Homework Problems

7B-1 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-2 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-3 :

Entropy Changes Associated with the Evaporator of a Refrigerator

4 pts

A refrigerator using R-134a as the refrigerant removes 200 kJ of heat from its refrigerated space. This heat is absorbed by a mixture of saturated liquid and saturated vapor that enters the evaporator at a pressure of 175 kPa. The R-134a leaves the evaporator as a saturated vapor. Determine...
a.) ... the entropy change of the R-134a,
b.) ... the entropy change of the refrigerated space assuming its temperature remains constant at -10°C,
c.) ... the entropy change of the universe for this process.

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Because the R-134a enters the evaporator as a saturated mixture at 175 kPa and leaves as a saturated vapor at 175 kPa, the temperature remains constant at Tsat. Consequently, you can calculate the entropy change of the refrigerant directly from the definition of entropy.

The saturation temperature of R-134a at 175 kPa is -13.4oC.

The entropy change of the universe is the sum of the entropy change of the R-134a and of the refrigerated space.

ΔSuniv = 0.010 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-4 :

Power Output of an Isentropic Turbine

4 pts

An isentropic turbine produces shaft work as R-134a at 275 psia and 225°F expands through the turbine and exits at 35°F. Assuming the turbine is adiabatic, determine the power output of the turbine per lbm of R-134a.

Given values of two intensive properties at the inlet, T and P, you can look up values for Hin and Sin in thermodynamic tables for R-134a.

The turbine is isentropic, so Sout = Sin.

Knowing the values of two intensive properties at the outlet, P and Sout, you can look up the value of Hout.

Ws = 32.2 Btu/lbm.

7B-5 :

Entropy Change for an Isobaric Heating Process

4 pts

A piston-and-cylinder device holds 9.7 lbm of water at 250 psia in a volume of 3.6 ft3. Heat is added to the water until the temperature reaches 425oF. During this process, the pressure in the cylinder remains constant because the piston moves freely. Determine the change in the entropy on the water in Btu/oR.

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You know the values of two intensive properties for both the initial and final states: P for both states and specific V for the initial state and T for the final state. Use them to determine all the other important properties for each state, such as x and S.

The quality at the initial condition is 0.193 lbm vap/lbm.

The specific entropy at the final condition is 1.545 Btu/lbm-oR.

ΔSH2O = 7.69 Btu/oR.

7B-6 :

Isentropic Compression of Water Vapor

3 pts

An isentropic compressor takes in H2O at 105 kPa and delivers it at 450 kPa. If the temperature of the feed is 180°C, determine the temperature and specific enthalpy of the H2O at the compressor outlet.

Because you know the values of two intensive variables for the water entering the compressor, the state is completely determined and you can look up the values of any other intensive variables. In this case, the variable of interest is the entropy.

The specific entropy of the water entering the compressor is 7.7251 kJ/kg-K. (Reference state: S=0 kJ/kg-K at the triple point, 0.01°C, 0.61173 kPa).

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

At the outlet of the compressor, the specific enthalpy of the water is 3195.4 kJ/kg. (Reference state: U=0 kJ/kg at the triple point, 0.01°C, 0.61173 kPa).

7B-7 :

Isentropic Expansion of R-134a

4 pts

R-134a expands reversibly and adiabatically in a piston-and-cylinder device from 80 psia and 120°F to 5 psia. If the cylinder contains 17.4 lbm of R-134a, what are the total work done by the system during this process and final temperature of the R-134a ?

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Because you know the values of two intensive variables for the water entering the compressor, the state is completely determined and you can look up the values of any other intensive variables. In this case, the variable of interest is the entropy.

Because the expansion process is both reversible and adiabatic, the process is isentropic:
Sout = Sin.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

The total work done by the R-134a on the surroundings during this process is Wtotal = -413 Btu.

7B-8 :

Vaporization of Water in a Rigid Tank

3 pts

An electric resistance heater is used to heat the contents of an insulated rigid tank that holds 23.1 kg of H2O. Initially, the pressure in the tank is 150 kPa and 44% of the mass inside the tank is saturated liquid and the remaining 56% is saturated vapor. Power is applied to the heater until all of the H2O in the tank is saturated vapor. Determine the entropy change of the H2O and the total electrical work done by the water in the tank during this process.

Because you know the values of two intensive variables for the water entering the compressor, the state is completely determined and you can look up the values of any other intensive variables. In this case, the variables of interest are V, S and U.

