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Argon is compressed in a polytropic process

03/12/2021 Client: muhammad11 Deadline: 2 Day

ME 3310/5310 Thermodynamics I - Summer 2015 WRIGHT STATE UNIVERSITY

Department of Mechanical and Materials Engineering

Homework #3 Assigned 5/28/2015

Due 6/18/2015

Worth 90 Points Problems involving Specific Heats, ∆u, ∆h Problem 4-57 (10 pnts) – Consider the cases of neon and argon behaving as Ideal gases so that their specific heats are functions of temperature only. The temperature of the neon gas is increases from 20 °C to 180 °C. (a) Calculate the change in the specific internal energy of the neon, in kJ/kg. (b) Repeat this calculation for argon. Why is the value for neon larger than that of argon? Answers: (a) neon: 98.9 kJ/kg, (b) argon: 50.0 kJ/kg Bob Problem 1 (10 pnts) – Consider the cases of argon and neon behaving as Ideal gases so that their specific heats are functions of temperature only. (a) Calculate the change in the specific enthalpy of argon, in kJ/kg, when it is cooled from 400 °C to 100 °C during a constant pressure process. (b) Repeat this calculation for neon. Why is the value for neon larger than that of argon? Answers: (a) argon: 156.1 kJ/kg, (b) neon: 309.0 kJ/kg Bob Problem 2 (5 pnts) – A granite block is heated from 50 °F to 80 °F. The change in internal energy of the granite block is ∆U = 145.8 Btu. Determine the mass of the block (lbm). Use an average specific heat value for granite obtained from the tables in the text. Answer: 20 lbm Problem 4-60 (10 pnts) – Consider hydrogen (H2) behaving as an Ideal gas so that its specific heats are functions of only temperature. Determine the specific internal energy change ∆u of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A-2c), (b) the cv value at the average temperature (Table A-2b), and (c) the cv value at room temperature (Table A-2a). Answers: (a) 6194 kJ/kg, (b) 6233 kJ/kg, (c) 6110 kJ/kg Moving Boundary Work and Polytropic Process Problems Problem 4-23 (5 pnts) – A piston-cylinder device initially contains 0.25 kg of nitrogen gas at 130 kPa and 180 °C. The nitrogen gas is now expanded isothermally to a pressure of 80 kPa (the gas System performs moving boundary work ON the piston). First, derive the following expression for the “pdV” moving boundary work:

pdV 2out 1 1 1

V W p V ln

V  

=    

Next, compute the value for the moving boundary work done BY the System ON the piston during this process. Assume that the nitrogen behaves as an Ideal gas during the entire process. Answer: 16.3 kJ

Problem 4-7 (15 pnts) Modified – A piston-cylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120 °C. The nitrogen gas is now expanded polytropically to a state of 100 kPa and 100 ° C (the gas System performs moving boundary work ON the piston). (a) First, determine the polytropic exponent, n. Second, derive the following expression for the “pdV” moving boundary work:

pdV 2 2 1 1out p V p V

W n 1 −

= −

(c) Finally, compute the value for the moving boundary work done BY the System ON the piston during this process. Assume that the nitrogen behaves as an Ideal gas during the entire process. Hint: To find the polytropic exponent, n, first compute the volume at state 2 (0.08637 m3). Next, write down the general expression for a polytropic process from State 1 to State 2. Take the natural logarithm of both sides of this equation and solve for n, algebraically. Then plug numbers into the equation to compute the value of n….if you don’t follow these steps you are more likely to make a mistake. Answer: (a) n = 1.249, (c) Moving “pdV” Boundary Work = 1.86 kJ. Bob Problem 3 (20 pnts) – A piston-cylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120 °C. The nitrogen gas is now expanded polytropically to a pressure of 100 kPa with a polytropic exponent having a value equal to the specific heat ratio (this process is called an isentropic expansion---we will learn this in Chapter 7). Assume that the nitrogen behaves as an Ideal gas. Hint: the specific heat ratios for gases can be found in Table A-2a

(a) Derive the following expression for the final temperature at State 2:

2 2 2

p V T

mR =

where m is the mass of the System and R is the specific gas constant for nitrogen.

(b) Use the polytropic relations to derive the following expression for the volume at State 2,

1 k

1 2 1

2

p V V

p  

=    

,

and compute this volume using the given information.

(c) Compute the mass of the System and then compute the final temperature, T2, in Celsius.

(d) Compute the “pdV” moving boundary work output done BY the System ON the piston.

Answers: (b) 0.08443 m3, (c) 0.07802 kg, 91.6 °C, (d) 1.64 kJ

Bob Problem 4 (15 pnts) – Argon gas is compressed in a polytropic process having n=1.2 from 120 kPa and 30 °C to 1200 kPa in a piston-cylinder device. The mass of argon is 0.4377 g and the initial volume of the piston-cylinder is 230 cm3. (a) Compute the final State temperature, T2, of the argon (°C), (b) compute the “pdV” moving boundary work done BY the piston ON the System (J) during this process, and (c) compute the final State volume (cm3). Assume that the argon always behaves as an Ideal gas. Answers: (a) 171.7 °C, (b) 64.53 J, (c) 33.8 cm3

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