A diesel engine piston which has an area of 45 cm² moves 5 cm during part of suction stroke. 300 cm³ of fresh air is drawn in from the atmosphere. The pressure in the cylinder during suction stroke is 0.9 × 10^5 N/m² and the atmospheric pressure is 1.013 × 10^5 N/m². The difference between the

Question

A diesel engine piston which has an area of 45 cm² moves 5 cm during part of suction stroke. 300 cm³ of fresh air is drawn in from the atmosphere. The pressure in the cylinder during suction stroke is [latex]0.9 imes 10^5[/latex] N/m² and the atmospheric pressure is [latex]1.013 imes 10^5[/latex] N/m². 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.

Accepted Answer

Area of diesel engine piston

= 45 [latex]cm ^2=45 imes 10^{-4}[/latex] [latex]m ^2[/latex]

Amount of fresh air drawn in from atmosphere

= 300 [latex]cm ^3=300 imes 10^{-6}[/latex] [latex]m ^3[/latex]

The pressure inside the cylinder during suction stroke

[latex]=0.9 imes 10^5[/latex] [latex]N / m ^2[/latex]

Atmospheric pressure [latex]=1.013 imes 10^5[/latex] [latex]N / m ^2[/latex]

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 = [latex]\int_{ ext {Piston }} p d V+\int_{\begin{array}{c} ext { Free air } \ ext { boundary }\end{array}} p d V[/latex]

[latex]=\left[0.9 imes 10^5 imes 45 imes 10^{-4} imes \frac{5}{100}-1.013 imes 10^5 imes 300 imes 10^{-6} ight][/latex]

[latex]=[20.25-30.39]=- 1 0 . 1 4[/latex] Nm or J.

Question 2.18: A diesel engine piston which has an area of 45 cm² moves 5 c...

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