Question

Hot engine oil at \(150^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(2 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.5-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine (a) the local convection heat transfer coefficient at $0.2 \mathrm{~m}$ from the leading edge and the average convection heat transfer coefficient, and \((b)\) repeat part \((a)\) using the Churchill and Ozoe (1973) relation.

Short answer

Based on the calculations, the local convection heat transfer coefficient at 0.2 m for hot engine oil flowing over a flat plate is 177.92 W/m²K and the average convection heat transfer coefficient is 76.37 W/m²K. When using the Churchill and Ozoe (1973) relation, the local convection heat transfer coefficient at 0.2 m becomes 251.56 W/m²K, while the average convection heat transfer coefficient increases to 100.63 W/m²K.

Step-by-step solution

Determine the properties of the engine oil at the film temperature

The film temperature (\(T_f\)) is the average of the surface temperature (\(T_s\)) and the temperature of the fluid (\(T_\infty\)). Calculate the film temperature: $$ T_f = \frac{T_s + T_\infty}{2} $$ Using given values: $$ T_f = \frac{50^{\circ} \mathrm{C} + 150^{\circ} \mathrm{C}}{2} = 100^{\circ} \mathrm{C} $$ At this temperature, engine oil has the following properties (from oil property tables): - Density (\(\rho\)): 852 kg/m³ - Specific Heat Capacity (\(c_p\)): 2,164 J/kg·K - Thermal Conductivity (\(k\)): 0.145 W/m·K - Dynamic Viscosity (\(\mu\)): 2.94 × 10⁻² kg/m·s

Calculate the Reynolds number and Grashof number

The Reynolds number (\(Re_x\)) is calculated using the formula: $$ Re_x = \frac{\rho Vx}{\mu} $$ At \(x = 0.2 \mathrm{~m}\): $$ Re_{0.2} = \frac{852(2)(0.2)}{2.94 \times 10^{-2}} = 116009.52 $$ The Grashof number (\(Gr_x\)) is calculated using the formula: $$ Gr_x = \frac{g(\rho^2)x^3\beta(T_\infty - T_s)}{\mu^2} $$ Where \(g\) is the acceleration due to gravity, and \(\beta\) is the coefficient of volume expansion. For engine oil, \(\beta \approx \frac{1}{T_f+273}\). $$ Gr_{0.2} = \frac{(9.81)(852^2)(0.2^3)(\frac{1}{373})(100)}{(2.94 \times 10^{-2})^2} = 2.365 \times 10^9 $$

Determine the local convection heat transfer coefficient using the Nusselt number

The Nusselt number is the ratio of convective to conductive heat transfer across a boundary and can be used to determine the local convection heat transfer coefficient (\(h_{L,0.2}\)). For laminar flow over a flat plate with constant temperature: $$ Nu_{L,0.2} = 0.332(Re_{0.2})^{1/2}(Pr)^{1/3} $$ Where \(Pr\) is the Prandtl number, which is equal to the ratio of kinematic viscosity to thermal diffusivity. $$ Pr = \frac{\mu c_p}{k} $$ Using the given engine oil properties: $$ Pr = \frac{(2.94 \times 10^{-2})(2164)}{0.145} = 43.98 $$ Now calculate the Nusselt number: $$ Nu_{L,0.2} = 0.332(116009.52)^{1/2}(43.98)^{1/3} = 243.77 $$ Finally, calculate the local heat transfer coefficient: $$ h_{L,0.2} = \frac{k \times Nu_{L,0.2}}{0.2} = \frac{0.145 \times 243.77}{0.2} = 177.92 \: \mathrm{W/m^2K} $$

Calculate the average convection heat transfer coefficient

Now we can calculate the average convection heat transfer coefficient (\(h_{avg}\)). $$ Nu_{avg} = 0.664Re_{L}^{1/2}Pr^{1/3} $$ $$ Nu_{avg} = 0.664(116009.52)^{1/2}(43.98)^{1/3} = 261.96 $$ $$ h_{avg} = \frac{k \times Nu_{avg}}{L} = \frac{0.145 \times 261.96}{0.5} = 76.37 \: \mathrm{W/m^2K} $$

Repeat steps 3 and 4 using the Churchill and Ozoe (1973) relation

The Churchill and Ozoe (1973) relation provides an approximation for the Nusselt number: $$ Nu = 0.037 Re^{4/5} Pr^{1/3} $$ Calculate the Nusselt number for the local heat transfer coefficient: $$ Nu_{L,0.2} = 0.037(116009.52)^{4/5}(43.98)^{1/3} = 344.89 $$ Calculate the local heat transfer coefficient: $$ h_{L,0.2} = \frac{k \times Nu_{L,0.2}}{0.2} = \frac{0.145 \times 344.89}{0.2} = 251.56 \: \mathrm{W/m^2K} $$ Calculate the Nusselt number for the average heat transfer coefficient: $$ Nu_{avg} = 0.037(116009.52)^{4/5}(43.98)^{1/3} = 344.89 $$ Calculate the average heat transfer coefficient: $$ h_{avg} = \frac{k \times Nu_{avg}}{L} = \frac{0.145 \times 344.89}{0.5} = 100.63 \: \mathrm{W/m^2K} $$ Therefore, using the Churchill and Ozoe (1973) relation, the local convection heat transfer coefficient at \(0.2 \mathrm{~m}\) is \(251.56 \mathrm{W/m^2K}\) and the average convection heat transfer coefficient is \(100.63 \mathrm{W/m^2K}\).