Problem 6.3 7–20 Consider a hot automotive engine, which can be approximated as a 0.5-m-high, 0.40-m-wide, and 0.8-m-long rectangular block. The bottom surface of the block is at a temperature of 80°C and has an emissivity of 0.95. The ambient air is at 20°C, and the road surface is at 25°C. Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of 80 km/h. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block

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

no he didnt use it

explanation:

he said no