Question

A 12 -m-long and 12-mm-inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with \(T_{i}=25^{\circ} \mathrm{C}, V=0.8 \mathrm{~m} / \mathrm{s}\). A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe so that the fluid exits at \(75^{\circ} \mathrm{C}\). Assuming fully developed flow and taking the average fluid properties to be $\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, \)\mu=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, and \(\mathrm{Pr}=10\), determine: (a) The required surface heat flux \(\dot{q}_{s}\), produced by the heater (b) The surface temperature at the exit, \(T_{s}\) (c) The pressure loss through the pipe and the minimum power required to overcome the resistance to flow.

Short answer

Question: Determine (a) the required surface heat flux produced by the heater, (b) the surface temperature at the exit, and (c) the pressure loss through the pipe and the minimum power required to overcome the resistance to flow in the given scenario. Answer: (a) The required surface heat flux is 6000 W/m². (b) The surface temperature at the exit is 116.04°C. (c) The pressure loss through the pipe is 1335.2 Pa, and the minimum power required to overcome the resistance to flow is 0.1205 W.

Step-by-step solution

Find the required surface heat flux

First, let's apply the conservation of energy principle: \(\dot{m} c_p (T_{o} - T_i) = \dot{q}_s A_s\) We are given the mass flow rate and temperature difference, so we can easily find the required heat flux, \(\dot{q}_s\). The mass flow rate, \(\dot{m}\), is given by \(\rho V A_f\), where \(A_f\) is the cross-sectional area of the pipe. The surface area, \(A_s\), is the product of the pipe's perimeter and its length. Using the given data: \(A_f = \frac{\pi D^2}{4} = \frac{\pi (0.012)^2}{4} = 1.131 \times 10^{-4} \mathrm{m^2}\) \(A_s = \pi D L = \pi (0.012)(12) = 0.452 \mathrm{m^2}\) \(\dot{m} = \rho V A_f = 1000 \times 0.8 \times 1.131 \times 10^{-4} = 0.09048 \mathrm{kg/s}\) Now plug these results into the conservation of energy equation: \(0.09048 \times 4000 \times (75 - 25) = \dot{q}_s \times 0.452\) \(\dot{q}_s = 6000 \mathrm{W/m^2}\)

Find the surface temperature at the exit

We can use the Nusselt number to find the relationship between the heat transfer rate and the temperature difference between the surface and the fluid: \(Nu = \frac{hD}{k}\) Since the flow is fully developed and Prandtl number is given, we can assume \(Nu = 3.66\) (a constant value for fully developed, constant property laminar flow in a pipe; using Dittus-Boelter or Sieder-Tate equations can give a more accurate result if the variation of properties is considered). Now, we can find the convection heat transfer coefficient, \(h\): \(h = \frac{Nu \times k}{D} = \frac{3.66 \times 0.48}{0.012} = 146.4 \mathrm{W/(m^2K)}\) We can relate the surface heat flux to the surface temperature at the exit and the exit temperature of the fluid: \(\dot{q}_s = h (T_s - T_o)\) \(6000 = 146.4 (T_s - 75)\) \(T_s = 116.04 ^\circ \mathrm{C}\)

Find the pressure loss and the minimum power required to overcome resistance to flow

To find the pressure loss, we use the Darcy-Weisbach equation: \(\Delta P = f \frac{L}{D} \frac{1}{2} \rho V^2\) In this equation, \(f\) is the Darcy friction factor, which can be obtained using the Hagen-Poiseuille equation for laminar flow: \(f = \frac{16}{Re}\) The Reynolds number, \(Re\), can be calculated as follows: \(Re = \frac{D\dot{m}}{\mu A_f} = \frac{0.012 \times 0.09048}{2 \times 10^{-3} \times 1.131 \times 10^{-4}} = 480\) Now, let's find the Darcy friction factor: \(f = \frac{16}{480} = 0.0333\) Finally, we will plug in the given values into the Darcy-Weisbach equation to find the pressure loss: \(\Delta P = 0.0333 \times \frac{12}{0.012} \times \frac{1}{2} \times 1000 \times (0.8)^2 = 1335.2 \mathrm{Pa}\) The minimum power required to overcome the resistance to flow can be calculated as follows: \(P_{min} = \frac{\Delta P \dot{V} A_f}{\rho} = \frac{1335.2 \times 0.8 \times 1.131 \times 10^{-4}}{1000} = 0.1205 \mathrm{W}\) So, the required surface heat flux is \(\dot{q}_s = 6000 \mathrm{W/m^2}\), the surface temperature at the exit is \(T_s = 116.04^\circ \mathrm{C}\), and the pressure loss is \(\Delta P = 1335.2 \mathrm{Pa}\). The minimum power required to overcome the resistance to flow is \(P_{min} = 0.1205 \mathrm{W}\).