Question

Atmospheric air with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and a temperature of \(20^{\circ} \mathrm{C}\) is in parallel flow over a flat heater surface which is maintained at \(80^{\circ} \mathrm{C}\). The surface area of the heater is \(0.30 \mathrm{~m}^{2}\). The drag force induced by the airflow on the heater is measured to be \(0.2 \mathrm{~N}\). Using the momentum-heat transfer analogy, determine the electrical power needed to maintain the prescribed heater surface temperature of \(80^{\circ} \mathrm{C}\). Evaluate the air properties at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

Answer: The electrical power needed to maintain the heater surface temperature at 80°C is approximately 1007.65 W.

Step-by-step solution

Find the Reynolds number of the flow

First, let's find the Reynolds number (\(Re\)) of the flow based on the air properties at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\): At these conditions, the dynamic viscosity of air, \(\mu\), is around \(1.9 \times 10^{-5} \mathrm{~kg/m.s}\). The density of air, \(\rho\), is about \(1.15 \mathrm{~kg/m^3}\). We can calculate the Reynolds number using the formula: \(Re = \frac{\rho v L}{\mu}\) Assuming the heat transfer length, \(L\), equals the largest length of the heater, we get: \(Re = \frac{1.15 \times 10 \times 0.3}{1.9 \times 10^{-5}} \approx 1.81 \times 10^4\)

Calculate Nusselt number using Reynolds number

We will use the Sieder-Tate (or Prandtl number, \(Pr\)), which is about \(0.72\) at \(50^{\circ} \mathrm{C}\). The Nusselt number (\(Nu\)) can be calculated using the formula: \(Nu = 0.023 Re^{0.8} Pr^{0.3} \approx 0.023 \times (1.81 \times 10^4)^{0.8} \times 0.72^{0.3} \approx 158.31\)

Calculate heat transfer coefficient

Now we can determine the heat transfer coefficient (\(h \)) using the Nusselt number and the thermal conductivity (\(k\)) of air (around \(0.028 \mathrm{~W/m.K}\) at \(50^{\circ} \mathrm{C}\)) using the formula: \(h = \frac{Nu \times k}{L} \approx \frac{158.31 \times 0.028}{0.3} \approx 14.74 \mathrm{~W/m^2.K}\)

Use the momentum-heat transfer analogy to find the heat transfer rate

The momentum-heat transfer analogy states that, for flat-plate boundary layer flows under constant heat flux, the friction force (\(F\)) is proportional to the heat transfer rate (\(q\)). \(F = \frac{1}{2}\rho v^{2}AC_f\) The electrical power needed to maintain the prescribed heater surface temperature of \(80^{\circ} \mathrm{C}\) can be determined using: \(q = hA(T_{heater}-T_{air})\) Solve for the heat transfer coefficient in terms of the drag force and substitute into the second equation to find the power required. \(h = \frac{2F}{v^{2}AC_f}\) Therefore, \(q = \frac{2F}{v^{2}AC_f}A(T_{heater}-T_{air})\)

Calculate the electrical power needed

Using the known values, we can now determine the electrical power needed to maintain the heater surface temperature: \(q = \frac{2\times0.2}{(10)^{2}\times 0.30\times 0.00574}\times 0.30\times (80-20) \approx 1007.65 \mathrm{~W}\) So, the electrical power needed to maintain the heater surface temperature at \(80^{\circ} \mathrm{C}\) is approximately \(1007.65 \mathrm{~W}\).