Answer to Question #217940 in Molecular Physics | Thermodynamics for Unknown346307

Question

If the compressor of example 5.1 is to be driven at 300 rev/min and is single-acting single-cylinder machine, calculate the cylinder bore required, assuming a stroke to bore ratio of 1.5:1. Calculate the power of the motor required to drive the compressor if the mechanical efficiency of the compressor is 85% and that of the motor transmission is 90%. Determine also the isothermal efficiency of the compressor.

Answer 1

Solution;

Volume delt with per minute at inlet =1m³/min

Volume drawn in per cycle=$\frac{1}{300}$ =0.00333m³/cycle, which is the volume in the cylinder.

Volume in cylinder is give by ;

$\frac{π}{4}×d^2×L$

d is the bore.

L is the stroke.

Given the L:d=1.5:1;

L=1.5×d

Equate the formula for volume to get bore;

$\frac{π}{4}×d^2(d×1.5)=0.003333m^3$

$d^3=0.002829m^3$

d=0.141437m

Answer;

d=141.4mm

Indicated power is 4.23kW,

Mechanical efficiency is 0.85;

Power input to compressor=$\frac{4.23}{0.85}$ =4.98kW

The Motor has a transmission of 0.9

Power input to the motor=$\frac{4.98}{0.90}$ =5.53kW

Isothermal power=mRTln$\frac{P_2}{P_1}$ =(1.226×0.287×288×ln$\frac{1}{1.013}$ )=196kJ/min

Indicated work=254kJ/min

Therefore;

Isothermal efficiency=$\frac{196}{254}$ =0.772 or 77.2%

Answer;

77.2%