PC1431 Experiment P2 Laboratory Report

Name: Ang Zhi Ping (U066463H)Lab Group: C06ADate: 30th October 2006Experiment: P2 – Heat Engine

Aims of Experiment:

• Understanding the mechanism of a simple heat engine• Verifying the principle of energy conservation by comparison of

mechanical work done by the heat engine on objects (of varying masses), and the work done as measured from the engine’s P-V diagram

• Comparing the effect of an increase in the temperature difference between heat reservoirs on the engine’s performance

Equipment list:

• Heat engine, consisting of a hollow cylinder with a near frictionless movable graphite piston.

• Air chamber with rubber tubing• Various masses• Heat reservoirs of various temperatures (Melting ice temperature,

room temperature, preheated water 60oC-70oC and hot water 80oC-90oC)

• Low pressure gauge and rotary motion sensors, science workshop interface and with software

Diagrammatic setup:

Air filled cylinder

Graphite piston

Mass

Air chamber with hose

Water acting as temperature reservoir

Figure A

Experiment observations and results:

Section E: Procedure Part I

i. Transition ab: Add the mass to the platform.

The graphite piston is observed to be quickly compressed under the additional weight of the mass upon adding the mass. This process whereby the piston is quickly compressed is fast, thus, there is little time for the air in the cylinder to reach thermal equilibrium with the surrounding. The heat transferred out of the system is therefore approximately zero, and the process ab is approximately an adiabatic compression process.

ii. Transition bc: Place the can in the hot reservoir:

The piston (with mass) is observed as steadily rising, corresponding to expansion of the air within cylinder. As the piston rises with constant speed, the net force acting on the piston is zero. Thus, the force exerted by the pressure of air in cylinder is approximately constant. This corresponds to bc being approximately an isobaric expansion.

iii. Transition cd: Remove mass from the platform:

Upon removing the mass, the forces acting on the piston (refer to Figure B for force diagram) becomes unbalanced. The air pressure in the cylinder pushes up the piston as it is observed to move slightly higher within a short time. As with (i), the duration of this process is short, compared to the time the air within the cylinder to reach thermal equilibrium with the surrounding. The heat transferred out of the system is approximately zero, and this process can be modeled as an adiabatic expansion process.

iv. Transition da: Place the can in the cold reservoir:

The piston (with mass) is observed as steadily falling, corresponding to compression of the air within cylinder. As the piston falls with constant speed, the net force acting on the piston is zero. Thus, the force exerted by pressure of air in cylinder is approximately constant. Therefore, this process can be modeled as an isothermal compression.

Is the upper part of piston exposed to atmospheric pressure?

Air chamber

Wpiston

Patm

A

Pgas

AA: Cross section area of piston

Pgas

,Patm

: Pressures due to air in cylinder and atmosphere

Wpiston

: Weight of piston

Figure B

Section G: Calculations and discussion

1. Calculation work done by engine from P-V diagram

The net work done by engine in one cycle is computed by the area enclosed by the P-V curve (with a multiplicative factor of A, area of piston). The pressure of the air in cylinder during the period within the hot/cold reservoirs is assumed to be constant. The adiabats during which the mass is placed/removed is assumed to be parallel too. Since the piston moves equal distance during isobaric compression/expansion, the area in theP-V graph is calculated by the length of the base of P-V loop multiplied by the vertical height of the loop.

Correspondingly, the mechanical work done in raising the masses can be calculated by the formula mgy, where m is the mass and y is the vertical distance moved by the mass. This distance is the same as the length of the base of the P-V loop.

2. Tabulation of results

Nominal measured lukewarm water temperature = 64.9oCNominal measured hot water temperature = 84.3oCNominal measured water room temperature = 24.5oCA, area of piston = π(0.01625)2 m2 = 8.30 x 10-4 m2

g, acceleration due to gravity = 9.80 ms-2

Conditions of cycle

Temperatureof

reservoirs

Massused,m/kg

Distance moved by mass m,

y/m

Cold reservoir pressure,

Pc/kPa

Hot reservoir pressure,

Ph/kPa

Work done obtained from P-V diagram,

WPV=A(Ph-Pc)y/mJ

Mechanical work done,

Wm=mgy/mJ

Percent error,(WPV-

Wm)/Wmx100%

Cold reservoir: room temperature (~25oC)

Hot reservoir: Lukewarm (~65oC)

0.050 0.016 0.21 0.84 8.4 7.8 7.7

Cold reservoir: room temperature (~25oC)

Hot reservoir: Lukewarm (~65oC)

0.100 0.015 0.21 1.44 15 15 0.0

Cold reservoir: room temperature (~25oC)

Hot reservoir: Lukewarm (~65oC)

0.150 0.014 0.21 2.03 21 21 0.0

Cold reservoir: Melting point of ice

(~0oC)Hot reservoir: Hot

water (~85oC)

0.050 0.042 0.16 0.93 27 21 29

Cold reservoir: Melting point of ice

(~0oC)Hot reservoir: Hot

water (~85oC)

0.100 0.041 0.16 1.53 47 40 18

Cold Reservoir: Melting point of ice

(~0oC)Hot reservoir: Hot

water (~85oC)

0.150 0.041 0.16 2.10 66 60 10

3.

The results obtained from using a room temperature reservoir and a lukewarm reservoir yields close agreement between the work done calculated from the P-V graph and the mechanical energy calculated from mgy. But the results obtained from using 2 reservoirs of greater temperature difference yields considerable experimental error. This might be due to the following reasons:

• When using reservoirs of larger temperature difference, the masses moves through a larger distance. Thus, the speeds of the masses are not constant at all times during the process. This shows that the pressures during these times are not really constant. Thus the processes involved are not isobaric.

• Using reservoirs of larger temperature difference, the pressure difference due to the two reservoirs is greater than the corresponding measurements using room temperature and lukewarm reservoirs. Thus, this increase in pressure difference will cause air in the cylinder to leak at a faster rate. This leads to inaccurate measurement of the P-V diagram.

• Friction between piston and cylinder might account for the disagreement between calculated mechanical energy and area in P-V diagram.

4.

The work done in Part III is greater for any given masses corresponding to Part II. This means that given a heat engine operating two heat reservoirs of larger temperature difference is capable of doing more work in a thermodynamic cycle

Conclusion

• The heat engine is a device which converts thermal energy partly into useful work energy and expels remaining thermal energy into a cold reservoir

• Heat engines rarely operate to their maximum possible efficiency due to leakage of thermodynamic medium from the engine and friction between moving parts in the engine

• A heat engine operating between two heat reservoirs with a larger temperature difference is capable of doing more useful work.

Appendix

Graph 1: Plots of Run #2 (m=50g), Run #3 (m=100g) and Run #7(m=150g) For Part II;Graph 2: Plots for Run #1 (m=50g),Run #6 (m=100g) and Run#8 (m=150g) For part III.