The amount of heat lost to the surroundings.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 8.8, Problem 49P

(a)

To determine

The amount of heat lost to the surroundings.

(a)

Expert Solution

Answer to Problem 49P

The amount of heat lost to the surroundings is 2563 kJ.

Explanation of Solution

Express the final volume of the air.

νa2=νa1−νw (I)

Here, initial volume of air is νa1 and volume of water is νw.

Express the mass of air in the large tank.

ma=P1νa1RTa1 (II)

Here, initial pressure is P1, universal gas constant of air is R and initial temperature of air is Ta1.

Express the pressure of air at the final state.

Pa2=maRTa2νa2 (III)

Here, final temperature of air is Ta2.

Express the mass of water.

mw=ρwνw (IV)

Here, density of water is ρw.

Express the amount of heat lost to the surroundings.

Qout=−[mwcw(T2−Tw1)+macva(T2−Ta1)] (V)

Here, specific heat of water is cw, final temperature is T2, initial temperature of water is Tw1, specific heat constant of air is cva and initial temperature of air is Ta1.

Conclusion:

Refer Table A-2, “ideal gas specific heats of various gases”, and write the properties of air at room temperature.

R=0.287 kPa⋅m3/kg⋅Kcp=1.005 kJ/kg⋅Kcva=0.718 kJ/kg⋅K

Refer Table A-3 (a), “properties of common liquids, solids and foods”, and write the properties of water at room temperature.

ρw=997 kg/m3cw=4.18 kJ/kg⋅K

Substitute 0.04 m3 for νa1 and 15 L for νw in Equation (I).

νa2=0.04 m3−15 L=0.04 m3−15 L[m31000 L]=0.04 m3−0.015 m3=0.025 m3

Substitute 100 kPa for P1, 0.04 m3 for νa1, 0.287 kPa⋅m3/kg⋅K for R and 22°C for Ta1 in Equation (II).

ma=(100 kPa)(0.04 m3)(0.287 kPa⋅m3/kg⋅K)(22°C)=(100 kPa)(0.04 m3)(0.287 kPa⋅m3/kg⋅K)(22+273) K=(100 kPa)(0.04 m3)(0.287 kPa⋅m3/kg⋅K)(295 K)=0.04724 kg

Substitute 0.04724 kg for ma, 0.287 kPa⋅m3/kg⋅K for R, 44°C for Ta2 and 0.025 m3 for νa2 in Equation (III).

Pa2=(0.04724 kg)(0.287 kPa⋅m3/kg⋅K)(44°C)0.025 m3=(0.04724 kg)(0.287 kPa⋅m3/kg⋅K)(44+273) K0.025 m3=(0.04724 kg)(0.287 kPa⋅m3/kg⋅K)(317 K)0.025 m3=171.9 kPa

Substitute 997 kg/m3 for ρw and 15 L for νw in Equation (IV).

mw=(997 kg/m3)(15 L)=(997 kg/m3)[15 L×m31000 L]=(997 kg/m3)(0.015 m3)=14.96 kg

Substitute 14.96 kg for mw, 4.18 kJ/kg⋅K for cw, 44°C for T2, 85°C for Tw1, 0.04724 kg for ma, 0.718 kJ/kg⋅K for cva and 22°C for Ta1 in Equation (V).

Qout=−[(14.96 kg)(4.18 kJ/kg⋅K){(44°C−85°C) K}+(0.04724 kg)(0.718 kJ/kg⋅K){(44°C−22°C) K}]=−[(14.96 kg)(4.18 kJ/kg⋅K)(−41 K)+(0.04724 kg)(0.718 kJ/kg⋅K)(22 K)]=2563 kJ

Hence, the amount of heat lost to the surroundings is 2563 kJ.

(b)

To determine

The exergy destruction during the process.

(b)

Expert Solution

Answer to Problem 49P

The exergy destruction during the process is 318 kJ.

Explanation of Solution

Express entropy change in water.

ΔSw=mwcwlnTw1T2 (VI)

Express entropy change in air.

ΔSa=ma[cplnTa1T2−RlnPa1P2] (VII)

Here, initial pressure of air is Pa1.

Express net internal energy of water.

ΔUw=mwcw(T1w−T2) (VIII)

Express net internal energy of air.

ΔUa=macva(T1a−T2) (IX)

Express the exergy destruction during the process.

Xdest=ΔUw−T0ΔSw+ΔUa−T0ΔSa (X)

Here, surrounding temperature is T0.

Conclusion:

Substitute 14.96 kg for mw, 4.18 kJ/kg⋅K for cw, 44°C for T2 and 85°C for Tw1 min Equation (VI).

ΔSw=(14.96 kg)(4.18 kJ/kg⋅K)ln85°C44°C=(14.96 kg)(4.18 kJ/kg⋅K)[ln358 K317 K]=7.6059 kJ/K

Substitute 0.04724 kg for ma, 1.005 kJ/kg⋅K for cp, 44°C for T2, 22°C for Ta1, 0.287 kPa⋅m3/kg⋅K for R, 100 kPa for Pa1 and 171.9 kPa for P2 in Equation (VII).

ΔSa=(0.04724 kg)[(1.005 kJ/kg⋅K)ln22°C44°C−(0.287 kPa⋅m3/kg⋅K)ln100 kPa171.9 kPa]=(0.04724 kg)[(1.005 kJ/kg⋅K)ln295 K317 K−(0.287 kPa⋅m3/kg⋅K)(−0.5417)]=0.003931 kJ/K

Substitute 14.96 kg for mw, 4.18 kJ/kg⋅K for cw, 44°C for T2 and 85°C for Tw1 min Equation (VIII).

ΔUw=(14.96 kg)(4.18 kJ/kg⋅K)[(85°C−44°C) K]=2564 kJ

Substitute 0.04724 kg for ma, 0.718 kJ/kg⋅K for cva and 22°C for Ta1, and 44°C for T2 in Equation (IX).

ΔUa=(0.04724 kg)(0.718 kJ/kg⋅K)[(22°C−44°C) K]=−0.7462 kJ

Substitute 2564 kJ for ΔUw, 22°C for T0, 7.6059 kJ/K for ΔSw, −0.7462 kJ for ΔUa and 0.003931 kJ/K for ΔSa in Equation (X).

Xdest=[(2564 kJ)−(22°C)(7.6059 kJ/K)+(−0.7462 kJ)−(22°C)(0.003931 kJ/K)]=[(2564 kJ)−(22+273) K(7.6059 kJ/K)+(−0.7462 kJ)−(22+273) K(0.003931 kJ/K)]=[(2564 kJ)−(295 K)(7.6059 kJ/K)+(−0.7462 kJ)−(295 K)(0.003931 kJ/K)]=318 kJ

Hence, the exergy destruction during the process is 318 kJ.

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

Steam enters an adiabatic turbine steadily at 7 MPa, 500°C, and 45 m/s and leaves at 100 kPa and 75 m/s. If the power output of the turbine is 5 MW and the isentropic efficiency is 77 percent, determine the mass flow rate of steam through the turbine.

Steam enters an adiabatic turbine at 8 MPa, 550°C, and (13 ) m/s and leaves at 60 kPa, 150°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (1) the mass flow rate of the steam flowing through the turbine and (2) the isentropic efficiency of the turbine.

Steam enters an adiabatic turbine steadily at 7 MPa, 500°C, and 45 m/s and leaves at 100 kPa and 75 m/s. If the power output of the turbine is 5 MW and the isentropic efficiency is 77 percent, determine the temperature at the turbine exit.

Answer 2