Liquid ethanol is a flammable fluid and can release vapors that form explosive mixtures at temperatures above its flashpoint at \(16.6^{\circ} \mathrm{C}\). In a chemical plant, liquid ethanol \(\left(c_{p}=2.44 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, \rho=789 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is being transported in a pipe with an inside diameter of \(5 \mathrm{~cm}\). The pipe is located in a hot area with the presence of ignition source, where an estimated \(20 \mathrm{~kW}\) of heat is added to the ethanol. Your task, as an engineer, is to design a pumping system to transport the ethanol safely and to prevent fire hazard. If the inlet temperature of the ethanol is \(10^{\circ} \mathrm{C}\), determine the volume flow rate that is necessary to keep the temperature of the ethanol in the pipe below its flashpoint.

Step by step solution

01

Calculate the mass flow rate

In order to find the volume flow rate, we need to determine the mass flow rate first. We'll use the energy balance equation to do this: $$ Q = m\dot c_p(T_{out} - T_{in}) $$ We need to solve for \(m\dot\), the mass flow rate. In this case, the temperature at the inlet, \(T_{in}\) is \(10^{\circ} \mathrm{C}\). The temperature we want to stay below is the flashpoint, which is \(16.6^{\circ} \mathrm{C}\). Therefore, we set \(T_{out} = 16.6^{\circ} \mathrm{C}\). We are also given the heat added, \(Q = 20 \text{ kW}\), and the heat capacity, \(c_p = 2.44 \text{ kJ/kg} \cdot \text{K}\). Plugging in the values, we have: $$ 20\text{ kW} = m\dot\cdot2.44\text{ kJ/kg K}(16.6-10) $$

02

Solve for the mass flow rate

Next, we solve for the mass flow rate \(m\dot\): $$ m\dot = \frac{20\text{ kW}}{2.44\text{kJ/kg K}(16.6-10)} $$ Calculating the value, we obtain: $$ m\dot = 14.077\text{ kg/s} $$

03

Calculate the volume flow rate

Now that we have the mass flow rate, we can find the volume flow rate using the density of ethanol, which is given as \(\rho=789 \text{ kg/m}^3\). The volume flow rate, \(Q_v\), can be found using the formula: $$ Q_v = \frac{m\dot}{\rho} $$ Plugging in the values, we have: $$ Q_v = \frac{14.077\text{ kg/s}}{789 \text{ kg/m}^3} $$

04

Solve for the volume flow rate

Finally, we can solve for the volume flow rate: $$ Q_v = 0.01784 \text{ m}^3\text{/s} $$ The necessary volume flow rate to keep the temperature of ethanol in the pipe below its flashpoint is approximately \(0.01784 \text{ m}^3\text{/s}\).

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