Question

Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of $20 \mathrm{~cm} / \mathrm{s}\( over a \)5.0\(-m-long and \)1.0$-m-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are $\rho=880 \mathrm{~kg} / \mathrm{m}^{3}\(, \)\mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$.

Short answer

Answer: The rate of heat transfer from the oil to the plate is 1314 W.

Step-by-step solution

Calculate the flow properties

We are given the temperature, flow velocity, and most of the fluid properties except the specific heat, \(c_{p}\). We can find it by converting it into the same unit as the other properties, which is given as: $$ c_{p} = 2.0 \times 10^3 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} $$

Calculate the Reynolds number

The Reynolds number, \(Re\), is a dimensionless quantity that characterizes the flow regime. It can be calculated as: $$ Re = \frac{\rho u L}{\mu} $$ Where \(u\) is the flow velocity, \(L\) is the characteristic length (in this case, the length of the plate), and \(\rho\) and \(\mu\) are the density and dynamic viscosity of the oil, respectively. Using the given values, we find: $$ Re = \frac{880\,\mathrm{kg/m^3} \cdot 0.2\,\mathrm{m/s} \cdot 5\,\mathrm{m}}{0.005\,\mathrm{kg/m\cdot s}} = 176000 $$

Calculate the Nusselt number

The Nusselt number, \(Nu\), is another dimensionless quantity that relates the convective heat transfer coefficient to the thermal conductivity of the fluid. For laminar flow over a flat plate with constant wall temperature, the Nusselt number can be approximated by: $$ Nu = 0.664\,Re^{1/2}Pr^{1/3} $$ Where \(Pr\) is the Prandtl number, which is given by: $$ Pr=\frac{c_{p} \mu}{k} $$ Using the given fluid properties, we can calculate the Prandtl number as: $$ Pr = \frac{2\times 10^3\,\mathrm{J/kg\cdot K} \cdot 0.005\,\mathrm{kg/m\cdot s}}{0.15\,\mathrm{W/m\cdot K}} = 6.67 $$ And then calculate the Nusselt number: $$ Nu = 0.664 \times (176000)^{1/2} \times (6.67)^{1/3} = 219.1 $$

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient, \(h\), can now be calculated using the Nusselt number: $$ h = \frac{Nu \times k}{L} = \frac{219.1 \times 0.15\,\mathrm{W/m\cdot K}}{5\,\mathrm{m}} = 6.57\,\mathrm{W/m^2\cdot K} $$

Compute the rate of heat transfer

Finally, we can calculate the rate of heat transfer, \(q\), using the convective heat transfer equation: $$ q = hA(T_\mathrm{fluid} - T_\mathrm{plate}) = 6.57\,\mathrm{W/m^2\cdot K} \times 5\,\mathrm{m} \times 1\,\mathrm{m} \times (60\,^{\circ}\mathrm{C} - 20\,^{\circ}\mathrm{C}) = 1314\,\mathrm{W} $$ So, the rate of heat transfer from the oil to the plate is 1314 W.