Work Done in an Adiabatic Process

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1. WELCOME TO OUR PRESENTATION 1

2. Work Done In An Adiabatic Process Presented By: Imran Hossain Chowdhury Presented To: Mirza Mahbubur Rahman Senior Lecturer, Department of EEE 2

3. Overview Work Done . Adiabatic Process . Work Done D uring an Adiabatic P rocess Adiabatic Expansion & Compression A Problem & Solution Conclusion 3

4. Work Done Work is a measure of change of energy. The net work done equals the change in Kinetic Energy. Work is the integral of the scalar product (dot-product) of two vectors: Force and Displacement. "Displacement" means the change in position of the point at which the force is applied. The SI unit of the amount of Work done by a Force of one Newton acting over a displacement of one meter, and is called the joule (J), or Newton-meter (N-m). 4

5. Adiabatic Process A process in which there is no heat transfer to or from the system is known as adiabatic process, i.e., dQ=0 When expansion happens, temperature falls When gas is compressed, temperature rises. 5

6. Work Done During An Adiabatic Process 6

7. Adiabatic Expansion & Adiabatic Compression Adiabatic Expansion: Adiabatic expansion is a situation whereby an external work acts upon a system at the expense of utilizing internal energy of the gas and results in lowering the temperature of the molecules of gas. Adiabatic Compression: Adiabatic compression is a situation whereby an external work acts upon a system at the produce of utilizing internal energy of the gas and results in increasing the temperature of the molecules of gas. 7

8. A Problem & Solution Problem : In a motorcycle engine, after combustion occurs in the top of the cylinder, the piston is forced down as the mixture of gaseous products undergoes an adiabatic expansion. Find the average power involved in this expansion when the engine is running at 4000 rpm, assuming that the gauge pressure immediately after combustion is 15.0atm, the initial volume is 50.0cm 3 , and the volume of the mixture at the bottom of the stroke is 250cm 3 . Assume that the gases are diatomic and that the time involved in the expansion is one-half that of the total cycle. 8

9. A Problem & Solution Solution: Here, Initial volume of the gas, V i =50cm 3 =50×10 -6 m 3 Final Volume of the gas, V f =250cm 3 =250×10 -6 m 3 Adiabatic gas constant, γ=1.40 Initial pressure of the gas, P i =15atm+1atm =16atm =16×101325 Pa =1621200Pa 9

10. A Problem & Solution In an adiabatic process, The final pressure P f is defined as, The work done W for adiabatic process is defined as, 10

11. A Problem & Solution First we have to find out final pressure P f . To obtain P f , substitute 1621200Pa for P i , 50×10 -6 m 3 for V i , 250×10 -6 m 3 for V f and 1.40 for γ in the equation = 170325Pa 11

12. A Problem & Solution Now, we have to find out work done W by the gas during the expansion. For this, we need to substitute 1621200Pa for P i ,170325Pa for P f , 50×10 -6 m 3 for V i , 250×10 -6 m 3 for V f and 1.40 for γ in the equation = 96J 12

13. A Problem & Solution The above process happens 4000 times per minute, but the actual time to complete the process is one half of the cycle, or 1/8000 of a minute. Thus the average power P involved in this expansion when the engine is running at 4000 rpm will be, P=W/t Here, =96J/0.0075s W=96J =12800W t =(1/8000)min =(60/8000)s =0.0075s So, the average power involved in this expansion when the engine is running at 4000rpm would be 12800W . 13

14. Conclusion At the end of the presentation we came to know- Concept of adiabatic expansion & compression. Analysis of adiabatic process Use of work done formula in adiabatic process. 14

15. End Of Presentation 15