This document contains solved problems from chapter 12 on positive displacement machines. Problem 12.1 calculates the indicated power and delivery temperature for air compression in a single-stage reciprocating compressor under isentropic, isothermal and polytropic processes. Problem 12.2 calculates the bore size required for the compressor running at 1000 rpm with a stroke to bore ratio of 1.2:1. Problem 12.3 calculates various parameters like bore, stroke, volumetric efficiency and indicated power for a single-stage single-acting air compressor running at 1000 rpm.

1. Assignment of Power Plant-I Designed by Sir Engr. Masood Khan
SOLVED PROBLEMS OF CHAPTER # 12
TITLE: POSITIVE DISPLACEMENT MACHINES

2. Positive Displacement Machines Designed by Sir Engr. Masood Khan
PROBLEM: 12.1:
Air is to be compressed in a single-stage-reciprocating
compressor from 1.013 bar and 15 C to 7 bar. Calculate the
indicated power required for free air delivery of 0.3
m3
/min., when the compression is as follows:
(i) isentropic;
(ii) reversible isothermal;
(iii) Polytropic, with n = 1.25.
What will be the delivery temperature in each case?
GIVEN:
Working Fluid = Air
Single Stage Compression
Initial Pressure P1 = 1.013 bar
Delivery Pressure P2 = 7 bar
Initial Temperature T1 = 150
C=288K
Free Air Delivery = FAD = 0.3m3
/min
REQUIRED:
1. Indicated Power for Compression.
1.1. Insentropic =
1.2. Reversible Isothermal =
1.3. Polytropic, n = 1.25.
2. Delivery Temperature.
SOLUTION:
As induction & FAD are same,
So FAD = Volume Induced, V = 0.3 m3
/min
(1.1) For Insentropic Process:
Pi = (r/r-1) P1V {(P2/P1)r-1
/r
-1} (1)
Putting values, we get: Pi = 1.31 KW
Now T2 =? As (T2/T1) = (P2/P1)r-1
/r
Putting the values, we have: T2 = 227.3 C
(1.2) For Isothermal Process
Pi = P1V1 ln(P2/P1) = 0.98 KW
T2 = T1 = 15C
(1.3) For Polytropic Process,
Pi = (n/n-1) P1V1 [(P2/P1)n-1
/n
–1] = 1.2 KW
(2) Delivery Temperature.
T2 = T1 (P2/P1)n-1
/n
= 423.9K = 150.9C
PROBLEM: 12.2:
The compressor of problem 12.1 is to run at 1000 rpm. If
the compressor is sigle acting and has a stroke / bore ratio
of 1.2/1, calculate the bore size required.
GIVEN: As Problem # 1
Speed = N = 1000 rpm = 1000 / 60 rps.
Stroke to Bore Ratio = l/d = 1.2
REQUIRED:
Bore Size = d =?
SOLUTION:
As Cylinder Volume = Vol. Induced /cycle.
(Πd2
/ 4)x L = V / Cycle.---------------------------(A)
As V/cycle = volume per unit time / Cycle per unit
time = Vo
/ N.
From Previous Problem, Vo
= 0.3 / 60 m3
per second.
So, V/cycle = Vo
/ N = 0.3 /1000 = 0.0003 m3
So Πd2
L / 4 = 0.0003
L /d = 1.2 => L = 1.2 d
Πd2
x1.2d / 4 = 0.0003 =>1.2 d3
= 4 x 0.0003 / π
So, Bore Size = d = 0.0683m = 68.3 mm

