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Refrigerant-134a enters a compressor as a saturated vapor at \(160 \mathrm{kPa}\) at a rate of \(0.03 \mathrm{m}^{3} / \mathrm{s}\) and leaves at 800 kPa. The power input to the compressor is \(10 \mathrm{kW}\). If the surroundings at \(20^{\circ} \mathrm{C}\) experience an entropy increase of \(0.008 \mathrm{kW} / \mathrm{K},\) determine \((a)\) the rate of heat loss from the compressor, \((b)\) the exit temperature of the refrigerant, and \((c)\) the rate of entropy generation.

Short Answer

Expert verified
Answer: The key steps are: 1. Calculate the mass flow rate using the volumetric flow rate and specific volume at the inlet. 2. Use the energy balance equation to find the exit temperature and rate of heat loss. 3. Apply the entropy balance equation to determine the rate of entropy generation. 4. Evaluate specific enthalpy and specific entropy properties of the refrigerant at the exit. 5. Calculate the refrigerant's exit temperature. 6. Determine the rate of heat loss from the compressor using the energy balance equation. 7. Calculate the rate of entropy generation using the entropy balance equation and given entropy increase of the surroundings.

Step by step solution

01

Mass flow rate calculation

Calculate the mass flow rate using the given volumetric flow rate and the specific volume of the refrigerant at the inlet. We'll use the inlet pressure to find the specific volume for the saturated vapor. Then, we can use the mass flow rate for further calculations.
02

Energy balance equation

Use the energy balance equation to find the exit temperature and the rate of heat loss from the compressor. To do this, we'll use the given power input and the properties of the refrigerant. We can then rearrange the energy balance equation to solve for the rate of heat loss.
03

Entropy balance equation

Utilize the entropy balance equation to find the rate of entropy generation. We are given the value of the surroundings' entropy increase, which can be used along with the entropy change for the refrigerant to find the rate of entropy generation.
04

Evaluate properties

To find the exit temperature of the refrigerant, we'll need to evaluate its properties, such as specific enthalpy and specific entropy. We can find these properties using the pressure at the exit of the compressor and the conditions for a saturated vapor.
05

Calculate the exit temperature

Using the specific enthalpy and specific entropy values found in Step 4, we can determine the exit temperature of the refrigerant.
06

Calculate the rate of heat loss

Now that we have the mass flow rate, specific enthalpy values, and power input, we can substitute them into the energy balance equation from Step 2 to find the rate of heat loss from the compressor.
07

Calculate the rate of entropy generation

Finally, we will use the mass flow rate, specific entropy values, and the given entropy increase of the surroundings to calculate the rate of entropy generation by substituting these values into the entropy balance equation from Step 3.

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Most popular questions from this chapter

The energy used to compress air in the United States is estimated to exceed one-half quadrillion \(\left(0.5 \times 10^{15}\right)\) kJ per year. It is also estimated that 10 to 40 percent of the compressed air is lost through leaks. Assuming, on average, 20 percent of the compressed air is lost through air leaks and the unit cost of electricity is \(\$ 0.13 / \mathrm{kWh}\), determine the amount and cost of electricity wasted per year due to air leaks.

The temperature of an ideal gas having constant specific heats is given as a function of specific entropy and specific volume as \(T(s, v)=A v^{1-k} \exp \left(s / c_{v}\right)\) where \(A\) is a constant. For a reversible, constant volume process, find the expression for heat transfer per unit mass as a function of \(c_{v}\) and \(T\) using \(Q=\int T d S .\) Compare this result with that obtained by applying the first law to a closed system undergoing a constant volume process.

A unit mass of an ideal gas at temperature \(T\) undergoes a reversible isothermal process from pressure \(P_{1}\) to pressure \(P_{2}\) while losing heat to the surroundings at temperature \(T\) in the amount of \(q .\) If the gas constant of the gas is \(R,\) the entropy change of the gas \(\Delta s\) during this process is \((a) \Delta s=R \ln \left(P_{2} / P_{1}\right)\) \((b) \Delta s=R \ln \left(P_{2} / P_{1}\right)-q / T\) \((c) \Delta s=R \ln \left(P_{1} / P_{2}\right)\) \((d) \Delta s=R \ln \left(P_{1} / P_{2}\right)-q / T\) \((e) \Delta s=0\)

A refrigerator with a coefficient of performance of 4 transfers heat from a cold region at \(-20^{\circ} \mathrm{C}\) to a hot region at \(30^{\circ} \mathrm{C}\). Calculate the total entropy change of the regions when \(1 \mathrm{kJ}\) of heat is transferred from the cold region. Is the second law satisfied? Will this refrigerator still satisfy the second law if its coefficient of performance is \(6 ?\)

Steam expands in an adiabatic turbine from \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) to \(0.1 \mathrm{MPa}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\). If steam leaves the turbine as saturated vapor, the power output of the turbine is \((a) 2058 \mathrm{kW}\) \((b) 1910 \mathrm{kW}\) \((c) 1780 \mathrm{kW}\) \((d) 1674 \mathrm{kW}\) \((e) 1542 \mathrm{kW}\)

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