A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?

Q = A k0 (T 1 2 – T 2 2)/3 δ

Q = 2A k0 (T 1 3 – T 2 3)/3 δ

Q = A k0 (T 1 3 – T 2 3)/3 δ

Q = A k0 (T 1 2 – T 2 2)/3 δ

A plane wall of thickness δ has its surfaces maintained at temperatu

1.	A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?
A.	Q = 2A k0 (T 1 3 – T 2 3)/3 δ
B.	Q = A k0 (T 1 3 – T 2 3)/3 δ
C.	Q = A k0 (T 1 2 – T 2 2)/3 δ
D.	Q = A k0 (T 1 – T 2)/3 δ
Answer» C. Q = A k0 (T 1 2 – T 2 2)/3 δ

Discussion

No Comment Found

Related MCQs

“If β is less than zero, then with respect to the relation k = k0 (1 + β t), conductivity depends on surface area”.
The temperatures on the two sides of a plane wall are t1 and t2 and thermal conductivity of the wall material is prescribed by the relationK = k0 e (-x/δ)Where, k0 is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?
The unit of thermal conductivity doesn’t contain which parameter?
If β is greater than zero, then choose the correct statement with respect to given relationk = k0 (1 +β t)
The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?
With respect to the equation k = k0 (1 +β t) which is true if we put β = 0?
A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?
The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relationK = (1.45 + 0.5 * 10-5 t2) KJ/m hr degWhere, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
‚Äö√Ñ√∂‚àö√ë‚àö‚à´If ‚âà√≠‚Äö√¢¬ß is less than zero, then with respect to the relation k = k<sub>0</sub> (1 + ‚âà√≠‚Äö√¢¬ß t), conductivity depends on surface area‚Äö√Ñ√∂‚àö√ë‚àöœÄ.$#
The unit of thermal conductivity doesn‚Äö√Ñ√∂‚àö√ë‚àö¬•t contain which parameter?#

A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?

Discussion

No Comment Found

Related MCQs

Reply to Comment