.2 W/m⋅K$− 15 C +36 C2( 25 W/m ⋅ K)( 0.003 m)(b) Ts,2 ⎦ calm 0.2 W/m⋅K1+2( 25 W/m ⋅ K)( 0.003 m)⎤ = = 22.1 C$ 0.2 W/m ⋅K$− 15 C +36 C( 65 W/m2⋅ K)( 0.003m)Ts,2 ⎤⎦= = 10.8 Cwindy 0.2 W/m ⋅K1+2(c) ( )( 65 W/m ⋅K)( 0.003m)( 0.003/0.2 + 1/ 25)( 0.003/ 0.2 + 1/ 65)T∞′ = 36 C − 36 + 15 C =−56.3 C