EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
9th Edition
ISBN: 8220106796979
Author: CENGEL
Publisher: YUZU
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Chapter 13.3, Problem 94RP

A spring-loaded piston–cylinder device contains a mixture of gases whose pressure fractions are 25 percent Ne, 50 percent O2, and 25 percent N2. The piston diameter and spring are selected for this device such that the volume is 0.1 m3 when the pressure is 200 kPa and 1.0 m3 when the pressure is 1000 kPa. Initially, the gas is added to this device until the pressure is 200 kPa and the temperature is 10°C. The device is now heated until the pressure is 500 kPa. Calculate the total work and heat transfer for this process.

Chapter 13.3, Problem 94RP, A spring-loaded pistoncylinder device contains a mixture of gases whose pressure fractions are 25

Expert Solution & Answer
Check Mark
To determine

The total work done and heat transfer for the process.

Answer to Problem 94RP

The total work done for the process is 118kJ.

The heat transfer for the process is 569kJ.

Explanation of Solution

Write the expression to calculate the partial pressure of Ne (Ne).

PNe=yNePm (I)

Here, mixture pressure is Pm and mole fraction of Ne is yNe.

Write the expression to calculate the partial pressure of O2 (PO2).

PO2=yO2Pm (II)

Here, mixture pressure is Pm and mole fraction of O2 is yO2.

Write the expression to calculate the partial pressure of N2 (PN2).

PN2=yN2Pm (III)

Here, mixture pressure is Pm and mole fraction of N2 is yN2.

Write the expression to calculate the mass of N2 (mN2).

mNe=PNeνmRNeT (IV)

Write the expression to calculate the mass of O2 (mO2).

mO2=PO2νmRO2T (V)

Write the expression to calculate the mass of N2 (mN2).

mN2=PN2νmRN2T (VI)

Write the expression to calculate the total mass of each component (mtotal).

mtotal=mNe+mN2+mO2 (VII)

Write the expression to calculate the mass fraction of Ne(mfNe)

mfNe=mNemm (VIII)

Write the expression to calculate the mass fraction of O2(mfO2)

mfO2=mO2mm (IX)

Write the expression to calculate the mass fraction of N2(mfN2)

mfN2=mN2mm (X)

Write the expression to calculate the mole number of Ne (NNe).

NNe=mNeMNe (XI)

Here, molar mass of Ne is MNe and mass of Ne is mNe.

Write the expression to calculate the mole number of N2 (NN2).

NN2=mN2MN2 (XII)

Here, molar mass of N2 is MN2 and mass of N2 is mN2.

Write the expression to calculate the mole number of CO2 (NCO2).

NCO2=mCO2MCO2 (XIII)

Here, molar mass of CO2 is MCO2 and mass of CO2 is mCO2.

Write the expression to calculate the total number of moles (Nm).

Nm=NCO2+NN2 (XIV)

Write the expression to calculate the apparent molecular weight of the mixture (Mm).

Mm=mmNm (XV)

Write the expression to calculate the constant volume specific heat of the mixture (cv).

cv=mfNecvNe+mfO2cv,O2+mfN2cvN2 (XVI)

Here, mole fraction of Ne,N2,andO2 is mfNe,mfN2,andmfO2, specific heat of Ne is cv,Ne, specific heat of N2 is cv,N2, and specific heat of O2 is cv,O2.

Write the expression to calculate the apparent gas constant of the mixture (R).

R=RuMm (XVII)

Here, universal gas constant is Ru.

Write the expression for the mass contained in the system (m).

m=P1ν1RT1 (XVIII)

Write the expression to calculate the final temperature (T2)

T2=P2ν2mR (XIX)

Write the expression to calculate the work done during the process.

Wout=P1+P22(ν2ν1) (XX)

Conclusion:

From Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the values of molar masses for Ne,O2 and N2 as 20.18kg/kmol,32.0kg/kmol and 28.0kg/kmol respectively.

From Table A-2a, “Ideal-gas specific heats of various common gases”, obtain the following properties for Ne,O2 and N2.

For Ne,

cv=0.6179kJ/kgK

For O2,

cv=0.658kJ/kgK

For N2,

cv=0.743kJ/kgK

Substitute 0.25 0for yNe and 200kPa for Pm in Equation (I).

PO2=(0.25)200kPa=50kPa

Substitute 0.50 for yO2 and 200kPa for Pm in Equation (II).

PO2=(0.50)200kPa=100kPa

Substitute 0.25 for yN2 and 200kPa for Pm in Equation (III).

