BASIC CONCEPTS OF THERMODYNAMICS 53dharm
M-therm/th2-2.pm5
Since the free air boundary is contracting, the work done by the system is negative, and
the surroundings do positive work upon the system.
+ Example 2.17. A piston and cylinder machine containing a fluid system has a stirring
device as shown in Fig. 2.36. The piston is frictionless, and it is held down against the fluid due
to atmospheric pressure of 101.3 kPa. The stirring device is turned 9500 revolutions with an
average torque against the fluid of 1.25 Nm. Meanwhile the piston of 0.65 m diameter moves out
0.6 m. Find the net work transfer for the system.
Solution. Refer Fig. 2.36.Fig. 2.36
Work done by the stirring device upon the system,
W 1 = 2πNT
where T = torque = 1.25 Nm
N = number of revolutions = 9500
W 1 = 2π × 9500 × 1.25 = 74622 Nm = 74.622 kJ
This is negative work for the system.
Work done by the system upon the surroundings
W 2 = (pA). L
where, p = Pressure = 101.3 kPa
A = Area of the piston = π/4 × (0.65)^2 = 0.3318 m^2 , and
L = Distance moved by the piston = 0.6 m
W 2 = 101.3 × 0.3318 × 0.6 = 20.167 kJ
This is a positive work for the system.
Hence, the net work transfer for the system
Wnet = W 1 + W 2 = – 74.622 + 20.167 = – 54.455 kJ. (Ans.)
Example 2.18. A diesel engine piston which has an area of 45 cm^2 moves 5 cm during part
of suction stroke. 300 cm^3 of fresh air is drawn in from the atmosphere. The pressure in the
cylinder during suction stroke is 0.9 × 105 N/m^2 and the atmospheric pressure is 1.013 × 105 N/
m^2. The difference between the suction and atmospheric pressure is accounted for flow resistance
in the suction pipe and inlet valve. Find the net work done during the process.