The amount of ice added.

Question

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Answer 1

Question

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

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Answer 2

Question

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

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Answer 3

Question

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

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Answer 4

Question

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

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Answer 5

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

Answer 6

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

Answer 7

Chapter 8, Problem 190RQ

a)

To determine

The amount of ice added.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

The initial volume of the water (v) is 0.02 m3.

The initial quality of the water mixture (x) is 0.1

The initial temperature of the water (Tw)1 is 100°C.

The temperature of the ice (Tice) is −18°C.

The final temperature of the mixture (T2) is 100°C

The melting temperature of the ice (Tw) is 0°C.

The heat of fusion of ice at 1 atm (hf) is 333.7 kJ/kg.

Calculation:

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

vf=0.001043 m3/kgvg=1.6720 m3/kgsf=1.3072 kJ/kg⋅Ksfg=6.0470kJ/kg⋅Khf=419.17 kJ/kghfg=2256.4 kJ/kg

Refer the Table A-4, “Saturated water-Temperature table” the obtain the following properties at temperature of is 100°C.

h2=419.17 kJ/kgs2=1.3072 kJ/kg⋅K

Calculate the initial entropy of the refrigerant (s1).

s1=sf+x1sfgs1=1.3072 kJ/kg⋅K+(0.1)6.0470kJ/kg⋅K=1.9119 kJ/kg⋅K

Calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfgh1=419.17 kJ/kg+(0.1)2256.4 kJ/kg=644.81 kJ/kg

Calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf)v1=0.001043 m3/kg+(0.1)(1.6720 m3/kg−0.001043 m3/kg)=0.16814 m3/kg

Calculate the mass of the stream (msteam).

msteam=ν1v1msteam=0.02 m30.16814 m3/kg=0.119 kg

Refer the Table A-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure at room temperature for liquid and ice as 4.18 kJ/kg⋅K and 2.11 kJ/kg⋅K respectively.

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Wb,in, Eout=0, and ΔEsystem=ΔU in Equation (I).

Wb,in=ΔUΔH=0ΔHice+ΔHwater=0

[mcp,ice(0°C−(T1)solid)solid+mhif+mcp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0mice[cp,ice(0°C−(T1)solid)solid+hif+cp,liquid((T2)liquid−0°C)liquid]ice+[m(h2−h1)]water=0{mice[(2.11 kJ/kg⋅K)(0°C−(−18° C))solid+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(100°C−0°C)]ice+[0.119 kg(419.17 kJ/kg−644.81 kJ/kg)]steam}=0mice=0.034 kg(1000 g1 kg)mice=34 g

Thus, the amount of ice added is 34 g.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K

Thus, the generation during the process is 0.0188 kJ/K.

Answer 13

b)

To determine

The entropy generation during the process.

Answer 14

b)

Expert Solution

Explanation of Solution

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Substitute Sin=0, Sout=0 and ΔSsystem=ΔSice+ΔSwater in Equation (II).

0−0+Sgen=ΔSice+ΔSsteamSgen=(micecp,iceln(TmeltingT1)solid+micehifTmelting+micecp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steamSgen=mice(cp,iceln(TmeltingT1)solid+hifTmelting+cp,liquidln(T2T1)liquid)ice+msteam(s2−s1)steam

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln(0°C−18°C)solid+333.7 kJ/kg0°C+[(4.18 kJ/kg⋅K)ln(100°C0°C)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}

Sgen={0.034 kg((2.11 kJ/kg⋅K)ln((0+273.15) K(−18+273.15) K)solid+333.7 kJ/kg0+273.15 K+[(4.18 kJ/kg⋅K)ln((100+273.15) K(0+273.15) K)liquid])ice+[(0.119 kg)(1.3072 kJ/kg⋅K−1.9119 kJ/kg⋅K)]steam}=0.0907 kJ/K−0.0719 kJ/K=0.0188 kJ/K