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Thermodynamics Problem Solver-Ralph Pike 2012-04-27 REA's Thermodynamics Problem Solver Each Problem Solver is an insightful and essential study and solution guide chock-full of clear, concise problem-solving gems. Academia.edu is a platform for academics to share research papers. Thermodynamics Problems And Solutions Pdf. It uses more everyday language in its listing, and the analyzer is now better organized and easier to follow. The latest version of MBSA also features an updated, more user-friendly interface than previous versions. MBSA 2.1, the most current version, is available from Microsoft (see the. The First Law of Thermodynamics Problems and Solutions - Free download as PDF File (.pdf), Text File (.txt) or read online for free. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system. Known: Heat (Q) = +3000 Joule Work (W) = +2500 Joule Wanted: the change in.

1. Two kg of air at 500kPa, 80°C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100kPa and 5°C.

For this process, determine

1)the maximum work

2)the change in a availability and

3)the irreversibility.

Take, Cv = 0.718 KJ/kg K, R = 0.287 KJ/kg K.

2.A reversible heat engine receives 3000 KJ of heat from a constant temperature source at 650 K . If the surroundings is at 295 K,

determine

i)the availability of heat energy

ii)Unavailable heat.


A = Q1 –T0 (ΔS)

=3000 –295 (3.17)

=2064.85 KJ.

Unavailable heat (U.A) = T0 (ΔS)

=295 (3.17)

=935.15 KJ.

Result:

1)The availability of heat energy (A) = 2064.85 KJ.

2)Unavailable heat (U.A) = 935.15 KJ.

3. Air in a closed vessel of fixed volume 0.15 m3 exerts pressure of 12 bar at 250 °C. If the vessel is cooled so that the pressure falls to 3.5 bar , determine the final pressure, heat transfer and change of entropy.

Given Data:

Leap motioneffective curriculum ideas. V1 = 0.15 m3

p1 = 12 bar = 1200 KN/m2 p2 = 3.5 bar = 350 KN/m2

T1 = 250°C = 273+250 = 523 K

To find:

1)The final pressure,

2)Heat transfer

3)Change of entropy.

Solution:


Result:

1)The final pressure, T2 = 152.54 K.

2)Heat transfer, Q = - 446.78 KJ.

3) Change of entropy,-1.06KJ/K. ΔS =

4. A domestic food freezer maintains a temperature of - 15°C. The ambient air is at 30°C. If the heat leaks into the freezer at a continuous rate of 1.75 KJ/s, what is the least power necessary to pump the heat out continuously?

Given Data:

TL = - 15°C = 273 –15 = 258 K

TH = 30°C = 273 + 30 = 303 K

QS = 1.75 KW

To find:

Least power, (W)

Solution:


Result:

Least power necessary to pump heat, W = 0.305 KW.

5. A refrigerator working on reversed Carnot cycle requires 0.5 KW per KW of cooling to maintain a temperature of -15°C. Determine the following:

a)COP of the refrigerator

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b)Temperature at which heat is rejected and

Amount of heat rejected to the surroundings per KW of cooling.

Given Data:

W = 0.5 KW

Q2 = 1 KW

T2 = -15°C = 273 –15 = 258 K.

To find:

1)COP of the refrigerator (COP)

2)Temperature at which heat is rejected (T1)

3)Amount of heat rejected to the surroundings per KW of cooling (Q1)

Solution:



Result:

1)COP of the refrigerator (COP) = 2

2)Temperature at which heat is rejected (T1) = 387 K

Amount of heat rejected to the surroundings per KW of cooling (Q1) = 1.5 KW.


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 :

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)

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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

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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

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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

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High temperature (TH) = 800 Kelvin

Low temperature (TL) = 300 Kelvin

Wanted: heat discharged by the engine

Solution :

Carnot (ideal) efficiency :

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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 :

Efficiency (e) = 40% = 40/100 = 0.4

High temperature (TH) = 727oC + 273 = 1000 K

Wanted : Low temperature

Solution :

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.

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 :

Work was done by Carnot engine :

W = e Q1

W = (1/3)(600) = 200 Joule