Question

Determine the power input for a compressor that compresses helium from \(110 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) to \(400 \mathrm{kPa}\) and \(200^{\circ} \mathrm{C} .\) Helium enters this compressor through a \(0.1-\mathrm{m}^{2}\) pipe at a velocity of \(9 \mathrm{m} / \mathrm{s}\).

Short answer

Question: Calculate the power input for a compressor compressing helium from an initial pressure of 110 kPa and initial temperature of 20°C to a final pressure of 1000 kPa and final temperature of 200°C, given the area of the entering pipe is 0.1 m² and the velocity of the helium is 9 m/s. Answer: The power input for the compressor can be calculated using the steady flow energy equation, which can be simplified to: \(\dot W_{in} = \dot m(h_2 - h_1)\) for this problem. After finding the mass flow rate, initial and final enthalpies of helium, and performing the calculations, the power input for the compressor can be found in units of kilowatts (kW).

Step-by-step solution

Calculate mass flow rate of helium

We're given the cross-sectional area of the pipe (\(A = 0.1\,\text{m}^2\)) and the velocity of helium (\(V = 9\,\text{m/s}\)). To find the mass flow rate (\(\dot m\)), we can use the equation: \(\dot m = \rho AV\) But first, we need to find the density (\(\rho\)) of helium at the given initial temperature (\(T_1 = 20^\circ\mathrm{C}\)) and pressure (\(P_1 = 110\, \mathrm{kPa}\)). To do this, we can use the ideal gas equation for helium: \(\rho = \frac{P_1}{RT_1}\) Where R is the specific gas constant for helium (\(R = 2.0769\, \mathrm{kJ/kg\cdot K}\)) and \(T_1\) must be converted to Kelvin ( \(T_1 = 20 + 273.15 = 293.15\,\mathrm{K}\)). Now, we can calculate \(\rho\) and then the mass flow rate \(\dot m\).

Compute initial and final enthalpies of helium

To calculate the power input of the compressor, we will use the steady flow energy equation, which in this case, can be simplified to: \(\dot W_{in} = \dot m(h_2-h_1)\) We need to find the initial and final enthalpies (\(h_1\) and \(h_2\)) of the helium. To do that, we can use the ideal gas definition for helium as follows: \(h_1 = C_pT_1\) and \(h_2 = C_pT_2\) Where \(C_p\) is the specific heat at constant pressure for helium (\(C_p = 5.193\, \mathrm{kJ/kg\cdot K}\)). The initial temperature is given as \(T_1 = 293.15\,\mathrm{K}\). The final temperature is \(T_2 = 200 + 273.15 = 473.15\,\mathrm{K}\)

Calculate the power input of the compressor

Now that we have all the variables needed, we can proceed to calculate the power input to the compressor, remember to convert the units accordingly: \(\dot W_{in} = \dot m(h_2 - h_1)\) By substituting the values from the previous steps and converting units, we can calculate the power input \(\dot W_{in}\) for the compressor. After performing the calculations, don't forget to include the proper units for the power input, which is \(\mathrm{kW}\) in this case.

Key concepts

Mass Flow Rate

The mass flow rate is a measure of the amount of mass moving through a particular point per unit time. It's critical in many engineering calculations, especially when dealing with fluid dynamics and thermodynamics. For example, when considering a compressor like in the given exercise, the mass flow rate of helium can directly influence the power input required for the compressor to function. It's calculated using the cross-sectional area through which the gas flows, the velocity of the gas, and the density of the gas at that point. Density, in turn, can be derived using the ideal gas law when the gas behaves ideally, which states that the density of a gas is proportional to its pressure and inversely proportional to its temperature (when measured in absolute units like Kelvin).The formula given in the solution, \( \dot m = \rho AV \) combines these variables to determine how much mass of helium flows through the pipe every second. This value is crucial for assessing the power requirements for the compressor.

Steady Flow Energy Equation

The steady flow energy equation is pivotal in understanding energy changes within a flow process that operates without accumulation or depletion of mass and energy over time. This equation provides a framework for energy balance, which is essential in processes such as compression in turbines, pumps, and compressors.In relation to compressors, this energy equation essentially states that the power input (or work done) on the fluid is equal to the change in enthalpy of the fluid as it flows through the compressor. The simplified form of the equation, provided in the exercise, \( \dot W_{in} = \dot m(h_2-h_1) \) showcases how the power input, \( \dot W_{in} \) is a function of mass flow rate, \( \dot m \) and the change in enthalpy from the inlet (\( h_1 \) ) to the outlet (\( h_2 \) ).Understanding this relationship helps us realize that the thermodynamic properties of the fluid and its flow rate are directly tied to the operational performance of the compressor.

Enthalpy of Helium

Enthalpy is a property of a substance that describes the total heat content. For ideal gases like helium, enthalpy is mainly dependent on temperature and can be calculated using the specific heat capacity at constant pressure. We use the formula \( h = C_pT \) where \( C_p \) is the specific heat at constant pressure, and \( T \) is the temperature in Kelvin.In the exercise, the enthalpy change of helium, from its initial state to its final state as it passes through the compressor, is essential for understanding the energy transformation occurring within the compressor. With known specific heat capacity and the temperatures at initial and final states, you can calculate the enthalpies, \( h_1 \) and \( h_2 \) respectively. This calculation forms the basis for determining the compressor's power requirement through the enthalpy change.Hence, knowing the enthalpy values at various states of helium allows for a comprehensive analysis of the thermodynamic cycle it undergoes, which is a fundamental concept for any process involving heat and work interactions within fluids.