Question

In a chemical process plant, liquid isobutane at \(50^{\circ} \mathrm{F}\) is being transported through a 30 -ft-long standard \(3 / 4\)-in Schedule 40 cast iron pipe with a mass flow rate of \(0.4 \mathrm{lbm} / \mathrm{s}\). Accuracy of the results is an important issue in this problem; therefore, use the most appropriate equation to determine (a) the pressure loss and \((b)\) the pumping power required to overcome the pressure loss. Assume flow is fully developed. Is this a good assumption?

Short answer

Answer: The pressure loss is 1.791 atm. The pumping power required to overcome the pressure loss is given by \(P = \frac{17908\times3.185\times10^{-4}}{\eta}\,\mathrm{W}\), where \(\eta\) is the pump efficiency. It is reasonable to assume that the flow is fully developed in this case.

Step-by-step solution

Calculate the properties of the fluid

First, we need to know the properties of isobutane at \(50^{\circ}F\) (approximately \(10^{\circ}C\)). Using a reference table or an online resource, we can obtain the following properties: - Density (\(\rho\)): \(551.6\,\mathrm{kg/m^3}\) - Dynamic viscosity (\(\mu\)): \(2.65\,\times 10^{-4}\,\mathrm{Pa\,s}\) Note that these values may vary slightly depending on the reference sources and their interpolation methods.

Calculate the velocity of the flow

Use the mass flow rate and fluid properties to calculate the flow velocity, using the formula: $$ v = \frac{\dot{m}}{A\rho} $$ where \(\dot{m}=0.4\,\mathrm{lbm/s}\) is the mass flow rate, \(A\) is the cross-sectional area of the pipe, and \(\rho=551.6\,\mathrm{kg/m^3}\) is the density of the fluid. We need to convert the mass flow rate from lbm/s to kg/s by using a conversion factor of \(\frac{1\,\mathrm{lbm}}{2.205\,\mathrm{kg}}\): $$ \dot{m} = 0.4\,\mathrm{lbm/s} \times \frac{1\,\mathrm{kg}}{2.205\,\mathrm{lbm}} = 0.1814\,\mathrm{kg/s} $$ Next, we need to find the cross-sectional area (\(A\)) of the pipe. The Schedule 40 cast iron pipe has an internal diameter (\(D\)) of \(3/4\)-in which converts to \(0.01905\,\mathrm{m}\). Thus: $$ A = \frac{\pi D^2}{4} = \frac{\pi(0.01905)^2}{4} = 2.846\times10^{-4}\,\mathrm{m^2} $$ Now, we calculate the flow velocity: $$ v = \frac{0.1814}{2.846\times 10^{-4}\times 551.6} = 1.119\,\mathrm{m/s} $$

Calculate the Reynolds number

Now we will calculate the Reynolds number (\(Re\)) to determine the type of flow (laminar or turbulent) by using the formula: $$ Re=\frac{\rho vD}{\mu} $$ where \(v=1.119\,\mathrm{m/s}\), \(D=0.01905\,\mathrm{m}\), and \(\mu=2.65\times10^{-4}\,\mathrm{Pa\,s}\). Thus: $$ Re=\frac{551.6\times1.119\times0.01905}{2.65\times10^{-4}}=46455 $$ Since \(Re>4000\), we conclude that the flow is turbulent.

Calculate the pressure loss

Using the Darcy-Weisbach equation for pressure loss in turbulent flow, we get: $$ \Delta P = f \frac{L}{D} \frac{1}{2} \rho v^2 $$ First, we need to find the friction factor (\(f\)) using the Moody chart and the relative roughness (\(\epsilon/D\)) of the pipe. For a cast iron pipe, \(\epsilon = 0.00026\,\mathrm{m}\). Thus: $$ \frac{\epsilon}{D} = \frac{0.00026}{0.01905} = 0.01365 $$ Using the Moody chart with \(Re=46455\) and \(\frac{\epsilon}{D}=0.01365\), we estimate \(f\approx0.023\). Also, the pipe is \(30\,\mathrm{ft}\) long, which is \(9.144\,\mathrm{m}\). Now we can calculate the pressure loss: $$ \Delta P = 0.023 \times \frac{9.144}{0.01905} \times \frac{1}{2} \times 551.6 \times (1.119)^2 = 17908\,\textrm{Pa}=1.791\,\textrm{atm} $$

Calculate the pumping power

Now we will calculate the pumping power required to overcome the pressure loss using the equation: $$ P = \Delta P \times \frac{\dot{V}}{\eta} $$ where \(\dot{V} = A\times v\) is the volume flow rate, and \(\eta\) is the pump efficiency. Since the pump efficiency is not given, we cannot find the exact value of the required pumping power. However, we can express the pumping power (\(P\)) in terms of the pump efficiency: $$ \dot{V} = 2.846\times10^{-4}\,\mathrm{m^2} \times 1.119\,\mathrm{m/s} = 3.185\times10^{-4}\,\mathrm{m^3/s} $$ So, $$ P = \frac{17908\times3.185\times10^{-4}}{\eta}\,\mathrm{W} $$ To answer the last part of the question about whether the assumption of fully developed flow is good: the flow becomes fully developed at some distance from the entrance of the pipe. This distance is usually about \(10D\) for turbulent flow. In our case, \(10D = 10\times0.01905 = 0.1905\,\mathrm{m}\). Since the pipe is \(9.144\,\mathrm{m}\) long, we can safely assume that the flow is fully developed in this case.