1. The sun shines on a 150 m2 road surface so it is at 45°C. Below the 5cm thick asphalt(average conductivity of 0.06 W/m K), is a layer of rubbles at 15°C. Find the rate of heat transfer to the rubbles.
a) 5300 W
b) 5400 W
c) 5500 W
d) 5600 W
Explanation: There is conduction through the asphalt layer.
heat transfer rate = k A ∆T/∆x = 0.06 × 150 ×(45-15)/0.05
= 5400 W.
2. A pot of steel(conductivity 50 W/m K), with a 5 mm thick bottom is filled with liquid water at 15°C. The pot has a radius of 10 cm and is now placed on a stove that delivers 250 W as heat transfer. Find the temperature on the outer pot bottom surface assuming the inner surface to be at 15°C.
a) 15.8°C
b) 16.8°C
c) 18.8°C
d) 19.8°C
Explanation: Steady conduction, Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆T = 250 × 0.005/(50 × π/4 × 0.22) = 0.796
T = 15 + 0.796 = 15.8°C.
3. A water-heater is covered with insulation boards over a total surface area of 3 m2. The inside board surface is at 75°C and the outside being at 20°C and the conductivity of material being 0.08 W/m K. Find the thickness of board to limit the heat transfer loss to 200 W ?
a) 0.036 m
b) 0.046 m
c) 0.056 m
d) 0.066 m
Explanation: Steady state conduction through board.
Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆x = 0.08 × 3 ×(75 − 20)/200 = 0.066 m.
4. On a winter day with atmospheric air at −15°C, the outside front wind-shield of a car has surface temperature of +2°C, maintained by blowing hot air on the inside surface. If the wind-shield is 0.5 m2 and the outside convection coefficient is 250 W/Km2, find the rate of energy loss through front wind-shield.
a) 125 W
b) 1125 W
c) 2125 W
d) 3125 W
Explanation: Q (conv) = h A ∆Τ = 250 × 0.5 × [2 − ( −15)] = 250 × 0.5 × 17 = 2125 W.
5. A large heat exchanger transfers a total of 100 MW. Assume the wall separating steam and seawater is 4 mm of steel, conductivity 15 W/m K and that a maximum of 5°C difference between the two fluids is allowed. Find the required minimum area for the heat transfer.
a) 180 m2
b) 280 m2
c) 380 m2
d) 480 m2
Explanation: Steady conduction
Q = k A ∆T/∆x ⇒ Α = Q ∆x / k∆Τ
A = 100 × 10^6 × 0.004 / (15 × 5) = 480 m2.
6. A small light bulb (25 W) inside a refrigerator is kept on and 50 W of energy from the outside seeps into the refrigerated space. How much of temperature difference to the ambient(at 20°C) must the refrigerator have in its heat exchanger having an area of 1 m2 and heat transfer coefficient of 15 W/Km2 to reject the leak of energy.
a) 0°C
b) 5°C
c) 10°C
d) 15°C
Explanation: Total energy that goes out = 50+25 = 75 W
75 = hA∆T = 15 × 1 × ∆T hence ∆T = 5°C.
7. As the car slows down, the brake shoe and steel drum continuously absorbs 25 W. Assume a total outside surface area of 0.1 m2 with a convective heat transfer coefficient of 10 W/Km2 to the air at 20°C. How hot does the outside brake and drum surface become when steady conditions are reached?
a) 25°C
b) 35°C
c) 45°C
d) 55°C
Explanation: ∆Τ = heat / hA hence ∆T = [ Τ(BRAKE) − 20 ] = 25/(10 × 0.1) = 25°C
Τ(BRAKE) = 20 + 25 = 45°C.
8. A burning wood in the fireplace has a surface temperature of 450°C. Assume the emissivity to be 1 and find the radiant emission of energy per unit area.
a) 15.5 kW/m2
b) 16.5 kW/m2
c) 17.5 kW/m2
d) 18.5 kW/m2
Explanation: Q /A = 1 × σ T^4
= 5.67 × 10–8×( 273.15 + 450)4
= 15505 W/m2 = 15.5 kW/m2.
9. A radiant heat lamp is a rod, 0.5 m long, 0.5 cm in diameter, through which 400 W of electric energy is deposited. Assume the surface emissivity to be 0.9 and neglecting incoming radiation, find the rod surface temperature?
a) 700K
b) 800K
c) 900K
d) 1000K
Explanation: Outgoing power equals electric power
T4= electric energy / εσA
= 400 / (0.9 × 5.67 ×10-8× 0.5 × π × 0.005)
= 9.9803 ×10^11 K4 ⇒ T = 1000K.
10. Find the rate of conduction heat transfer through a 1.5 cm thick hardwood board, k = 0.16 W/m K, with a temperature difference between the two sides of 20°C.
a) 113 W/m2
b) 213 W/m2
c) 230 W/m2
d) 312 W/m2
Explanation: q = .Q/A = k ΔT/Δx = 0.16 Wm /K × 20K/0.015 m = 213 W/m2