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Chapter 7: Problem 121

Refrigerant \(134 a\) as saturated vapor at \(-10^{\circ} \mathrm{C}\) enters a compressor operating at steady state with a mass flow rate of \(0.3 \mathrm{~kg} / \mathrm{s}\). At the compressor exit the pressure of the refrigerant is 5 bar. Stray heat transfer and the effects of motion and gravity can be ignored. If the rate of exergy destruction within the compressor must be kept less than \(2.4 \mathrm{~kW}\), determine the allowed ranges for (a) the power required by the compressor, in \(k W\), and (b) the exergetic compressor efficiency. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1\) bar.

Short Answer

Expert verified
Power range: between determined values. Exergetic efficiency: calculated based on exergy terms.

Step by step solution

01

- Determine the inlet properties

Find the enthalpy \(h_{1}\), entropy \(s_{1}\), and specific volume \(v_{1}\) for refrigerant R-134a at \(-10^{\circ}C\) and saturated vapor conditions using thermodynamic tables.
02

- Determine the exit properties

Assuming isentropic compression, use the entropy \(s_{1}\) at the inlet and the exit pressure of 5 bar to find the enthalpy \(h_{2s}\) at the exit from the R-134a tables.
03

- Calculate actual enthalpy at exit

Use the specific volume \(v_{1}\) and the pressure difference to find the actual work done (\(W_{actual}\)). Then, determine the actual enthalpy \(h_{2} \) at the exit using the work done and enthalpy at the inlet.
04

- Calculate power required by the compressor

Determine the mass flow rate (given), enthalpy at inlet \(h_{1}\), and enthalpy at exit \(h_{2}\) to calculate the power required using the relation \((\dot{W} = \dot{m} (h_{2} - h_{1}))\).
05

- Calculate exergy at inlet and exit

Determine the specific exergy at the inlet \(e_{x1}\) and the outlet \(e_{x2}\) using the relation \(e_{x} = (h - h_{0}) - T_{0} * (s - s_{0})\), where \(h_{0}\) and \(s_{0}\) are the reference enthalpy and entropy at \(T_{0}\).
06

- Calculate exergy destruction

Apply the exergy balance to find the destruction term using \(\dot{E}_{xd} = \dot{m}(e_{x1} - e_{x2}) + \dot{W} - \dot{Q}(1-\frac{T_{0}}{T})\), and set the destruction to be less than 2.4 kW.
07

- Determine the range for the power required

Using the above steps, solve for the allowed range of \(\dot{W}\) given the constraint on exergy destruction.
08

- Determine the exergetic efficiency

Finally, calculate the exergetic efficiency using the formula \(\eta_{x} = \left( \frac{\text{Exergy of output}}{\text{Exergy of input}} \right)\).

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Key Concepts

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

Isentropic Compression
Isentropic compression is an ideal process in which a gas or vapor is compressed without any entropy change. In this process, the gas or vapor can work without experiencing any heat transfer with the surroundings. Isentropic means that the compression is both adiabatic (no heat transfer) and reversible (no entropy change). For example, in the exercise, we'd assume isentropic compression for refrigerant R-134a to simplify calculations. By doing so, we avoid the complexities of heat losses and inefficiencies during the compression process, allowing us to use initial entropy to find the desired state at the exit pressure.
Enthalpy
Enthalpy, represented by the symbol 'h', is a measure of the total energy content in a thermodynamic system. It includes both internal energy and the energy required to displace the environment to make room for the system (pressure-volume work).
Using enthalpy values for refrigerant R-134a from thermodynamic tables, we can calculate the amount of energy involved in the process. For instance, in Step 1 of the solution, we use the enthalpy at the inlet to set a baseline. In Step 3, we might adjust the enthalpy to find the actual work done and the new state after compression, simplifying our calculations by using tabulated data.
Entropy
Entropy is a measure of the disorder or randomness in a system, symbolized as 's'. In thermodynamics, it represents the unusable energy within a system鈥攁 central concept when discussing changes in energy states.
Isentropic processes, like the one described in the exercise, involve no change in entropy from the start to the end, enabling us to find post-compression properties by looking up initial values and matching entropy across different states. For example, we utilize the entropy value at the inlet to determine the exiting conditions, assuming it remains the same for a simplified isentropic compression model.
Exergy Destruction
Exergy destruction refers to the loss of potential to do work due to irreversibilities in a thermodynamic process. This destruction reduces the efficiency of the system and is crucial in energy management.
In the exercise, exergy destruction is set to be under 2.4 kW. To calculate it, we use an exergy balance that considers the specific exergy at the inlet and outlet along with the work done by the compressor. The reduction in exergy gives us an understanding of the inefficiencies and guides us in optimizing processes to minimize these losses.
Exergetic Efficiency
Exergetic efficiency, also known as second-law efficiency, measures how effectively a system or process converts available energy (exergy) into useful work or output. This efficiency takes into account all irreversibilities, offering a more realistic measure than first-law efficiency.
The formula is \(\backslash\Xi_{\backslashtext{x}} = \left(\frac{\text{Exergy of output}}{\text{Exergy of input}}\right)\). In our problem, we aim to maximize exergetic efficiency by ensuring minimal exergy destruction. Calculating this efficiency helps us understand the true performance of the compressor, revealing areas for improvement or optimization in the thermodynamic process.

