Question

Consider air flowing over a 1-m-long flat plate at a velocity of $3 \mathrm{~m} / \mathrm{s}$. Determine the convection heat transfer coefficients and the Nusselt numbers at \(x=0.5 \mathrm{~m}\) and \(0.75 \mathrm{~m}\). Evaluate the air properties at \(40^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

Answer: The convection heat transfer coefficients and the Nusselt numbers at x=0.5m are: \(h_{0.5} = 8.16~W/m^2K\) and \(Nu_{0.5} = 145.91\), and at x=0.75m are: \(h_{0.75} = 6.35~W/m^2K\) and \(Nu_{0.75} = 169.50\).

Step-by-step solution

Determine air properties

Using a thermodynamic property table or an online calculator, find the properties of air at the given temperature (40°C) and pressure (1 atm). We will need the kinematic viscosity (\(\nu\)), thermal conductivity (k), and Prandtl number (Pr) for our calculations. For air at 40°C and 1 atm, we have: \(\nu = 2.06 \times 10^{-5} m^2/s\) \(k = 0.0281 W/mK\) \(Pr = 0.707\)

Calculate the Reynolds number

The Reynolds number (Re) will help us determine if the flow is laminar or turbulent. We can find the Reynolds number using the following formula: \(Re_x = \frac{U_{\infty} x}{\nu}\) Where \(U_{\infty}\) is the free-stream velocity (3 m/s), \(x\) is the length along the flat plate (either 0.5m or 0.75m), and \(\nu\) is the kinematic viscosity. For x=0.5m: \(Re_{0.5} = \frac{3 \times 0.5}{2.06 \times 10^{-5}} = 72,820\) For x=0.75m: \(Re_{0.75} = \frac{3 \times 0.75}{2.06 \times 10^{-5}} = 109,223\)

Determine if the flow is laminar or turbulent

If the Reynolds number is less than 5×10^5, the flow is considered laminar. If it is greater than 5×10^5, the flow is considered turbulent. In this case, the Reynolds numbers are less than 5×10^5 for both x=0.5m and x=0.75m, so the flow is laminar.

Calculate the local Nusselt number

For laminar flow over a flat plate, we can calculate the local Nusselt number using the following correlation: \(Nu_x = 0.332 \times Re_x^{1/2} \times Pr^{1/3}\) For x=0.5m: \(Nu_{0.5} = 0.332 \times (72,820)^{1/2} \times (0.707)^{1/3} = 145.91\) For x=0.75m: \(Nu_{0.75} = 0.332 \times (109,223)^{1/2} \times (0.707)^{1/3} = 169.50\)

Calculate the local convection heat transfer coefficient

We can find the local convection heat transfer coefficient (h) using the following equation: \(h_x = \frac{Nu_x \times k}{x}\) For x=0.5m: \(h_{0.5} = \frac{145.91 \times 0.0281}{0.5} = 8.16~W/m^2K\) For x=0.75m: \(h_{0.75} = \frac{169.50 \times 0.0281}{0.75} = 6.35~W/m^2K\) In conclusion, the convection heat transfer coefficients and the Nusselt numbers at x=0.5m are: \(h_{0.5} = 8.16~W/m^2K\) and \(Nu_{0.5} = 145.91\), and at x=0.75m are: \(h_{0.75} = 6.35~W/m^2K\) and \(Nu_{0.75} = 169.50\).