Heat Transfer Miscellaneous
 Consider one dimensional steady state heat conduction across a wall (as shown in figure below) of thickness 30 mm and thermal conductivity 15 W/mK. At x = 0, a constant heat flux, q'' = 1 × 10^{5} W/m^{2} is applied. On the other side of the wall, heat is removed from the wall by convection with a fluid at 25°C and heat transfer coefficient of 250 W/m^{2}K. The temperature (in °C), at x = 0 is _______.

View Hint View Answer Discuss in Forum
Correct Option: A
 A material P of thickness 1 mm is sandwiched between two steel slabs, as shown in the figure below. A heat flux 10 kW/m^{2} is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel is 10 W/mK. Considering onedimensional steady state heat conduction for the configuration, the thermal conductivity (k, inW/mK) of material P is ________.

View Hint View Answer Discuss in Forum
Correct Option: A
 A plane wall has a thermal conductivity of 1.15 W/mK. If the inner surface is at 1100°C and the outer surface is at 350°C, then the design thickness (in meter) of the wall to maintain a steady heat flux of 2500 W/m^{2} should be

View Hint View Answer Discuss in Forum
Q = KA dt dx q = Q = 2500 W/m² A 2500 = K 6dt dx 2500 = 1.15 × = (1100  350) x
x = 0.345 m.Correct Option: A
Q = KA dt dx q = Q = 2500 W/m² A 2500 = K 6dt dx 2500 = 1.15 × = (1100  350) x
x = 0.345 m.
 A cylindrical uranium fuel rod of radius 5 mm in a nuclear is generating heat at the rate of 4 × 107 W/m^{3}. The rod is cooled by a liquid (convective heat transfer coefficient 1000 W/ m^{2}K) at 25°C. At steady state, the surface temperature (in K) of the rod is

View Hint View Answer Discuss in Forum
Given data: r = 0.005 m
d = 2r = 2 × 0.005 = 0.010 mq_{6} = 4 × 10^{7} W m^{3} h_{0} = 1000 W m^{2}K
T_{0} = 25°c
For steady state,
Rate of heat generation in the rod} = Rate of convection H.T. from rod surface to fluid q_{G} × volume of rod = h_{0} A(T_{s} – T_{0})q_{G} × π d²l = h_{0} × πdl (T_{s}  T_{0}) 4 d.q_{G} (T_{s}  T_{0}) = h_{0} 4 h_{0} .010 × 4 × 10^{7} = 1000 (T_{s}  25) m^{2}4
T_{s} – 25 = 100
T_{s} = 125° c or 398 KCorrect Option: B
Given data: r = 0.005 m
d = 2r = 2 × 0.005 = 0.010 mq_{6} = 4 × 10^{7} W m^{3} h_{0} = 1000 W m^{2}K
T_{0} = 25°c
For steady state,
Rate of heat generation in the rod} = Rate of convection H.T. from rod surface to fluid q_{G} × volume of rod = h_{0} A(T_{s} – T_{0})q_{G} × π d²l = h_{0} × πdl (T_{s}  T_{0}) 4 d.q_{G} (T_{s}  T_{0}) = h_{0} 4 h_{0} .010 × 4 × 10^{7} = 1000 (T_{s}  25) m^{2}4
T_{s} – 25 = 100
T_{s} = 125° c or 398 K
 Heat is generated uniformly in a long solid cylindrical rod (diameter =10 mm) at the rate of 4 × 10^{7} W/m^{3}. The thermal conductivity of the rod material is 25 W/mK. Under steady state conditions, the temperature ditference between the centre and the surface of the rod is 1°C.

View Hint View Answer Discuss in Forum
Correct Option: A