EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 3.8, Problem 29P

A piston–cylinder device contains 0.85 kg of refrigerant-134a at −10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

FIGURE P3–30

Chapter 3.8, Problem 29P, A pistoncylinder device contains 0.85 kg of refrigerant-134a at 10C. The piston that is free to move

(a)

Expert Solution
Check Mark
To determine

The final pressure of the refrigerant R-134a.

Answer to Problem 29P

The final pressure of the refrigerant R-134a is 90.4kPa_

Explanation of Solution

The final pressure is equal to the initial pressure of the refrigerant R-134a.

P2=P1=Patm+mpg(πD24) (I)

Here, atmospheric pressure is Patm, mass of piston-cylinder device is mp, acceleration due to gravity is g, and diameter of a piston is D.

Conclusion:

Substitute 88 kPa for Patm, 12 kg for mp, 9.81m/s2 for g, and 25 cm for D in Equation (I).

P2=88kPa+(12kg)(9.81m/s2)(π(25cm)24)=88kPa+(12kg)(9.81m/s2)(π(25cm)24)(1kN1000kgm/s2)=90.4kPa

Thus, the final pressure of the refrigerant R-134a is 90.4kPa_.

(b)

Expert Solution
Check Mark
To determine

The change in the volume of the cylinder.

Answer to Problem 29P

The change in the volume of the cylinder is 0.02142m3_.

Explanation of Solution

Convert the unit of initial pressure from kPa to MPa.

P1=90.4kPa=90.4kPa(1MPa1000kPa)=0.0904MPa

Write the formula of interpolation method of two variables at 10°C.

y2=(x2x1)(y3y1)(x3x1)+y1 (II)

Here, the variables denote by x and y are pressure and specific volume.

Calculate the initial volume of cylinder.

ν1=mv1 (III)

Here, the initial state specific volume is v1.

Calculate the final volume of cylinder.

ν2=mv2 (IV)

Here, the final state specific volume is v2.

Calculate the change in the volume of cylinder.

Δν=ν2ν1 (V)

Conclusion:

Refer to Table A-13, obtain the values of below variables as in Table (I) at 10°C.

Pressure, MPa (x)Specific volume, m3/kg (y)
0.060.35048
0.0904?
0.100.20743

Substitute 0.06 for x1, 0.0904 for x2, 0.10 for x3, 0.35048 for y1, and 0.20743 for y3 in Equation (II).

y2=(0.09040.06)(0.207430.35048)(0.100.06)+0.35048=0.2418

Thus, the specific volume of refrigerant R-134a at the initial state of 90.4 kPa and 10°C is 0.2418m3/kg.

Refer to Table A-13, obtain the values of below variables as in Table (II) at 10°C.

Pressure, MPa (x)Enthalpy, kJ/kg (y)
0.06248.60
0.0904?
0.10247.51

Substitute 0.06 for x1, 0.0904 for x2, 0.10 for x3, 248.60 for y1, and 247.51 for y3 in Equation (II).

y2=(0.09040.06)(247.51248.60)(0.100.06)+248.60=247.77

Thus, the enthalpy of refrigerant R-134a at the initial state of 90.4 kPa and 10°C is 247.77kJ/kg.

Apply spreadsheet and solve the final state specific volume at 15°C and 0.0904 MPa using interpolation method.

Refer to Table A-13, obtain the values of below variables as in Table (III) at 15°C and 0.06 MPa.

Temperature, °C (x)Specific volume, m3/kg (y)
100.37893
15?
200.39302

Substitute 10 for x1, 15 for x2, 20 for x3, 0.37893 for y1, and 0.39302 for y3 in Equation (II).

y2=(1510)(0.393020.37893)(2010)+0.37893=0.386

Similarly, solve final state specific volume at 15°C and 0.10 MPa using interpolation method as 0.2294m3/kg.

Now use interpolation method again to solve the final state specific volume at 15°C. Refer to Table A-13, obtain the values of below variables as in Table (IV) at 15°C and 0.0904 MPa.

Pressure, MPa (x)Specific volume, m3/kg (y)
0.060.386
0.0904?
0.100.2294

Substitute 0.06 for x1, 0.0904 for x2, 0.10 for x3, 0.386 for y1, and 0.2294 for y3 in Equation (II).

y2=(0.09040.06)(0.22940.386)(0.100.06)+0.386=0.267

Thus, the final state specific volume at 15°C and 90.4 kPa is 0.267m3/kg.

Apply the above steps to calculate the enthalpy at 15°C and 90.4 kPa using interpolation method as 268.19kJ/kg.

Substitute 0.85 kg for m and 0.2418m3/kg for v1 in Equation (III).

ν1=0.85kg(0.2418m3/kg)=0.20553m3

Substitute 0.85 kg for m and 0.267m3/kg for ν2 in Equation (IV).

ν2=0.85kg(0.267m3/kg)=0.22695m3

Substitute 0.22695m3 for ν2 and 0.20553m3 for ν1 in Equation (V).

Δν=0.22695m30.20553m3=0.02142m3

Thus, the change in the volume of the cylinder is 0.02142m3_.

(c)

Expert Solution
Check Mark
To determine

The change in the enthalpy of the refrigerant R-134a.

