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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flashpoint

The flashpoint of a substance, like liquid ethanol, is a critical property in the context of safety and engineering. It is the lowest temperature at which a liquid can form an ignitable vapor mixture in the air. Below this temperature, the concentration of vapor in the air is insufficient to promote combustion.
Understanding the flashpoint is essential, especially in environments where ignition sources are present. At temperatures above the flashpoint, even a small ignition source can trigger a fire or explosion. Therefore, it is vital to maintain the temperature of flammable liquids below their flashpoints to prevent hazardous situations.

• For ethanol, the flashpoint is relatively low at 16.6°C, making it more hazardous in hot environments.
• The temperature must be controlled to stay above room temperature and below the flashpoint in facilities handling ethanol.
• Engineers must ensure that pumping and cooling systems are effectively managing these temperatures.

This knowledge helps create safer industrial processes and reduces the risk of accidents significantly.

Volume Flow Rate

The volume flow rate is an important aspect of fluid dynamics and engineering, representing the volume of fluid moving through a specific section of a pipe, per unit time. In this context, controlling the volume flow rate ensures safe transport of ethanol by keeping the temperature below its flashpoint.
The formula to determine the volume flow rate (\[ Q_v = \frac{\dot{m}}{\rho} \] ), highlights the relationship between mass flow rate and density.
Here’s a breakdown of the variables and their impact:

• \(\dot{m}\) is the mass flow rate, indicating the mass of ethanol passing through the pipe per second.
• \(\rho\) is the density of ethanol, a constant value that affects how the mass converts into volume.
• The result is the volume flow rate \(Q_v\), showing how much cubic meter of ethanol is flowing through the pipe every second.

Engineers must accurately calculate this rate to ensure the ethanol remains below the critical temperature, thus preventing fire hazards.

Energy Balance Equation

The energy balance equation is a fundamental tool in heat and mass transfer, crucial for solving problems like controlling ethanol temperature in transport pipes. It accounts for the heat added and subtracts it from what happens inside the system.
The equation format: \[ Q = \dot{m}c_p(T_{out} - T_{in}) \] helps determine how many energetic resources needed or lost within the system.

• \(Q\) represents the rate of heat transfer into the ethanol, directly influencing the temperature change.
• \(\dot{m}\) indicates the mass flow rate, a variable influencing how much ethanol moves through the system.
• \(c_p\) stands for the specific heat capacity of ethanol, showing how much energy is required to change the temperature of one kilogram by one degree Celsius.
• \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures, respectively, playing a part in maintaining control over the system.

By balancing these variables, engineers can design ways to transport ethanol while maintaining temperatures safely under the threshold, reducing risks of reaching the flashpoint.