Question

Air flows in a pipe under fully developed conditions with an average velocity of \(1.25 \mathrm{~m} / \mathrm{s}\) and a temperature of $20^{\circ} \mathrm{C}\(. The pipe's inner diameter is \)4 \mathrm{~cm}$, and its length is \(4 \mathrm{~m}\). The first half of the pipe is kept at a constant wall temperature of \(100^{\circ} \mathrm{C}\). The second half of the pipe is subjected to a constant heat flux of \(200 \mathrm{~W}\). Determine \((a)\) the air temperature at the \(2 \mathrm{~m}\) length, \((b)\) the air temperature at the exit, \((c)\) the total heat transfer to the air, and \((d)\) the wall temperature at the exit of the tube. Evaluate the properties of air at \(80^{\circ} \mathrm{C}\).

Short answer

Question: Determine the air temperature at 2 m length, air temperature at the exit, total heat transfer to the air, and wall temperature at the exit of the tube for a fully developed flow in a pipe with average velocity 1.25 m/s, inner diameter of 4 cm, and length of 4 m. The first half of the pipe has a constant wall temperature of 100°C, and the second half of the pipe has a constant heat flux of 200 W. The properties of air must be evaluated at 80°C. Answer: Follow the given step-by-step solution to calculate the air temperature at different points along the pipe, the total heat transfer to the air, and the wall temperature at the exit using the convective heat transfer formula, temperature formulas, and the provided information. Once all the calculations are performed, you will have the required results for air temperature at 2 m length, air temperature at the exit, total heat transfer to the air, and wall temperature at the exit of the tube.

Step-by-step solution

Calculate the inner area of the pipe

The area can be calculated using the formula, Area = \(\pi \times (\frac{d}{2})^2\) where \(d = 0.04 \thinspace m\) is the diameter of the pipe. Area = \(\pi \times (\frac{0.04}{2})^2 = \pi \times (0.02)^2 \approx 1.256 \times 10^{-3} \thinspace m^2\)

Calculate the volumetric flow rate (Q) of air

Volumetric flow rate (Q) is calculated using the formula, Q = Area \(\times\) velocity Substitute the given values to calculate Q, Q = \(1.256 \times 10^{-3} \thinspace m^2 \times 1.25 \thinspace \frac{m}{s} = 1.57 \times 10^{-3} \thinspace \frac{m^3}{s}\)

Evaluation of air properties at 80°C

The properties of air at 80°C should be obtained from references or tables, such as the one available in "Fundamentals of Heat and Mass Transfer" by Incropera, DeWitt, Bergman, and Lavine. At 80°C, the properties of air are: Density (\(\rho\)) = 1.067 kg/m³ Specific heat (cp) = 1009 J/(kg·K) Thermal conductivity (k) = 0.0314 W/(m·K) Dynamic viscosity (μ) = 2.207 x 10⁻⁵ kg/(m·s)

Calculate the air temperature at 2 m length (first half of the pipe)

Since the first half of the pipe has constant wall temperature, we need to find the convective heat transfer coefficient (h) first. To calculate h, use the formula for Nusselt number (Nu = h D/k). Considering fully developed flow and using the Gnielinski correlation, we get: Nu = \(\frac{(f/8) \times (Re - 1000)}{1 + 12.7 \sqrt{(f/8)}(Pr^{\frac{2}{3}}-1)}\) In this formula, f is the Fanning friction factor calculated as \(f = 0.0791/Re^{0.25}\), Re is the Reynolds number (\({ρ \times V \times D}/{μ}\)), and Pr is the Prandtl number (\({cp × μ}/{k}\)). Substitute the given values to find Nu, and from there, find h. Then apply the steady-state energy balance equation considering a control volume that includes the first half of the pipe: \(m \times cp \times (T_2 - T_1) = h \times \text{circumference} \text{(} C=\pi \times D \text{)} \times L_1 \times (T_{w1} - T_2)\) Solve for \(T_2\) to find the air temperature at the 2 m length.

Calculate the air temperature at the exit (4 m length - second half of the pipe)

In the second half of the pipe, we have a constant heat flux. We can apply the steady-state energy balance equation again using a control volume that includes the second half of the pipe. The heat transfer to the air equals the heat flux multiplied by the area, \({Q_h = q_w \times A}\): \(m \times cp \times (T_3 - T_2) = q_w \times A_2\) (A_2 is the area of the second half of the pipe) From the given values, q_w = 200 W, and A_2 can be calculated as \(A_2 = \pi \times D \times L_2 = \pi \times 0.04 \times 2 \approx 0.251 \thinspace m^2\). Substituting the values, solve for \(T_3\) to find the air temperature at the exit.

Calculate the total heat transfer to the air

The total heat transfer to the air is the sum of heat transfer in the first half and the second half of the pipe. We already calculated the heat transfer in the second half of the pipe, which is the heat flux multiplied by the area. To calculate the heat transfer in the first half, we can use the heat transfer coefficient h and the surface temperature difference: \(Q_{total} = h \times C \times L_1 \times (T_{w1} - T_2) + q_w \times A_2\)

Calculate the wall temperature at the exit of the tube

To find the wall temperature at the exit, we can use the convective heat transfer formula: \(q_w = h \times (T_{w3} - T_3)\) Solve for \(T_{w3}\), which is the wall temperature at the exit of the tube. After solving all the equations, you will have the results for air temperature at 2 m length, air temperature at the exit, total heat transfer to the air, and wall temperature at the exit of the tube.