3. Positive Displacement Machines Designed by Sir Engr. Masood Khan
PROBLEM: 12.3:
A single-stage, single-acting air compressor running at
1000 rpm delivers air at 25 bar. For this purpose the
induction and free air conditions can be taken as 1.013 bar
and 15 C, and the FAD as 0.25 m3
/min. The clearance
volume is 3% of the swept volume and the stroke/bore ratio
is 1.2/1. Calculate:
(i) the bore and stroke;
(ii) the volumetric efficiency;
(iii) the indicated power;
(iv) The isothermal efficiency.
Take the index of compression and re-expansion as 1.3.8
GIVEN:
Speed = N = 1000 rpm.
Delivery Pressure = P2 = 25 bar
Induction & FAD are:
P = P1 = 1.013 bar, T= T1 = 15 C
FAD = Vo
= 0.25 m3
/min
Vc = 0.03 Vs, L /d = 1.2
REQUIRED:
Bore = d =? Stroke = L =?
Volumetric Efficiency = ηv =?
Isothermal Effeiency = ηiso =?
Indicated Power = Pi =?
SOLUTION:
We know that:Swept Volume = Volume of Stroke
Vs = π d2
L / 4, L = 1.2d So, Vs = π d2
1.2 d / 4
d = (Vs x 4 / 1.2 π)1/3
----------(1)
Vo
= Va – Vd = 0.25 m3
/ min
(Va – Vd) / cycle = V’
/ N = 0.25/1000
Va – Vd = 0.25 x 10-3
m3
/cycle.
Va = Vs + Vc = Vs + 0.03 Vs = 1.03 Vs
Now Vd = Vc(P2/P1)1/n
-----------------(2)
So, Vd = 0.3533 Vs
From (2) Va – Vd = 0.25x10-3
m3
/cycle
Vs = 0.3694 x 10-3
m3
/ cycle
Putting in (1): d = 73.16 mm
L/d = 1.2 ⇒ L = 1.2 d = 87.8 mm
Volumetric Efficiency:
ηv = FAD / Vs ----------------------------(3)
FAD / cycle = (FAD/min) / (cycles / min)
= 0.25/1000 = 0.25 x 10-3
m3
/cycle
So, ηv = 0.676 = 67.6 %
Now Pi = (n/n-1) P1Vo
[(P2/P1)n-1
/n
–1] = 2 KW
Isothemal Efficiency:
ηiso = Isothermal Power / Indicated Power = Piso / Pin.
Piso = P1Vo
ln Pi / P1 = 1.353 KW
Putting values:ηiso = 67.67 %
PROBLEM 12.4:
The compressor of problem 12.3 has actual induction
conditions of 1 bar and 40 C, and the delivery pressure is
25 bars. Taking the bore and stroke as calculated in
problem 12.3, calculate the FAD Referred to 1.013 bar and
15 C and the indicated power required. Calculate also the
volumetric efficiency and compare it with that of 12.3.
GIVEN:
As in Problem 12.03
Induction Conditions
P1 = 1 bar T1 = 40C = 313K
P2 = 25 bar L/d = 1.2

4. Positive Displacement Machines Designed by Sir Engr. Masood Khan
REQUIRED:
(i). FAD =? (ii) Pi =? (iii) ηv =?
SOLUTION:
(i) Mass delivered or mass of FAD = Mass of volume induced
mo
(FAD) = mo
Vo
= (P x FAD) / R.T. = P1 V’
/ R T1
FAD = 0.227 m3
/min
(ii) Pi = (n/n-1)m’
RT1{(P2/P1)n-1/n
–1}
m’
= P1 V’
/ R T1 = 4.63 x 10-3
kg / sec.
So Pi = 1.985 KW
(iii) ηv = (FAD/cycle) / Vs ---------------------(A)
FAD/cycle = (FAD/unit time) / (cycles/unit time)
FAD/cycle = 0.227 x 10-3
m3
So equ: (A) ⇒ ηv = 61.4 %
PROBLEM 12.5:
A single-acting compressor is required to deliver air at 70
bar from an induction pressure of 1 bar, at the rate of 2.4
m3
/min measured at free air conditions of 1.013 bar and 15
C. The compression is carried out in two stages with an
ideal intermediate pressure and complete intercooling. The
clearance volume is 3% of the swept volume in each
cylinder and the compressor speed is 750 rpm. The index of
compression and re-expansion is 1.25 for both cylinders
and the temperature at the end of the induction stroke in
each cylinder is 32 C. The mechanical efficiency of the
compressor is 85%. Calculate:
(i) the indicated power required;
(ii) the saving in power over single-stage compression
between the same pressures;
(iii) the swept volume of each cylinder;
(iv) The required power output of the drive motor.
GIVEN:
Delivery Pressure = P2 = 70 bar
Induction Pressure = P1 = 1 bar
FAD = 2.4 m3
/min at P = 1.013 bar
T = 288 K Two Stage Compression
(i) Ideal Intermediate Pressure, Pi/P1 = P2/Pi
(ii) Complete Intercooling ⇒ T1 = T
(iii) Clearance Volume is 3% of Swept Volume
i.e. Vc
L
= 0.03 Vs
L
, Vc
H
= 0.03 Vs
H
As Intercooling is Complete & Clearance Ratio is same, we
can say that; Vs
L
/ Vs
H
= Pi / P1 = P2 / Pi
⇒ Pi
2
= P1 P2 => P2 / P1 = √(P2/P1)
Speed = N = 750 rpm, n = 1.25
Temp. at the end of Induction Stroke in each cylinder.
T1 = Ti = 32C = 305K, ηmech. = 85 %
REQUIRED:
(i) Indicated Power = Pi =?
(ii) The savings in power over the single compressor
b/w the same pressures.
(iii) Swept volume of each cylinder = Vs=?
(iv) Power output of the driver motor =?
DIAGRAM:
SOLUTION:
(i) As the intermediate pressure is ideal & Intercooling is
complete, thus minimum work conditions are,
Hence; Pi = 2x(n/n-1)mo
RT1{(P2/P1)n-1/2n
–1} -----(A)
But mo
=?
Mass induced = Mass of FAD
mo
= mo
FAD = P x FAD / R T = 2.94 kg/min
(A)⇒ Pi = 22.7 KW