PN2=(0.25)200kPa=50kPa

Substitute 50kPa for PNe, 0.1m3 for νm, 0.4119kpam3/kgK for RNe and 283K in Equation (IV).

mNe=(50kPa)×(0.1m3)(0.4119kpam3/kgK)(283K)=0.4289kg

Substitute 100kPa for PO2, 0.1m3 for νm, 0.2598kpam3/kgK for RO2 and 283K in Equation (V).

mO2=(100kPa)×(0.1m3)(0.2598kpam3/kgK)(283K)=0.1360kg

Substitute 50kPa for PN2, 0.1m3 for νm, 0.2968kpam3/kgK for RN2 and 283K in Equation (VI).

mN2=(50kPa)×(0.1m3)(0.2968kpam3/kgK)(283K)=0.05953kg

Substitute 0.4289kg for mNe, 0.1360kg for mO2, and 0.05953kg for mN2 in Equation (VII).

mtotal=0.4289kg+0.1360kg+0.05953kg=0.2384kg

Substitute 0.04289kg for mNe, and 0.2384kg for mm in Equation (VIII).

mfNe=0.04289kg0.2384kg=0.1799

Substitute 0.1360kg for mO2, and 0.2384kg for mm in Equation (IX).

mfO2=0.1360kg0.2384kg=0.5705

Substitute 0.05953kg for mN2, and 0.2384kg for mm in Equation (X).

mfN2=0.05953kg0.2384kg=0.2497

Substitute 0.04289kg for mNe and 20.18kg/mol for MNe in Equation (XI).

NNe=0.04289kg20.18kg/mol=0.002126kmol

Substitute 0.05953kg for mN2 and 28kg/mol for MN2 in Equation (XII).

NN2=0.05953kg28kg/mol=0.002126kmol

Substitute 45kg for mCO2 and 28kg/mol for MCO2 in Equation (XIII).

NO2=0.1360kg32kg/mol=0.00425kmol

Substitute 0.002126kmol for NNe, 0.00425kmol for NO2, and 0.002126kmol for NN2 in Equation (XIV).

Nm=0.002126kmol+0.00425kmol+0.002126kmol=0.008502kmol

Substitute 0.008502kmol for Nm and 2.987kmol for mm in Equation (XV).

Mm=100kg0.008502kmol=28.04kg/kmol

Substitute 0.6179kJ/kgK for cvNe, 0.743kJ/kgK for cv,N2, 0.658kJ/kgK for cv,O2, and 0.1799 for mfNe , 0.5705 for mfO2 and 0.2497 for mfN2 in Equation (XVI).

cv=(0.1799)(0.6179kJ/kgK)+(0.5705)(0.658kJ/kgK)+(0.2497)(0.743kJ/kgK)=0.672kJ/kgK

Substitute 8.134kJ/KmolK for Ru and 28.04kg/kmol for Mm in Equation (XVII).

R=8.134kJ/KmolK28.04kg/kmol=0.2964kJ/kgK

Substitute 200kPa for P1, 0.1m3 for ν1 , 0.2964kPam3/kgK for R and 283K for T1 in Equation (XVIII).

m=200kPa×0.1m30.2964kPam3/kgK×283K=0.2384kg

The pressure changes linearly with volume as shown is Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 13.3, Problem 94RP

Using the data form Prob. 13–94 obtain the value of final volume by linear interpolation.

Write the straight line equation for two points.

(y1y)(y2y)=(x1x)(x2x) (XXI)

Here, coordinates of the point 1 is (x1,y1), coordinates of the point 2 is (x2,y2) and intermediate point between 1 and 2 is (x,y).

Substitute 500 for y1 , 200 for y , 1000 for y2 , 0.1 for x , 1.0 for x2 and 0.1 for x2 in Equation (XXI).

(500200)(1000200)=(x10.1)(1.00.1)x1=0.4375

The final volume ν2 is 0.4375m3.

Substitute 500kPa for P1, 0.4375m3 for ν2 , 0.2964kPam3/kgK for R and 0.2384kg for m in Equation (XIX).

T2=500kPa×0.4375m30.2384kg×0.2483kPam3/kgK=3096K

Substitute 500kPa for P1 , 200kPa , 0.1m3 for ν1 and 0.4375m3 for ν2 in Equation (XX).

Wout=500kPa+200kPa2(0.4375m30.1m3)=118kPam3(1kJ1kPam3)=118kJ

Thus, the total work done for the process is 118kJ.

Write a energy balance on the system.

EinEout=ΔEsystem (XXII)

Here, input energy transfer and output energy transfer is EinandEout, and change in energy of a system is ΔEsystem.

The rate of change in energy of a system ΔE˙system is zero for steady state.

For given system the energy balance Equation (XXII) is expressed as follows:

Qin=Wout+mcv(T2T1) (XXIII).

The rate of change in energy of a system ΔE˙system is zero for steady state.

Substitute 118kJ for Wout, 0.2384kg for m , 0.672kJ/kgK for cv , 3096K for T2 and 283K for T1 in Equation (XXIII).

Qin=118kJ+(0.2384kg)(0.672kJ/kgK)(3096K283K)=569kJ

Thus, the heat transfer for the process is 569kJ.

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Chapter 13 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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