Principles of Energy ConversionHomework #3 January 30, 2018

Homework #3 (group) – Tuesday, February 13 by 4:00 pm

5290 exercises (individual) – Thursday, February 15 by 4:00 pm

extra credit (individual) – Thursday, February 15 by 4:00 pm

Readings for this homework assignment and upcoming lectures

1. Read Chapter 1 of Textbook (Weston)

2. Read Chapter 2 of Textbook (Weston)

3. Review lecture notes:

• Part 5. The Rise of Heat Engines• Part 6. Review of Engineering Thermodynamics, Parts A and B

4. Review Thermodynamics textbook as needed

Homework Submission

For this assignment, the homework is to be worked as a group assignment and submitted as a groupin class or by dropping off at my office (room 905).

PLEASE include the course number (MEEM4200 -or- MEEM5290) in the subject line of any emailcorrespondence.

MEEM 5290 problems are always to be worked and submitted individually.

Extra credit are always to be worked and submitted individually.

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Principles of Energy ConversionHomework #3 January 30, 2018

Homework #3 - due Tuesday, February 13 by 4:00 pm

1. Weston 1.11

Solution: Compare the heats of vaporization at 0.01, 10, and 1000 psia. Compare the saturatedliquid specific volumes at these pressures. What do you conclude about the influence of pressure onthese properties?

Thermodynamic properties from EES:heat of vaporization liquid specific

pressure (psia) temperature (F) hg − hf (Btu/lbm) volume, (ft3/lbm)0.01 -11.45 1220 0.0174110.0 193.2 981.9 0.016591000 544.7 649.9 0.02159

There is a large variation in the heat of vaporization of steam as pressure increases, but there islittle change in the liquid specific volume.

2. A 10-m2 rigid tank contains steam at 30 bar and 400 ◦C. It is left to cool down until its pressuredrops to 5 bar. Find (a) the final condition of the steam and (b) the heat transfer in kilojoules.

Solution:

1Q2 = U2 − U1 = m(u2 − u1)

(1)P1 = 30 barT1 = 400 ◦C

}superheated steam

from EES: v1 = 0.09935 m3/kg

m =∀v1

=10 m3

0.09935 m3/kg= 100.7 kg

u1 = 2933 kJ/kg

(2) P2 = 5 bar

• this is a constant volume process, v2 = v1 = 0.09935 m3/kg

• check for saturation, v2g = 0.3751 m3/kg > v2

• therefore, the steam is saturates with quality x2 = 0.2627

• the internal energy is u2 = 1145 kJ/kg

Part (a). saturated steam, P2 = 5 bar, x2 = 0.2627, u2 = 1145 kJ/kg

Part (b). 1Q2 = m(u2 − u1) = −179 975 kJ

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Principles of Energy ConversionHomework #3 January 30, 2018

3. A 1.2 kWe compressor moves R134a through a residential heat pump at a rate of 0.018 kg/s.The refrigerant enters the heat pump condenser at 800 kPa and 35 ◦C and exits at 800 kPa assaturated liquid. Determine (a) the COP of the heat pump, and (b) the rate of heat absorptionfrom the outside air.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

4. Liquid ammonia at 87 ◦F and 250 psia is throttled to 100 psia into a flash tank. The flow rateinto the flash tank is 100 lbm/hr. Ammonia vapor is drawn off of the top of the flash tank andliquid ammonia is discharged at the bottom. Determine the temperature [◦F] in the tank andthe mass and volume flow rates [lbm/hr, ft3/hr] of the two exit streams. (good problem to useEES)

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

5. Air expands from 10 bar and 1000 ◦C to 1 bar and 500 ◦C in an insulated turbine. Calculate

(a) the isentropic turbine efficiency,(b) the change in entropy during expansion, in kJ/kg K,(c) the specific work output of the turbine, in kJ/kg, and(d) the polytropic exponent.

Assume a constant specific heat of 1.005 kJ/kg K.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

6. Helium is compressed from 15 psia and 40 ◦F to 60 psia. The isentropic compressor efficiency is0.70. Determine (a) the helium exit temperature, in ◦F, (b) the specific work of the compressor,in Btu/lbm, and (c) the change in entropy, in Btu/lbm ◦R.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

7. 106 lbm/hr of steam at 250 psia and 1000 ◦F expands in a turbine to 1 psia. The turbinehas isentropic and mechanical efficiencies of 0.90 and 0.95, respectively. The turbine drives anelectric generator that has an efficiency of 0.96. What is the output power of the generator, inMWe?

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

8. An ideal gas cycle is composed of four processes with air as the working fluid:

1-2: Isentropic compression from 100 kPa and 27 ◦C to 1 MPa.2-3: Isobaric heat addition of 2800 kJ/kg.3-4: Isochoric heat rejection to 100 kPa.4-1: Isobaric heat rejection to state (1).

(a) Show the cycle on a P -v and T -s diagram.

(b) Calculate the maximum temperature in the cycle.

(c) Calculate the thermal efficiency.

