Summer 2018 - Q.1 [Preview] - Grad Plus

Summer 2018 – Q.1 [Preview]

Q. 1 a) Explain isothermal efficiency of a reciprocating compressor. What methods are adopted for increasing the isothermal efficiency? [5M]
Answer: It compares the actual work done on the gas with isothermal compression work.
Isothermal efficiency is defined as the ratio of isothermal work input to actual work input during compression.
niso = Isothermal work input/Actual work input
Following methods are adopted for increasing the isothermal efficiency
1. Water spray: The water is injected into the cylinder during compression of air to keep the temperature of air constant. But this method has certain disadvantages and thus become obsolete.
2. Water jacketing: The water is circulated around the cylinder through the water jackets which helps to cool the air during compression.
3. External fins: For the small capacity compressor, the effective cooling can be achieved by attaching fins of conducting material around the cylinder. the fins increase the surface area of the cylinder for heat transfer.
4. Intercooler: The intercooler is used between two stages of compression for cooling of compressed air after one stage, before entering the next stage. The water jackets are also used around the cylinder of each stage of the compressor.
5. By suitable cylinder proportion: If the compressor has a larger surface to volume ratio, the greater surface area will be available for the heat transfer and cooling will be more effective.
It is possible by choosing a cylinder with a large bore and short piston stroke. The large cylinder head dissipates heat in a much more effective way, which contains the hottest compressed air all the time.

Q. 1 b) A single stage double acting air compressor delivers air at 7 bar pressure. The amount of free air delivered is 2 m3/min at 300 rpm. The pressure and temperature at the end of the suction stroke are 1 bar and 27 0C. The ambient conditions are 1.03 bar and 20 0C. The clearance is 5% of stroke.
(1) Power required to run the compressor if mechanical efficiency is 85%.
(2) Diameter and stroke of the cylinder if both are equal.
Assume the index of compression and expansion as 1.3. [8M]
P1 = 1 bar, P2 = 7 bar, T1 = 270C = 300 K, Vf = 2 m3/min, Pf = 1.03 bar, Tf = 200C = 293 K, Vc = 0.05 Vs, N = 300 rpm, nmech = 85%, nc = ne = 1.3

EC-2 Mechanical Nagpur

Single cylinder double acting/em>

clearance volume, Vc = V3 = 0.05 Vs
The volume V4 after re-expansion of compressed air in clearance space

$latex V_4=V_3\left(\frac{P_2}{P_1}\right)^\frac1{n_e}=0.05V_s\times\left(\frac71\right)^\frac1{1.3}=0.2233V_s$

The total volume of cylinder,
V1 = Vs + Vc = 1.05 Vs
Effective swept volume,
V1 – V4 = 1.05 Vs – 0.2233 Vs = 0.8267 Vs
The volumetric efficiency

$latex \eta_{vol}=\frac{V_1-V_4}{V_s}=\frac{0.8267V_s}{V_s}=0.8267$

Volumetric efficiency = 82.67%

i) Diameter and stroke of the cylinder:

As per given condition, the piston diameter is equal to stroke i.e. d = L
we know,

$latex V_s=\left(\frac\pi4\right)d^2L=\left(\frac{\mathrm\pi}4\right)d^3$

Vs is calculated as follow,

free air delivered per cycle is given as

$latex V_f=\frac{P_1T_f}{P_fT_1}\left(V_1-V_4\right)$

thus volume of air inducted per minute at suction condition

$latex \dot{V_1}-\dot{V_4}=\frac{P_1T_f}{P_fT_1}{\dot V}_f=\frac{1.03\times300}{1\times293}\times2=2.109\frac{m^3}{min}$

It is the volume sucked per minute which can be expressed as

$latex \dot{V_1}-\dot{V_4}=\eta_{vol}\times V_s\times N\times k$

$latex 2.109=0.8267\times V_s\times300\times2$ (k=2, for double acting)

Vs = 0.00425 m3

$latex V_s=\left(\frac{\mathrm\pi}4\right)d^3$


$latex \begin{array}{l}d=\sqrt[3]{\frac{V_s\times4}\pi}=\sqrt[3]{\frac{0.00425\times4}{3.14159}}=0.1756m\end{array}$

Diameter = stroke = 17.56 cm

ii) Power required if the mechanical efficiency is 85%

Vs = 0.00425 m3
V1 = 1.05 Vs = 1.05 x 0.00425 = 0.004462 m3

$latex V_4=0.2233V_s$ = 0.2233 x 0.00425 = 0.00095 m3

Indicated work input per cycle is given by

$latex W_{in}=\frac{n_c}{n_c-1}P_1V_1\left[\left(\frac{P_2}{P_1}\right)^\frac{n_c-1}{n_c}-1\right]-\frac{n_e}{n_e-1}P_1V_4\left[\left(\frac{P_2}{P_1}\right)^\frac{n_e-1}{n_e}-1\right]$

$latex W_{in}=\frac{1.3}{1.3-1}1\times10^5\times0.004462\left[\left(\frac{7}1\right)^\frac{1.3-1}{1.3}-1\right]-\frac{1.3}{1.3-1}1\times10^5\times0.00095\left[\left(\frac{7}1\right)^\frac{1.3-1}{1.3}-1\right]$

Win = 862.64 J/cycle = 0.86264 kJ/cycle
The indicated power input to the compressor,
IP = Win x (N/60)x 2 (for double acting)
IP = 0.86264 x (300/60) x 2 = 8.626 kW
$latex BP=\frac{IP}{\eta_{mech}}$ = 8.626/0.85 = 10.148 kW
Power required =10.148 kW

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