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Chapter 4: Problem 71

Refrigerant \(134 \mathrm{a}\) at a flow rate of \(0.5 \mathrm{lb} / \mathrm{s}\) enters a heat exchanger in a refrigeration system operating at steady state as saturated liquid at \(0^{\circ} \mathrm{F}\) and exits at \(20^{\circ} \mathrm{F}\) at a pressure of \(20 \mathrm{lbf} / \mathrm{in}^{2}\) A separate air stream passes in counterflow to the Refrigerant \(134 \mathrm{a}\) stream, entering at \(120^{\circ} \mathrm{F}\) and exiting at \(77^{\circ} \mathrm{F}\). The outside of the heat exchanger is well insulated. Neglecting kinetic and potential energy effects and modeling the air as an ideal gas, determine the mass flow rate of air,

Short Answer

Expert verified
Mass flow rate of air is calculated using the energy balance equation and specific heat capacity Cp.

Step by step solution

01

- Analyze Refrigerant 134a Properties

Determine the enthalpies of Refrigerant 134a at the inlet and outlet conditions using the refrigerant tables. Since the refrigerant enters as a saturated liquid at 0掳F and leaves at 20掳F with a pressure of 20 lbf/in虏, first find the enthalpy of saturated liquid at 0掳F, denoted as h1. For the outlet, identify the enthalpy h2 at 20掳F and 20 lbf/in虏.
02

- Compute Energy Balance for Refrigerant 134a

Use the steady-flow energy equation for the refrigerant stream: en_{in} - en_{out} = 0given by Q_{in} - W_{out} = -m_{ref} (h2 - h1), where m_{ref} is the mass flow rate of the refrigerant. Since there is no work transfer (W_out = 0), it simplifies to: Q = m_{ref} (h1 - h2).
03

- Determine Enthalpies for Air Stream

Using the given temperatures for the air stream, calculate the enthalpies h4 (120掳F) and h3 (77掳F) using specific heat capacity (Cp) of air, with Cp assumed to be constant for ideal gas air at these temperatures. Use the formula: h = Cp * (T2 - T1) to find enthalpy changes.
04

- Energy Balance Equation for Air Stream

Apply energy balance on the air stream. Use the relation: m_{air} cp (T4 - T3) = m_{ref} (h1 - h2), where T4 and T3 are air temperatures, m_{air} is the mass flow rate of the air, and cp is the specific heat capacity of air.
05

- Solve for Mass Flow Rate of Air

Rearrange the energy balance equation to solve for the mass flow rate of the air:m_{air} = m_{ref} (h1 - h2) / (cp (T4 - T3)). Substitute the known values in the equation to complete the calculation.

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Key Concepts

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

enthalpy
Enthalpy is a measurement of energy in a thermodynamic system. It includes both the internal energy and the product of the pressure and volume of the system. Essentially, it represents the total heat content.
When working with refrigerants, like Refrigerant 134a in the exercise, enthalpy changes are crucial in determining energy transfers. In the given exercise, you need to find the enthalpy at two states: the inlet and outlet of the refrigerant.
For example:
  • **h1**: Enthalpy of Refrigerant 134a entering the heat exchanger as a saturated liquid at 0掳F.
  • **h2**: Enthalpy after leaving the exchanger at 20掳F and a pressure of 20 lbf/in虏.

Using refrigerant tables, you can find these enthalpies, which are vital for solving energy balance equations.
steady-flow energy equation
The steady-flow energy equation is used to analyze systems where fluid flows continuously. In steady-state physical systems, conditions remain constant over time.
The steady-flow energy equation for the refrigerant stream in the exercise can be written as:
  • \( Q_{in} - W_{out} = m_{ref} (h2 - h1) \)


Here, *Q* is the heat transfer, *W* is the work done, and *m_{ref}* is the mass flow rate of the refrigerant.
In our problem, work transfer is zero (\(W_{out}=0\)) since no work is being done. From our simplified energy equation:
  • \( Q = m_{ref} (h1 - h2) \)

This equation helps in determining the energy exchange between the refrigerant and the air stream.
specific heat capacity
Specific heat capacity (\( cp \)) is the amount of heat required to change the temperature of a unit mass of a substance by one degree. For gases, it is assumed constant over reasonable temperature ranges.
In the workout of the exercise, you'll use the specific heat capacity of air to find the enthalpy changes with this formula:
  • \( h = cp * (T4 - T3) \)

Where:\( h \) is the enthalpy, \( cp \) is the specific heat capacity of air, \( T4 \) and \( T3 \) are the outlet and inlet temperatures of air respectively.
These enthalpies are crucial for the energy balance on the air stream, which eventually helps solve for the mass flow rate.

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Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).

Steam enters a counterflow heat exchanger operating at steady state at \(0.07 \mathrm{MPa}\) with a specific enthalpy of \(2431.6 \mathrm{~kJ} / \mathrm{kg}\) and exits at the same pressure as saturated liquid. The steam mass flow rate is \(1.5 \mathrm{~kg} / \mathrm{min}\). A separate stream of air with a mass flow rate of \(100 \mathrm{~kg} / \mathrm{min}\) enters at \(30^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The ideal gas model with \(c_{p}=\) \(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) can be assumed for air. Kinetic and potential energy effects are negligible. Determine (a) the quality of the entering steam and (b) the rate of heat transfer between the heat exchanger and its surroundings, in \(\mathrm{kW}\).

Steam enters a well-insulated turbine operating at steady state at \(4 \mathrm{MPa}\) with a specific enthalpy of \(3015.4 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(10 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.07 \mathrm{MPa}\), specific enthalpy is \(2431.7 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(90 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(11.95 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in \(\mathrm{kW}\).

Oil enters a counterflow heat exchanger at \(450 \mathrm{~K}\) with a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\) and exits at \(350 \mathrm{~K}\). A separate stream of liquid water enters at \(20^{\circ} \mathrm{C}, 5\) bar. Each stream experiences no significant change in pressure. Stray heat transfer with the surroundings of the heat exchanger and kinetic and potential energy effects can be ignored. The specific heat of the oil is constant, \(c=2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). If the designer wants to ensure no water vapor is present in the exiting water stream, what is the allowed range of mass flow rates for the water, in \(\mathrm{kg} / \mathrm{s}\) ?

Refrigerant \(134 a\) enters an insulated compressor operating at steady state as saturated vapor at \(-20^{\circ} \mathrm{C}\) with a mass flow rate of \(1.2 \mathrm{~kg} / \mathrm{s}\). Refrigerant exits at 7 bar, \(70^{\circ} \mathrm{C}\). Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in \(\mathrm{m}^{3} / \mathrm{s}\), and (b) the power input to the compressor, in \(\mathrm{kW}\).
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