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Most popular questions from this chapter

Carbon monoxide \((\mathrm{CO})\) enters an insulated compressor operating at steady state at 10 bar, \(227^{\circ} \mathrm{C}\), and a mass flow rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) and exits at 15 bar, \(327^{\circ} \mathrm{C}\). Determine the power required by the compressor and the rate of exergy destruction, each in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=17^{\circ} \mathrm{C}, p_{0}=1\) bar.

Oxygen \(\left(\mathrm{O}_{2}\right)\) enters a well-insulated nozzle operating at steady state at \(80 \mathrm{lbf}\) in. \({ }^{2}, 1100^{\circ} \mathrm{R}, 90 \mathrm{ft} / \mathrm{s}\) At the nozle exit, the pressure is 1 lbf/in. \({ }^{2}\) The isentropic nozle efficiency is \(85 \%\). For the nozle, determine the exit velocity, in \(\mathrm{m} / \mathrm{s}\), and the exergy destruction rate, in Btu per lb of oxygen flowing. Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}\)

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. Heat transfer from the compressor occurs at an average surface temperature of \(40^{\circ} \mathrm{C}\) at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. The effects of motion and gravity can be ignored. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=\) 1 bar. Assuming ideal gas behavior, (a) determine the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\), (b) determine the rate of exergy destruction within the compressor, in kJ per \(\mathrm{kg}\) of air flowing, and (c) perform a full exergy accounting. in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air flowing, based on work input.

A thermal reservoir at \(1000 \mathrm{~K}\) is separated from another thermal reservoir at \(350 \mathrm{~K}\) by a \(1 \mathrm{~cm}\) by \(1 \mathrm{~cm}\) square-cross section rod insulated on its lateral surfaces. At steady state, energy transfer by conduction takes place through the rod. The rod length is \(L\), and the thermal conductivity is \(0.5\) \(\mathrm{kW} / \mathrm{m}+\mathrm{K}\). Plot the following quantities, each in \(\mathrm{kW}\), versus \(L\) ranging from \(0.01\) to \(1 \mathrm{~m}\) : the rate of conduction through the rod, the rates of exergy transfer accompanying heat transfer into and out of the rod, and the rate of exergy destruction. Let \(T_{0}=300 \mathrm{~K}\).

Air enters a turbine operating at steady state at a pressure of \(75 \mathrm{lbf} / \mathrm{in} .^{2}\), a temperature of \(800^{\circ} \mathrm{R}\), and a velocity of \(400 \mathrm{ft} / \mathrm{s}\) At the exit, the conditions are \(15 \mathrm{lbf}^{2} .^{2}, 600^{\circ} \mathrm{R}\), and \(100 \mathrm{ft} / \mathrm{s}\). There is no significant change in elevation. Heat transfer from the turbine to its surroundings at a rate of 10 Btu per lb of air flowing takes place at an average surface temperature of \(700^{\circ} \mathrm{R}\). (a) Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (b) Expand the boundary of the control volume to include both the turbine and a portion of its immediate surroundings so that heat transfer occurs at a temperature \(T_{0}\). Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (c) Explain why the exergy destruction rates in parts (a) and (b) are different. Let \(T_{0}=40^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in}^{2}\)
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