Answer:
Q_total = 1431 W
Explanation:
Given:-
- The dimension of the engine = ( 0.5 x 0.4 x 0.8 ) m
- The engine surface temperature T_s = 80°C
- The road surface temperature T_r = 25°C = 298 K
- The ambient air temperature T∞ = 20°C
- The emissivity of block has emissivity ε = 0.95
- The free stream velocity of air V∞ = 80 km/h
- The stefan boltzmann constant σ = 5.67*10^-8 W/ m^2 K^4
Find:-
Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation
Solution:-
- We will extract air properties at 1 atm from Table 15, assuming air to be an ideal gas. We have:
T_film = ( T_s + T∞ ) / 2 = ( 80 + 20 ) / 2
= 50°C = 323 K
k = 0.02808 W / m^2
v = 1.953*10^-5 m^2 /s
Pr = 0.705
- The air flows parallel to length of the block. The Reynold's number can be calculated as:
Re = V∞*L / v
= [ (80/3.6)*0.8 ] / [1.953*10^-5]
= 9.1028 * 10^5
- Even though the flow conditions are ( Laminar + Turbulent ). We are to assume Turbulent flow due to engine's agitation. For Turbulent conditions we will calculate Nusselt's number and convection coefficient with appropriate relation.
Nu = 0.037*Re^0.8 * Pr^(1/3)
= 0.037*(9.1028 * 10^5)^0.8 * 0.705^(1/3)
= 1927.3
h = k*Nu / L
= (0.02808*1927.3) / 0.8
= 67.65 W/m^2 °C
- The heat transfer by convection is given by:
Q_convec = A_s*h*( T_s - T∞ )
= 0.8*0.4*67.65*(80-20)
= 1299 W
- The heat transfer by radiation we have:
Q_rad = A_s*ε*σ*( T_s - T∞ )
= 0.8*0.4*0.95*(5.67*10^-8)* (353^4 - 298^4)
= 131.711 W
- The total heat transfer from the engine block bottom surface is given by:
Q_total = Q_convec + Q_rad
Q_total = 1299 + 131.711
Q_total = 1431 W
