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Chapter 4: Problem 62

Air, modeled as an ideal gas, is compressed at steady state from \(1 \mathrm{bar}, 300 \mathrm{~K}\), to \(5 \mathrm{bar}, 500 \mathrm{~K}\), with \(150 \mathrm{~kW}\) of power input. Heat transfer occurs at a rate of \(20 \mathrm{~kW}\) from the air to cooling water circulating in a water jacket enclosing the compressor. Neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{s}\).

Short Answer

Expert verified
The mass flow rate of the air is approximately 0.646 kg/s.

Step by step solution

01

- Write the Given Information

Initial pressure, \(P_1 = 1 \text{bar}\), initial temperature, \(T_1 = 300 \text{K}\). \Final pressure, \(P_2 = 5 \text{bar}\), final temperature, \(T_2 = 500 \text{K}\). \Power input, \(W = 150 \text{kW}\).\Heat transfer rate, \(Q = 20 \text{kW}\).
02

- Use the Steady Flow Energy Equation (SFEE)

Considering the assumptions provided, the Steady Flow Energy Equation (SFEE) simplifies to: \ \[\begin{equation}\ \dot{m} (\tilde{h}_2 - \tilde{h}_1) = Q - W \ \text{ where \ \tilde{h}_ is the specific enthalpy.} \end{equation}\]\
03

- Calculate the Change in Specific Enthalpy

For ideal gases, \ \[\begin{equation}\ \tilde{h} = c_p T \ \text{ where \ c_p is the specific heat capacity at constant pressure.} \end{equation}\]\For air, \ c_p \ is approximately \ 1.005 \text{kJ/kg K} \. Hence, \ \Delta{h} = 1.005 \text{(500 - 300)} \ = 201 \text{kJ/kg}\.
04

- Rearrange the SFEE to Solve for Mass Flow Rate

Rearrange to solve for \dot{m}\. \ \[\begin{equation}\ \dot{m} = \frac{W - Q}{\tilde{h}_2 - \tilde{h}_1} = \frac{150 - 20}{201} \ = \ 0.646 \text{kg/s}\ \end{equation}\]\.

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Key Concepts

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

Ideal Gas
In this exercise, air is treated as an ideal gas. The ideal gas assumption simplifies our equations and makes solving this problem more straightforward.
For an ideal gas, the relationship between pressure, volume, and temperature is given by the Ideal Gas Law:
\[ PV = nRT \]
Where:
  • \(P\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the gas constant
  • \(T\) is the temperature

Recognizing air as an ideal gas allows us to use specific properties like specific enthalpy to analyze energy changes.
Specific Enthalpy
In the context of this problem, specific enthalpy ( \(\tilde{h}\) is a critical parameter. It represents the total heat content of the system per unit mass.
For an ideal gas, specific enthalpy is determined using the equation:
\[ \tilde{h} = c_p T \]
Where:
  • \(c_p\) is the specific heat capacity at constant pressure
  • \(T\) is the absolute temperature

Given air鈥檚 specific heat \(c_p\) holds almost constant at about 1.005 kJ/kg K. In our exercise, the change in specific enthalpy (\Delta{h}) was found by calculating the difference between the initial and final state temperatures.
Mass Flow Rate
The goal of the exercise is to find the mass flow rate of the air compressed in the system. The mass flow rate ( \dot{m}) is the amount of mass passing through a section per unit time.
Using the Steady Flow Energy Equation (SFEE), which summarizes the energy changes in a steady-flow process, we have:
\[ \dot{m} ( \tilde{h}_2 - \tilde{h}_1 ) = Q - W \]
By rearranging the equation, we compute:
\[ \dot{m} = frac{W - Q}{ \tilde{h}_2 - \tilde{h}_1} \]
This incorporates the effects of power input and heat transfer, helping us determine the mass flow rate accurately.
Heat Transfer
Heat transfer in this problem occurs from the air to the cooling water around the compressor. The rate of heat transfer ( Q) is 20 kW, acting to remove heat from the air as it compresses.
Heat transfer plays a crucial role because it affects the energy balance of the system.
When analyzing the thermodynamic process, always consider the direction of heat transfer. Here, heat is leaving the air, which impacts the energy balance captured in the SFEE.
This decrease in energy represented by heat loss is subtracted in the equation to find out how much work is converted into increasing air pressure and temperature.
Power Input
Power input ( W) is provided to the compressor in the system to do work on the air. In this problem, 150 kW of power is added to compress the air.
Power input represents work done per unit of time. This energy helps increase the air pressure from 1 bar to 5 bars and raise its temperature from 300 K to 500 K.
The energy balance presenting how the supplied power transforms the air's properties is simplified in the SFEE.
Understanding power input and how it interacts with other energy forms like heat and enthalpy can help determine efficient compressor operations.