Assume the specific heats are constant at the room temperature value.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

9. A new inert gas appears to be well-suited for converting thermal energy from the exhaust heatof an engine manifold into mechanical energy via an Erickson engine. In order to design thecycle you need to determine the specific heats, the ideal gas constant and the entropy (all as afunction of temperature) of this gas. The gas can be assumed to behave as an ideal gas. Designa simple experiment which will, with accompanying analysis, provide the necessary propertydata.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

5290 exercises (individual) – Thursday, February 15 by 4:00 pm

10. Air at 800 kPa and 27 ◦C enters a steady-flow nozzle with a very low velocity and leaves at100 kPa. If the air undergoes an adiabatic expansion process through the nozzle, what is themaximum velocity of the air at the nozzle exit, in m/s?

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

11. Weston 1.24

Solution: Ten kilograms of compressed air is stored in a tank at 250 kPa and 50 ◦C. The tank isheated to bring the air temperature to 200 ◦C. What is the final tank pressure, and how much heatwas added?

1Q2 = m (e2 − e1) = m (u2 − u1) = mcp (T2 − T1)

1Q2 = 1520 kJ

The specific volume is unchanged since the tank volume and air mass remain constant during theheating process.

specific volume: v =RT1P1

= 0.3708 m3/kg

density: ρ =1

v= 2.7 kg/m3

Thus, the pressure after heating is:

P2 = ρRT2 = 366 kPa

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Principles of Energy ConversionHomework #3 January 30, 2018

12. A gas has a molecular mass of 30 and a specific heat at constant pressure of 0.25 Btu/lbm-R.The gas undergoes a non-flow polytropic compression during which its temperature increasesfrom 100 to 200 ◦F. The polytropic exponent is 1.3. Calculate the work done and the heattransfer in Btus per hour.

Solution:

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Principles of Energy ConversionHomework #3 January 30, 2018

extra credit (individual) – Thursday, February 15 by 4:00 pm

13. Air at 140 ◦F and 100 psia is confined in an uninsulated 10 ft3 vessel. A propeller is driveninside the vessel by a 50-W electric motor. After a period of 1 hour the air temperature droppedto 100 ◦F. Find the heat transfer in Btu per hour.

Solution:

Q u n i n s u l a t e d

W

a i r

T (t = 0 hr) = 140 ◦F = 600 ◦R

T (t = 1 hr) = 100 ◦F = 500 ◦RApplying First Law:

Q− W = m∆u

∆t−→ ∆t = 1 hr

Q = m∆u

∆t+ W = m

(cv∆T

∆t

)+ W

from Table A-20E, Moran & Shapiro, 5th ed.or Table A-2E, Cengel & Boles, 5th ed.

cv(T = 500 ◦F) = 0.179Btu

lbm ◦R

cv(T = 600 ◦F) = 0.182Btu

lbm ◦R

cv = 0.1805Btu

lbm ◦R

from EES:

The mass in the tank may be calculated using the ideal gas law:

m =P∀RT

=(100 lbf/in2)(144 in2/ft2)(10 ft3)

(0.250− 0.182 Btu/lbm ◦R)︸ ︷︷ ︸R=cp−cv

= 4.54 lbm

Q = (4.54 lbm)(0.18 Btu/lbm ◦R)(−40 ◦R/hr) + (−50W )(3.412 Btu/Whr)︸ ︷︷ ︸(-) because work is into air

Q = −32.78 Btu/hr− 170.61 Btu/hr = −203.39 Btu/hr︸ ︷︷ ︸(-) indicates thermal energy transferred from air

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Principles of Energy ConversionHomework #3 January 30, 2018

14. A 10-m2 rigid tank contains steam at 30 bar and 400 ◦C. It is left to cool down until its pressuredrops to 5 bar. Find (a) the final condition of the steam and (b) the heat transfer in kilojoules.

Solution:

1Q2 = U2 − U1 = m(u2 − u1)

(1)P1 = 30 barT1 = 400 ◦C

}superheated steam

from EES: v1 = 0.09935 m3/kg

m =∀v1

=10 m3

0.09935 m3/kg= 100.7 kg

u1 = 2933 kJ/kg

(2) P2 = 5 bar

• this is a constant volume process, v2 = v1 = 0.09935 m3/kg

• check for saturation, v2g = 0.3751 m3/kg > v2

• therefore, the steam is saturates with quality x2 = 0.2627

• the internal energy is u2 = 1145 kJ/kg

Part (a). saturated steam, P2 = 5 bar, x2 = 0.2627, u2 = 1145 kJ/kg

Part (b). 1Q2 = m(u2 − u1) = −179 975 kJ

15. An inventor claims to have built an engine that operates on a cycle, receives 1000 kJ at 500 ◦C,produces work, and rejects 350 kJ at 50 ◦C. Is this claim valid? Why or Why not?

Solution: For a cycle, the efficiency is:

ηth =W

Qin

=Qin − Qout

Qin

= 1− Qout

Qin

= 0.65 = 65%

The Carnot efficiency is:

ηcarnot = 1− TcoldThot

= 1− 50 + 273 K

500 + 273 K= 0.58 = 58%

The claimed efficiency is greater than the Carnot efficiency. Therefore the claim is not valid.

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Principles of Energy ConversionHomework #3 January 30, 2018

16. A long roll of 2-m-wide and 0.5-cm-thick manganese steel plate (ρ = 7854 kg/m3 and cp =0.434 kJ/kg◦C)coming off a furnace at 820 ◦C is to be quenched in an oil bath at 45 ◦C to atemperature of 51.1 ◦C. If the metal sheet is moving at a steady velocity of 10 m/min, determinethe required rate of heat removal from the oil to keep its temperature constant at 45 ◦C.

Solution:

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