Question

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

Answer 1

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

                     

Answer 2

Following are the calculation to the given question:

Calculating the average film temperature:

$\to T_f = \frac{T_s + T_{\infty}}{2} =\frac{80+20}{2}= 50^{\circ}\ C$  

At  $T_f = 50^{\circ}\ C$, obtain the properties of air:

$\to k=0.02735\ \frac{W}{m \cdot K} \\\\\to v=1.798 \times 10^{-5} \ \frac{m^2}{s}\\\\ \to Pr =0.7228 \\\\\to Re_{L} =\frac{VL}{v} =\frac{80 (\frac{5}{18}) \times 0.8}{1.798 \times 10^{-5}} = 988752.935$

Assume the engine block's bottom surface is a flat plate.

  For

$\to 5\times 10^{5} \leq Re_{L} \leq {10^7}, 0.6 \leq Pr \leq 60$

$\to Nu=0.0296 \ \ Re_{L}_{0.8} Pr^{\frac{1}{3}}$

          $= 0.0296(988752.935)^{0.8} (0.7228)^{\frac{1}{3}}\\\\ = 1660.99$

But,  

$\to Nu=\frac{hL}{k} \\\\ \to h= Nu \frac{k}{L}$

      $=\frac{1660.99 \times 0.02735}{0.8}\\\\ = 56.7850 \frac{W}{m^2 . K}$

$\to A_s= L \times w$  

         $= 0.8 \times 0.4\\\\ =0.32\ m^2$

$\to Q_{com} = h A_s (T_s -T_{\infty})$

              $= 56.7850 \times 0.32 \times (80-20)\\\\ = 1090.272\ W$

$\to Q_{rad} = \varepsilon A_s \sigma (T_{s}^{4} -T_{surr}^4)$

            $= (0.95)(0.32 )(5.67 \times 10^{-8}) [(80+273)^4-(25+273)^4]\\\\ = 131.710\ W$

Then the total rate of heat transfer is  

$\to Q_{total} = Q_{corw} + Q_{rad}$

              $=1090.272 +131.710 \\\\= 1221.982\ W$

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