The amount of ice added.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

! Required information A telephone cable is clamped at A to the pole AB. The tension in the left-hand portion of the cable is given to be T₁ = 815 lb. A 15° 25° B T₂ Using trigonometry, determine the required tension T₂ in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical. The required tension is lb.

What are examples of at least three (3) applications of tolerance fitting analysis.

The primary material used in the production of glass products is silica sand. True or False

Answer 2

Question

Chapter 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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 7.13, Problem 208RP

a)

To determine

The amount of ice added.

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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 7.13, Problem 208RP

a)

To determine

The amount of ice added.

Answer 8

a)

Expert Solution

Answer to Problem 208RP

The amount of ice added is 34 g.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (II)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the initial enthalpy of the refrigerant.

h1=hf+x1hfg (III)

Here, initial enthalpy is h1, saturated liquid enthalpy is hf, initial vapor quality is x1 and evaporated enthalpy is hfg.

Write the expression to calculate the initial specific volume of the mixture.

v1=vf+x1(vg−vf) (IV)

Here, initial specific volume of the mixture is v1 , specific volume of the liquid phase is vf, specific volume of the vapor phase is vg and initial vapor quality is x1.

Write the expression to calculate the mass of the stream msteam.

msteam=ν1v1 (V)

Here, the initial volume of a container is ν1.

Conclusion:

Substitute Wb,in for Ein, 0 for Eout and ΔU for ΔEsystem 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 (VI)

Here, change in enthalpy in ice is ΔHice, change in enthalpy in water is ΔHwater, heat of fusion of ice is hif, mass of the ice is mice, mass of the water is mwater, initial temperature is T1, final temperature is T2 , specific heat at constant pressure for liquid is cp,liquid, initial enthalpy is h1 and final enthalpy is h2.

From 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

From 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

Here, the final entropy is s2.

Substitute 419.17 kJ/kg for hf, 2256.4 kJ/kg for hfg and 0.1 for x1 in Equation (III).

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

Substitute 0.001043 m3/kg for vf, 1.6720 m3/kg for vg and 0.1 for x1 in Equation (IV).

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

Substitute 1.3072 kJ/kg⋅K for sf, 6.0470kJ/kg⋅K for sfg and 0.1 for x1 in Equation (II).

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

Substitute 0.16814 m3/kg for v1 and 0.02 m3 for ν1 in Equation (I).

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

From 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.

The melting temperature and the heat of fusion of ice at 1 atm as 0 C and 333.7 kJ/kg respectively.

Substitute 0.119 kg for msteam, 100°C for T2, 2.11 kJ/kg⋅K for cp,ice, −18° C for T1, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cp,liquid, 644.81 kJ/kg for h1 and 419.17 kJ/kg for h2 in Equation (VI).

{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 13

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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 14

b)

To determine

The entropy generation during the process.

Answer 15

b)

Expert Solution

Answer to Problem 208RP

The generation during the process is 0.0188 kJ/K.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (VII)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSice+ΔSwater for ΔSsystem in Equation (VII).

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 (VIII)

Substitute 0.119 kg for msteam,0.034 kg for mice, 2.11 kJ/kg⋅K for cp,ice, 0 °C for Tmelting, −18° C for T1,solid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅K for cp,liquid, 0 °C for T1,liquid, 100 °C for T2,liquid,1.3072 kJ/kg⋅K for s2 and 1.9119 kJ/kg⋅K for s1 in Equation (VIII).

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