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Most popular questions from this chapter

Figure P4.8 shows a mixing tank initially containing \(2000 \mathrm{lb}\) of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of \(0.8 \mathrm{lb} / \mathrm{s}\) and the other delivering cold water at a mass flow rate of \(1.2 \mathrm{lb} / \mathrm{s}\). Water exits through a single exit pipe at a mass flow rate of \(2.5 \mathrm{lb} / \mathrm{s}\). Determine the amount of water, in lb, in the tank after one hour.

Air expands through a turbine operating at steady state. At the inlet, \(p_{1}=150 \mathrm{lbf} / \mathrm{in}^{2}, T_{1}=1400^{\circ} \mathrm{R}\), and at the exit, \(p_{2}=14.8 \mathrm{lbf} / \mathrm{in.} .^{2}, T_{2}=700^{\circ} \mathrm{R}\). The mass flow rate of air entering the turbine is \(11 \mathrm{lb} / \mathrm{s}\), and \(65,000 \mathrm{Btu} / \mathrm{h}\) of energy is rejected by heat transfer. Neglecting kinetic and potential energy effects, determine the power developed, in hp.

Figure P4.101 shows a pumped-hydro energy storage system delivering water at steady state from a lower reservoir to an upper reservoir using off-peak electricity (see Sec. 4.8.3). Water is delivered to the upper reservoir at a volumetric flow rate of \(150 \mathrm{~m}^{3} / \mathrm{s}\) with an increase in elevation of \(20 \mathrm{~m}\). There is no significant change in temperature, pressure, or kinetic energy from inlet to exit. Heat transfer from the pump to its surroundings occurs at a rate of \(0.6 \mathrm{MW}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the pump power required, in MW. Assuming the same volumetric flow rate when the system generates on-peak electricity using this water, will the power be greater, less, or the same as the pump power? Explain.

Air with a mass flow rate of \(2.3 \mathrm{~kg} / \mathrm{s}\) enters a horizontal nozzle operating at steady state at \(450 \mathrm{~K}, 350 \mathrm{kPa}\), and velocity of \(3 \mathrm{~m} / \mathrm{s}\). At the exit, the temperature is \(300 \mathrm{~K}\) and the velocity is \(460 \mathrm{~m} / \mathrm{s}\). Using the ideal gas model for air with constant \(c_{p}=1.011 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the area at the inlet, in \(\mathrm{m}^{2}\). (b) the heat transfer between the nozzle at its surroundings, in kW. Specify whether the heat transfer is to or from the air.

A rigid, well-insulated tank of volume \(0.9 \mathrm{~m}^{3}\) is initially evacuated. At time \(t=0\), air from the surroundings at 1 bar, \(27^{\circ} \mathrm{C}\) begins to flow into the tank. An electric resistor transfers energy to the air in the tank at a constant rate for 5 minutes, after which time the pressure in the tank is 1 bar and the temperature is \(457^{\circ} \mathrm{C}\). Modeling air as an ideal gas, determine the power input to the tank, in \(\mathrm{kW}\).
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