TITLE.PM5

54 ENGINEERING THERMODYNAMICS

dharm
M-therm/th2-2.pm5

Solution. Area of diesel engine piston

= 45 cm^2 = 45 × 10–4 m^2

Fig. 2.37

Amount of fresh air drawn in from atmosphere

= 300 cm^3 = 300 × 10–6 m^3

The pressure inside the cylinder during suction stroke

= 0.9 × 10^5 N/m^2

Atmospheric pressure = 1.013 × 10^5 N/m^2

Initial and final conditions of the system are shown in Fig. 2.37.

Net work done = Work done by free air boundary + work done on the piston

The work done by the free air = – ve because boundary contracts

The work done by the cylinder on the piston = + ve because the boundary expands

∴ Net work done = pdV pdV

Piston Free air

boundary

zz+

=××× × − ×××L

NM

O

QP

0 9 10 45 10−−^5

100 1013 10 300 10

..^54 5 6

= [20.25 – 30.39] = – 10.14 Nm or J.(Ans.)

Example 2.19. The properties of a closed system change following the relation between

pressure and volume as pV = 3.0 where p is in bar V is in m^3. Calculate the work done when the

pressure increases from 1.5 bar to 7.5 bar.

Solution. Initial pressure, p 1 = 1.5 bar

Final pressure, p 2 = 7.5 bar

Relation between p and V, pV = 3.0

Work done, W :