5. Positive Displacement Machines Designed by Sir Engr. Masood Khan
(ii) Saving Power = {(indicated power of single stage)-
(indicated power of double stage)}
For single stage,
Pi = (n/n-1)mo
RT1{{P2/P1)n-1/n
–1} = 28.7 KW
Thus Saving Power = 28.7 – 22.7 = 6 KW.
(iii) Swept Volume Vs
L
& Vs
H
for LP Stage.
Va – Vd / cycle = (Va-Vd/unit time)/cycles /unit time
= Vo
Lp / N --------------------------------------------- (1)
Vo
Lp = m’
RT1/P1 = 2.57 m3
/min
(1)⇒Va – Vd / cycle = V’
Lp / N
Va – Vd = 3.43 x 10-3
m3
/cycles
Va = VsL + Vc = VsL + 0.03 VsL = 1.03 VsL
Vd =Vc(Pi/P1)1/n
= 01641 VsL
VsL = 0.00396 m3
For HP Stage, Vs
H
= Va
/
- Vc
/
Vs
H
= Va
/
- 0.03 Vs
H
Va
/
= 1.03 Vs
H
=>Vs
H
= Va
/
/ 1.03---------------------(2)
As point a & a/
touches the Isothermal line due to complete
Intercooling, thus; Pi Va
/
= P1 Va
Va
/
= Va P1/Pi = Va / (Pi/P1) ----------------------------(3)
Va = 1.03 Vs = 1.03 x 0.003962 = 4.08 x 10-3
m3
/cycle
Equ:(3)⇒ Va
/
= 0.0004877 m3
/ cycle
Now Vs
H
= Va
/
/1.03 = 0.00473 m3
So Swept Volume of HP stage = 0.000473 m3
(iv) Power output of motor = Shaft Power
Shaft Power = Indicated Power / ηmech.= 22.7 / 6.85
Power output of Motor = 26.71 KW
PROBLEM 12.7:
A single-cylinder, single-acting air compressor of 200 mm
bore by 250 mm stroke is constructed so that its clearance
can be altered by moving the cylinder head, the stroke
being unaffected.
(a) using the data below calculate:
(i) the free air delivery;
(ii) The power required from the drive motor.
Data Clearance volume set at 700 cm3
; rotational speed,
300rpm; delivery pressure, 5 bar; suction pressure and
temperature, 1 bar and 32 C; free air conditions, 1.013 bar
and 15 C; index of compression and re-expansion, 1.25;
mechanical efficiency, 80%.
To what minimum value can the clearance volume be
reduced when the delivery pressure is 4.2 bar, assuming
that the same driving power is available and that the suction
conditions, speed, value of index, and mechanical
efficiency, remain unaltered?
GIVEN:
Single Stage Compressor
d = 200 mm = 0.2 m, L = 250 mm = 0.25 m
N = 300 rpm, n = 1.25
Vc = 700 cm3
= 700 x 10-6
m3
P2 = 5 bar, P1 = 1 bar, P = 1.013 bar
T1 = 32C = 305K, T = 288K
REQUIRED:
(i) FAD =? (ii) Power required, P = ?
When ηmech. = 80 %
SOLUTION:
As Vs = (π/4) d2
L
⇒Vs = 7.85 x 10-3
m3
, Vc = 7 x 10-4
m3
Va = Vs + Vc = 7.85
Vo
= FAD = (Va-Vd)x(T/T1)x(P1/P)
Vd = Vc(P2/P1)1/n
= 2.54 x 10-3
m3

