Problems: (a) 20 The temperature distribution in a 3-dimensional cube (in Cartes
ID: 2074598 • Letter: P
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Problems: (a) 20 The temperature distribution in a 3-dimensional cube (in Cartesian coordinates) is given as T, Explain terms labeled 1, 2 and 3 in the equation. What type of differential equation is this? How many conditions are required to solve this problem? wall (b) 1s Using the energy balance write the boundary condition for the boundary shown in the figure a. Figure a (c) 2S The inner and outer surface temperatures of 5 mm thick glass window are 15 and 50c what is the heat loss through a 1 m x 3 m window? The thermal conductivity of glass is 1.4 W/m-K (d) 40 One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of 50 mm and a constant thermal conductivity of 5 W/m-K. For these conditions, the temperature distribution has the form T(x) -a+bx+cx. The surface at x 0 has at emperature of To 120 riences convection with a fluid for which T- 20 °C and h 500 W/m2-K. The surface at x Lis well insulated. 7,-120C nsulated tx) T-20°C 1.50mmExplanation / Answer
a Answer )
Part 1 is called Temperature profile of General Heat Conduction Equation
Part2 is called Internal Heat Conduction
Part3: Un Steady state heat Conduciton
The equaiton is second order non linear differential equation
The equation requires both the Initial conditions and Boundry conditions to solve it.
b Answer)
The boundry condition at the surface is
Heat tranfer by Conduction at surface = Heat transfer by convection to environment from surface
KA/L (T0 - Ts) = hA(Ts- T)
q”x=0=– k dT/dxx=0 =– h (T0– Tf)
q”x=L=– k dT/dxx=L=0
C Solution )
Heat Loss Q = KA/L (T0 - Ts)
Q = 1.4 * 3/0.005 (15-5)
Q = 8400 Watts =8.4 KW
d Ans)
Temperature Profile T=a+bx+cx2
L = 50 mm =0.05m
Tk = 5 W/m-K, h = 500 W/m2-K
To= 120oC,Tf= 20oC
a) An energy balance may be written on the wall in the form
k d2T/dx2+ qg =0
Boundary conditions may be written as
q”x=0=– k dT/dxx=0 =– h (T0– Tf)
q”x=L=– k dT/dxx=L=0
Integrating once yields
k [dT/dxx=L– dT/dxx=0]+qg =0
qg=(h/L) (T0– Tf) =500/0.05 (120-20)
qg= 106 Watt/m3
b) If T=a+bx+cx2
Boundry condition at x =0 T =120
on substituting a=120
Boundry Condition 2 at x=0 dT/dx=(h/k) (T0– Tf)
dT/dx = b +2cx
(h/k) (T0– Tf) = b
b =400/5 *100 = 104 m-1
Boundry Coondition 3 d2T/dx2 = – qg/ k
d2T/dx2= 2c = – qg/ k
c = – qg/ 2k = 106/ 10
c = 10-5 m2
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