Because the tank is rigid, the specific volume of the water in the tank remains constant. This gives you the value of a second intensive variable for the water in the final state. So, you can determine all of the other properties of water in the final state, especially S and U.

Because the tank is insulated, Q = 0. This allows you to determine the electric work required for this process.

Welec = -21,350 kJ and ΔSH2O = 54.06 kJ/K.

7B-9 :

Expansion of Water into a Vacuum

5 pts

A rigid tank is divided into two equal parts by a wall. One part of the tank contains 0.74 kg of water at 100 kPa and 25°C. The other part is a perfect vacuum. The wall is now removed and the water expands to fill the entire tank. A thermocouple indicates that the final equilibrium temperature of the H2O in the tank is 50°C. Determine the entropy change of water and the total heat transfer to the system during this process.

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Because you know the values of two intensive variables for the water in its initial state, the initial state is completely determined and you can look up the values of any other intensive variables. In this case, the variables of interest are V, S and U.

The specific volume of the water in the final state is twice the specific volume of the water in the initial state because the water initially filled half the tank and in the final state, it fills the whole tank. V2 = 2 V1.

Consider the water to be your system. Because the water is expanding against a perfect vacuum, the restraining force is zero! As a result, no work is done by the gas as it expands.

Q = 77.9 kJ and ΔSH2O = 0.250 kJ/K.

7B-10 :

Isobaric Condensation of R-134a

5 pts

The refrigerant R-134a is cooled in a piston-and-cylinder device from 190°F to 75°F. The piston moves freely and the initial pressure is 200 psia. Determine the specific entropy change of the refrigerant, the work per lbm and heat transfer per lbm during this process.

Because you know the values of two intensive variables for the R-134a in both its initial and final states, both states are completely determined and you can look up the values of any other intensive variables for each state. In this case, the variables of interest are V, S and U.

Because this process is isobaric, you can easily use the defining equation for direct evaluation for boundary work.

Once you have evaluated the boundary work, you can apply the 1st Law for closed systems to determine Q, assuming that boundary work is the only form of work that crosses the system boundary.

ΔSR134a = -0.171 Btu/lbm-°R. , Q = -10.2 Btu/lbm and Q = -99.9 Btu/lbm

7B-11 :

Entropy Change and Boundary Work for the Isobaric Expansion of Water

10 pts

Electrical work is done by an electrical resistance heater on H2O contained in an insulated piston-and-cylinder device. The cylinder initially contains 2.4 L of saturated liquid water at a pressure of 500 kPa. Assuming the piston moves freely, determine the total entropy change of the H2O during a process in which 1875 kJ of heat is transferred into the water. What is the boundary work for this process in kJ? (Hint: The solution is iterative. I strongly suggest you use Excel and its Solver add-in along with the SpreadsheetWorld Thermal-Fluid Properties add-in to solve this problem. Click here to get the add-in. Click here to download an Excel workbook that will show you how to do this.

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Because you know the values of two intensive variables for the water in its initial state, the initial state is completely determined and you can look up the values of any other intensive variables. In this case, the variables of interest are V, S and U.

This problem is very challenging. You must repetitively guess values of U2. For each guessed value of U2, you can determine V2 from the steam tables and Wb from the defining equation for boundary work. You can then test to see whether the left-hand side of the 1st Law equation is equal to the right hand side. That is, check to see if -(Welec + Wb) is equal to U2 - U1.

Excel, Solver and the SpreadsheetWorld Thermal-Fluid Properties add-in yielded U2 = 1025.84 kJ/kg. You can than use two intensive variables for the water in its final state, P and U, to look up the values of any other intensive variables. In this case, the variables of interest are V and S.

ΔSwater = 2.19 kJ/K and Wb = 82.6 kJ.

7B-12 :

Reversible Expansion of R-134a in a Piston-and-Cylinder Device

4 pts

The refrigerant R-134a expands reversibly from 2 MPa to 500 kPa in an insulated piston-and-cylinder device. The cylinder initially contains 15 L of R-134a with a quality of 1.0. Determine the final temperature of the R-134a in the cylinder and the work done by the refrigerant during this process.

Given values of two intensive properties at the inlet, P and x, you can calculate values for Vinit and Uinit based on data you will find in the thermodynamic tables for R-134a.

The key to solving this problem is to recognize that since the process is both reversible and adiabatic, it is also isentropic: Sfinal = Sinit.

m = 1.614 kg.

Tfinal = 15.74°C. and W = -46.0 kJ.