Answer to Problem 29P

The change in the enthalpy of the refrigerant R-134a is 17.35kJ/kg_.

Explanation of Solution

Calculate the total enthalpy change of refrigerant R-134a.

ΔH=m(h2h1) (VI)

Here, enthalpy at initial state and final state are h1andh2 respectively.

Conclusion:

Substitute 0.85 kg for m, 247.77kJ/kg for h1 and 268.19kJ/kg for h2 in equation (VI).

ΔH=0.85kg(268.19kJ/kg247.77kJ/kg)=17.35kJ/kg

Thus, the change in the enthalpy of the refrigerant R-134a is 17.35kJ/kg_.

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Students have asked these similar questions
A piston–cylinder device contains 0.85 kg of refrigerant- 134a at 210ºC. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15ºC. Determine: (a) the final pressure,(b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.
A piston-cylinder device contains 0.850 kg of refrigerant-134a at –10.0°C. The piston that is free to move and has a mass of 12.0 kg and a diameter of 25.0 cm. The local atmospheric pressure is 88.0 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15.0°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant- 134a. Assume the following: va-10°C = 0.2302 m³/kg, va-15°C = 0.2544 m³/kg, ha-10°C = 247.22 kJ/kg, h@15°C = 268.18 kJ/kg
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Chapter 3 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 3.8 - Does the amount of heat absorbed as 1 kg of...Ch. 3.8 - Does the reference point selected for the...Ch. 3.8 - What is the physical significance of hfg? Can it...Ch. 3.8 - Does hfg change with pressure? How?Ch. 3.8 - Is it true that it takes more energy to vaporize 1...Ch. 3.8 - What is quality? 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Using appropriate software,...Ch. 3.8 - Prob. 74PCh. 3.8 - Prob. 75PCh. 3.8 - A rigid tank whose volume is unknown is divided...Ch. 3.8 - Prob. 77PCh. 3.8 - Prob. 78PCh. 3.8 - Prob. 79PCh. 3.8 - Prob. 80PCh. 3.8 - Prob. 81PCh. 3.8 - Determine the specific volume of superheated water...Ch. 3.8 - Determine the specific volume of superheated water...Ch. 3.8 - Prob. 85PCh. 3.8 - Prob. 86PCh. 3.8 - Prob. 87PCh. 3.8 - Prob. 88PCh. 3.8 - Prob. 89PCh. 3.8 - Prob. 90PCh. 3.8 - Carbon dioxide gas enters a pipe at 3 MPa and 500...Ch. 3.8 - A 0.016773-m3 tank contains 1 kg of...Ch. 3.8 - What is the physical significance of the two...Ch. 3.8 - A 3.27-m3 tank contains 100 kg of nitrogen at 175...Ch. 3.8 - Prob. 95PCh. 3.8 - Refrigerant-134a at 400 psia has a specific volume...Ch. 3.8 - Nitrogen at 150 K has a specific volume of...Ch. 3.8 - A 1-m3 tank contains 2.841 kg of steam at 0.6 MPa....Ch. 3.8 - Prob. 102PCh. 3.8 - Prob. 103PCh. 3.8 - On a certain day, the temperature and relative...Ch. 3.8 - Prob. 105PCh. 3.8 - Consider two rooms that are identical except that...Ch. 3.8 - A thermos bottle is half-filled with water and is...Ch. 3.8 - Prob. 108RPCh. 3.8 - The combustion in a gasoline engine may be...Ch. 3.8 - A tank contains argon at 600C and 200 kPa gage....Ch. 3.8 - Prob. 111RPCh. 3.8 - Prob. 112RPCh. 3.8 - A rigid tank with a volume of 0.117 m3 contains 1...Ch. 3.8 - Prob. 114RPCh. 3.8 - Ethane at 10 MPa and 100C is heated at constant...Ch. 3.8 - Prob. 116RPCh. 3.8 - A 10-kg mass of superheated refrigerant-134a at...Ch. 3.8 - A 4-L rigid tank contains 2 kg of saturated...Ch. 3.8 - The gage pressure of an automobile tire is...Ch. 3.8 - Prob. 120RPCh. 3.8 - Steam at 400C has a specific volume of 0.02 m3/kg....Ch. 3.8 - A tank whose volume is unknown is divided into two...Ch. 3.8 - Prob. 123RPCh. 3.8 - Prob. 124RPCh. 3.8 - Prob. 125RPCh. 3.8 - A tank contains helium at 37C and 140 kPa gage....Ch. 3.8 - Prob. 127RPCh. 3.8 - Prob. 131RPCh. 3.8 - Consider an 18-m-diameter hot-air balloon that,...Ch. 3.8 - Prob. 134FEPCh. 3.8 - Water is boiled at 1 atm pressure in a coffeemaker...Ch. 3.8 - Prob. 136FEPCh. 3.8 - Prob. 137FEPCh. 3.8 - Water is boiled in a pan on a stove at sea level....Ch. 3.8 - Prob. 139FEPCh. 3.8 - Consider a sealed can that is filled with...Ch. 3.8 - A rigid tank contains 2 kg of an ideal gas at 4...Ch. 3.8 - The pressure of an automobile tire is measured to...
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