Thermodynamics Problems And Solutions Pdf
Thermodynamics and Chemistry Second Edition Version 5, May 2014 Howard DeVoe Associate Professor of Chemistry Emeritus University of Maryland, College Park, Maryland. First Law Of Thermodynamics Problems And Solutions Download Thermodynamics Problems With Solutions book pdf free download link or read online here in PDF. 4 Third Law 54 3. Thermodynamics Problems And Solutions Pdf engineering thermodynamics problems and solutions Substituting andmultiplying by the factor 109 for the density unity kg/km3, the. File Type PDF Thermodynamics Problems With Solutions Thermodynamics Problems With Solutions If you ally craving such a referred thermodynamics problems with solutions ebook that will offer you worth, acquire the categorically best seller from us currently from several preferred authors. Download Thermodynamics Problems With Solutions book pdf free download link or read online here in PDF. Read online Thermodynamics Problems With Solutions book pdf free download link book now. All books are in clear copy here, and all files are secure so don't worry about it. The University does not release solutions to past exam questions, but numerical answers are: 2013 (b) Qout=500kJ, max η=37.1% 2014 (a) 57.3%, (b) 54.0%. School of Engineering, University of Edinburgh Engineering Thermodynamics 2 and Thermodynamics (Chemical) 2. Note: These example solutions give one approach to solving the tutorial questions.
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1. Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II?
Known :
Process 1 :
Pressure (P) = 20 N/m2
Initial volume (V1) = 10 liter = 10 dm3 = 10 x 10-3 m3
Final volume (V2) = 40 liter = 40 dm3 = 40 x 10-3 m3
Process 2 :
Acces PDF Thermodynamics Problems And Solutions Thermodynamics Problems And Solutions As recognized, adventure as capably as experience more or less lesson, amusement, as competently as pact can be gotten by just checking out a ebook thermodynamics problems and solutions then it is not directly done, you could take on even more approximately this life, as regards the world. Thermodynamics – problems and solutions. The first law of thermodynamics. Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II? Known: Process 1: Pressure (P) = 20 N/m 2. Initial volume (V 1) = 10 liter = 10 dm 3 = 10 x 10-3 m 3.
Process (P) = 15 N/m2
Initial volume (V1) = 20 liter = 20 dm3 = 20 x 10-3 m3
Final volume (V2) = 60 liter = 60 dm3 = 60 x 10-3 m3
Wanted : The ratio of the work done by gas
Solution :
The work done by gas in the process I :
W = P ΔV = P (V2–V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10-3 m3) = 0.6 m3
The work done by gas in the process II :
W = P ΔV = P (V2–V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10-3 m3) = 0.6 m3
The ratio of the work done by gas in the process I and the process II :
0.6 m3 : 0.6 m3
1 : 1
2.
Based on the graph below, what is the work done by helium gas in the process AB?
Known :
Pressure (P) = 2 x 105 N/m2 = 2 x 105 Pascal
Initial volume (V1) = 5 cm3 = 5 x 10-6 m3
Final volume (V2) = 15 cm3 = 15 x 10-6 m3
Wanted : Work done by gas in process AB
Solution :
W = ∆P ∆V
W = P (V2 – V1)
W = (2 x 105)(15 x 10-6 – 5 x 10-6)
W = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5)
W = 2 Joule
3.
Based on the graph below, what is the work done in process a-b?
Known :
Initial pressure (P1) = 4 Pa = 4 N/m2
Final pressure (P2) = 6 Pa = 6 N/m2
Initial volume (V1) = 2 m3
Final volume (V2) = 4 m3
Wanted : work done I process a-b
Solution :
Work done by gas = area under curve a-b
W = area of triangle + area of rectangle
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = 10 Joule
4. Based on graph below, what is the work done in process A-B-C-A.
Solution :
Work (W) = Area of the triangle A-B-C
W = ½ (20-10)(6 x 105 – 2 x 105)
W = ½ (10)(4 x 105)
W = (5)(4 x 105
)W = 20 x 105
W = 2 x 106 Joule
5. An engine absorbs 2000 Joule of heat at a high temperature and exhausted 1200 Joule of heat at a low temperature. What is the efficiency of the engine?
Known :
Heat input (QH) = 2000 Joule
Heat output (QL) = 1200 Joule
Work done by engine (W) = 2000 – 1200 = 800 Joule
Wanted : efficiency (e)
Solution :
e = W / QH
e = 800/2000
e = 0.4 x 100%
e = 40%
6. An engine absorbs heat at 960 Kelvin and the engine discharges heat at 576 Kelvin. What is the efficiency of the engine.
Known :
High temperature (TH) = 960 K
Low temperature (TL) = 576 K
Wanted: efficiency (e)
Solution :
Efficiency of Carnot engine = 0.4 x 100% = 40%
7. Based on the graph below, work done by the engine is 6000 Joule. What is the heat discharged by engine each circle?
Known :
Work (W) = 6000 Joule
High temperature (TH) = 800 Kelvin
Low temperature (TL) = 300 Kelvin
Wanted: heat discharged by the engine
Solution :
Carnot (ideal) efficiency :
Heat absorbed by Carnot engine :
W = e Q1
6000 = (0.625) Q1
Q1 = 6000 / 0.625
Q1 = 9600
Heat discharged by Carnot engine :
Q2 = Q1 – W
Q2 = 9600 – 6000
Q2 = 3600 Joule
8. The efficiency of a Carnot engine is 40%. If heat absorbed at 727°C then what is the low temperature.
Known :
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Efficiency (e) = 40% = 40/100 = 0.4
High temperature (TH) = 727oC + 273 = 1000 K
Wanted : Low temperature
Solution :
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TL = 600 Kelvin – 273 = 327oC
9. Based on graph below, if the engine absorbs 800 J of heat, what is the work done by the engine.
Known :
High temperature (TH) = 600 Kelvin
Low temperature (TL) = 250 Kelvin
Heat input (Q1) = 800 Joule
Wanted: Work (W)
Solution :
The efficiency of Carnot engine :
Work was done by the engine :
W = e Q1
W = (7/12)(800 Joule)
W = 466.7 Joule
10. The high temperature of a Carnot engine is 600 K. If the engine absorbs 600 J of heat and the low temperature is 400 K, what is the work done by the engine.
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Known :
Low temperature (TL) = 400 K
High temperature (TH) = 600 K
Heat input (Q1) = 600 Joule
Wanted: Work was done by Carnot engine (W)
Solution :
The efficiency of the Carnot engine :
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Work was done by Carnot engine :
W = e Q1
W = (1/3)(600) = 200 Joule