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Chapter 1: Problem 58

The crankshaft of a single-cylinder air compressor rotates \(1800 \mathrm{rpm}\). The piston area is \(2000 \mathrm{~mm}^{2}\), and the piston stroke is \(50 \mathrm{~mm}\). Assume a simple "idealized" case where the average gas pressure acting on the piston during the compression stroke is \(1 \mathrm{MPa}\), and pressure during the intake stroke is negligible. The compressor is \(80 \%\) efficient. A flywheel provides adequate control of speed fluctuation. (a) What motor power \((\mathrm{kW})\) is required to drive the crankshaft? (b) What torque is transmitted through the crankshaft?

Short Answer

Expert verified
The motor power required to drive the crankshaft is 3.75 kW, and the torque transmitted through the crankshaft is 0.06 kNm

Step by step solution

01

Calculate Stroke Volume

Stroke Volume = Piston Area x Stroke Length = \(2000 \, \mathrm{mm}^{2} \) x \(50 \, \mathrm{mm} = 100000 \, \mathrm{mm}^{3}\) = 0.0001 \, \mathrm{m}^{3}\)
02

Calculate the Work done

Work \(W\) done during one cycle is the product of the change in pressure (assumed as an average) and the stroke volume. Here, \(W = P \Delta V\) = \(1 \, \mathrm{MPa} \) x \(0.0001 \, \mathrm{m}^{3}\) = \(100 \, \mathrm{J}\). That's the work done per cycle.
03

Calculate Cycles Per Second

Since the compressor runs at \(1800 \, \mathrm{rpm}\), its frequency of operation, \(f\), is \(f = 1800/ 60 = 30 \, \mathrm{Hz}\) or 30 cycles per second.
04

Calculate Theoretical Power

Power \(P\) is the work done per cycle multiplied by the number of cycles per second. So \(P = Wf = 100 \, \mathrm{J} \) x \(30 \, \mathrm{s}^{-1} = 3000 \, \mathrm{W} = 3 \, \mathrm{kW}\)
05

Calculate Actual Power

Actual Power \(P_{actual}\) required will be higher than the theoretical power due to inefficiency. If the compressor is \(80\%\) efficient, we can calculate actual power \(P_{actual} = P_{theoretical}/\text{efficiency} = 3 \, \mathrm{kW}/ 0.80 = 3.75 \, \mathrm{kW}\)
06

Calculate Angular Speed

Angular speed \(w\) (in radian per second) \(w = 2\pi f = 2\pi (30 \, \mathrm{s}^{-1}) = 60\pi \, \mathrm{rad/s}\)
07

Calculate Torque

Torque \(T\) can be calculated using formula \(T = P_{actual}/w = 3.75 \, \mathrm{kW}/60\pi \, \mathrm{rad/s}\ = 0.06 \, \mathrm{kNm}\)

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