1. Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering Summary of Energy Topics Chapter 1: Thermodynamics / Energy Introduction Chapter 2: Systems & Processes Chapter 3: Work, Energy, Temperature & Heat Chapter 4: Work Processes of Closed Systems Chapter 5: Thermodynamic Properties Chapter 6: Steam Tables Chapter 7: Ideal Gases Chapter 8: Conservation of Mass & Energy Chapter 9: 1 st Law of Thermodynamics Chapter 10: Steady Flow Energy Equation Chapter 11: Heat Engines and Reversibility Chapter 12: 2 nd Law of Thermodynamics Chapter 13: Entropy Chapter 14: General Energy

2. Chapter 4: Work Processes of Closed Systems NORMAL OR DISPLACEMENT WORK PROCESSES OF CLOSED SYSTEMS Work is normally Force times Distance. An equilibrium normal or displacement work process of a closed system can be shown on a p-V equilibrium diagram (Red area below process curve) Weights added to increase pressure thus increase volume, as temperature increases

3. Chapter 4: Work Processes of Closed Systems Special cases CONSTANT PRESSURE PROCESS CONSTANT VOLUME PROCESS Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering, dV=0  constant

4. Chapter 4: Work Processes of Closed Systems HYPERBOLIC PROCESS (where pV are constant, i.e. change at the same rate)

5. Chapter 4: Work Processes of Closed Systems POLYTROPIC PROCESS (associated to the compression or expansion of a gas) PV n =C

6. Chapter 4: Work Processes of Closed Systems EXAMPLE 4.1 A piston-cylinder device containing a gas has a mass of 60 kg and cross sectional area of 0.04 m 2, shown below. The local atmospheric pressure is 0.97 bar and the gravitational acceleration is 9.81 m/s 2. Determine the pressure inside the cylinder in MPa? Solution Assume friction to be negligible. The pressure in the piston is a combination of atmospheric pressure and the weight (pressure) of the piston. Pressure of the piston = Force or weight / Area = mg/A P piston = P atm + mg/A N.B. Remember: 1 bar = 100000 Pa or 10 5 Pa or 0.1 MPa (where M = 10 6 ) P atm = 0.97 bar = 0.097MPa So P piston = P atm + mg/A ; therefore P piston = 0.097 [MPa] + (60 [kg]. 9.81 [m/s 2 ]) / (0.04 [m 2 ]) P piston = 0.097 x10 6 [Pa] + 14715 [kg /ms 2 = Pa] = 111715 Pa = 0.11172 MPa

7. Chapter 4: Work Processes of Closed Systems EXAMPLE 4.2 Air at a pressure of 0.1 MPa is enclosed by a cylinder and piston. Its volume is reduced from 1 litre to 1/8 litre in a polytropic process with an exponent n of 1.4. The pressure is then increased at constant volume to 4.8 MPa. The air is brought back to its original volume in another polytropic process with the same exponent. It is then brought to the initial pressure in a constant volume process. Sketch the cycle on a p-V diagram. Calculate the work for each process and calculate the average power output if there are 3,000 cycles per minute. Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering (1)(2)(3)

8. Chapter 4: Work Processes of Closed Systems Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering 0.1 MPa

9. Chapter 4: Work Processes of Closed Systems Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering

10. Chapter 4: Work Processes of Closed Systems Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering

11. Chapter 4: Work Processes of Closed Systems Because the vol is constant – see special cases 3000

12. Chapter 4: Work Processes of Closed Systems SHEAR OR SHAFT WORK PROCESSES OF CLOSED SYSTEMS EXAMPLE 4.3 Suppose a mass of 10 kg descends at a steady rate of 2 m/s while doing work on the system. Take the acceleration due to gravity to be 9.81 m/s 2. What is the rate at which stirring work is done on the system? Dr. Owen Clarkin School of Mechanical & Manufacturing Engineering

13. Chapter 4: Work Processes of Closed Systems (Energy Dissipation Example of this application; to slow down fighter planes landing on navy ships – the wire turns the paddle wheel in water to decelerates the plane)