7B-13 :

Minimum Power Required for Adiabatic Compression of R-134a

3 pts

An adiabatic compressor is used to increase the pressure from 105 kPa to 1.25 MPa in a stream of R-134a with a volumetric flow rate of 1800 L/min. Determine the minimum power that must be supplied to the compressor.

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Given values of two intensive properties at the inlet, P and x, you can look up values for Vin , Sin and Hin in the thermodynamic tables for R-134a.

Because the compressor is isentropic, Sout = Sin.

Given values of two intensive properties at the outlet, P and S, you can calculate the value of Hout from data you will find in the thermodynamic tables for R-134a.

Ws = 8.46 kW.

7B-14 :

Reversible Compression of Steam in a Piston-and-Cylinder Device

4 pts

Superheated steam is reversibly compressed from 200 kPa to 1.4 MPa in a well-insulated piston-and-cylinder device. The cylinder initially contains 0.175 m3 of steam at 250°C. Determine the final temperature of the steam and the work done on the steam during this process.

Given values of two intensive properties at the inlet, T and P, you can look up values for V1 , S1 and U1 in the thermodynamic tables for steam.

The key to solving this problem is to recognize that since the process is both reversible and adiabatic, it is also isentropic: S2 = S1.

Given values of two intensive properties at the outlet, P and S, you can calculate the value of U2 from data you will find in the thermodynamic tables forsteam.

Wb = 66.5 kJ.

7B-15 :

Heat and Work in an Isothermal Expansion of Steam

3 pts

Saturated water vapor is contained in a piston-and-cylinder device that is positioned in a constant-temperature bath at 250oC. The 5.7 kg of steam inside the cylinder expands reversibly and isothermally to a final pressure of 1.2 MPa. Determine the heat transferred between the constant-temperature bath and the water inside the cylinder as well as the work done by the H2O inside the cylinder during this process.

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Given values of two intensive properties at the initial conditions, P and x, you can look up values for U1 and S1 in the thermodynamic tables for steam.

Given values of two intensive properties at the final conditions, T and P, you can calculate values for U2 and S2 based on data you will find in the thermodynamic tables for steam.

Because the process is reversible and isothermal, the water inside the cylinder behaves as a thermal reservoir and you can evaluate Q from the defining equation for entropy: Q = T ΔS.

Q = 475 kJ and Wb = 353 kJ.

7B-16 :

Isentropic Compression of High Quality Steam

5 pts

Steam is compressed isentropically to a final pressure of 5 MPa. The steam is initially at 25°C and has a quality of 91%. How much work does this process require, in kJ/kg, if the process takes place in... (a) A closed system? (b) An open system ? (c) Explain why the answers to parts (a) and (b) are not the same.

Given values of two intensive properties at the initial conditions, P and x, you can calculate values for U1 , H1 and S1 based on data you can find in the thermodynamic tables for steam.

Because the expansion process is isentropic: S2 = S1.

Given values of two intensive properties at the final conditions, P andS2, you can calculate values for U2 and H2 based on data you can find in the thermodynamic tables for steam. Then, apply the 1st Law to evaluate Wb and Ws for parts (a) and (b).

(a) Wb = 1510 kJ/kg , (b) Ws = 1890 kJ/kg , (c) Think about flow work !

7B-17 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-18 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-19 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-20 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-21 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-22 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-23 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-24 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-25 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.

7B-26 :

Rate of Entropy Increase for a Heat Transfer Process

3 pts

A hot reservoir at 944 K transfers heat to a cold reservoir at 298 K. If the rate of heat transfer is 25 kW,determine the rate at which the entropy of the two reservoirs combined changes (in kW/K) and determine if the second law is satisfied.

Because the temperature of a thermal reservoir is constant, you can calculate the rate of entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -0.0265 kW/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 0.0574 kW/K. No, this process does not violate the 2nd Law because ΔSuniv > 0.

7B-27 :

Entropy Change for a Heat Transfer Process

3 pts

Calculate the entropy change of each reservoir when 1000 kJ of heat is transferred directly from a hot reservoir at 1000 K to a cold reservoir at 400 K. Calculate the entropy change of the universe for this process. Does this process violate the principle of increasing entropy?

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Because the temperature of a thermal reservoir is constant, you can calculate the entropy change of the reservoir directly from the definition of entropy.

The entropy change of the hot reservoir is -1 kJ/K.

The entropy change of the universe is the sum of the entropy changes of the two reservoirs.

ΔSuniv = 1.5 kJ/K. No, this process does not violate the increase of entropy principle because ΔSuniv > 0.