6. Positive Displacement Machines Designed by Sir Engr. Masood Khan
Va – Vd = 6.01 x 10-3
m3
V = (Va – Vd)x(T/T1)x(P/P1) = 5.61 x 10-3
m3
/cycle
V/min = 300 x 5.61 x 10-3
FAD = 1.68 m3
/min
(ii) Power = (n/n-1)m’
R(T2 – T1) --------------(1)
Where mo
= PVo
/RT = 2.068 Kg/min
So, equ: (1)⇒ Indicated Power = Pi = 5.7 KN
As we know that
Power required to drive the motor
= Indicated Power / ηmech.= 7.2 KW
PROBLEM 12.8:
A single acting, single-cylinder air compressor running at
300 rpm is driven by an electric motor. Using the data
given below, and assuming that the bore is equal to the
stroke, calculate:
(i) the free air delivery;
(ii) the volumetric efficiency;
(iii) The bore and stroke.
Data Air inlet conditions, 1.013 bar and 15 C; delivery
pressure, 8 bar; clearance volume, 7% of the swept volume;
index of compression and re-expansion, 1.3; mechanical
efficiency of the drive between motor and compressor,
87%; motor power output, 23 kW.
GIVEN:
Single Acting Single Cylinder Compressor
N = 300 rpm, Power of the motor = 23 KW
Efficiency of Transmission = η = 87 %
P1 = 1.013 bar P2 = 8 bar
T1 = 288 K Vc = 0.07 Vs
L = d, n = 1.3
REQUIRED:
(i) FAD =?
(ii) Volumetric Efficiency = ηv =?
(iii) L =? (iv) d = ?
SOLUTION:
Motor power = Indicated Power / ηtrans.
23 x 103
= Indicated power / 0.87
Indicated Power = 20010 watts.
Now Pi = (n/n-1)P1(Va – Vd){(P2/P1)n-1/n
–1}
⇒Va – Vd = 4.47 m3
/ min = FAD
(ii) ηv = 1 – Vc/Vs{(P2/P1)1/n
–1} = 72.6 %
(iii)&(iv)
As Va = Vs + Vc = 1.07 Vs
Vd = Vc(P2/P1)1/n
= 0.343 Vs
Va – Vd = 0.727 Vs =>4.47/300 = 0.727 Vs
⇒ Vs = 0.0205
Now Πd3
/4 = Vs = 0.0205
⇒ d = 296 mm L = 296 mm.
PROBLEM 12.9:
A two-stage air compressor consists of three cylinders
having the same bore and stroke. The delivery pressure is 7
bar and the FAD is 4.2 m3
/min. Air is drawn in at 1.013
bar, 15 C and an intercooler cools the air at 38 C. The index
of compression is 1.3 for all three cylinders. Neglecting
clearance, calculate:
(i) the intermediate pressure;
(ii) the power required to drive the compressor;
(iii) The isothermal efficiency.
GIVEN:

7. Positive Displacement Machines Designed by Sir Engr. Masood Khan
Two Stages Air-Compressor
L = d P2 = 7 bar
FAD Conditions = V’
1 = 4.2 m3
/min
P1 = 1.013 bar T1 = 15C = 288K
Intercooling
T2 = 311K, n = 1.3, clearance = 0
REQUIRED:
(a) Intermediate pressure = Pi =?
(b) Power reqd. to drive the Motor =?
(c) Isothermal Efficiency = ηiso. =?
DIAGRAM:
SOLUTION:
We know that PVo
= mo
RT
mo
= PVo
/ RT = 0.086 Kg / sec
With Intercooler Pi = √ P1P2 = √ 2.66 bar
Now we know that
Pi / P1 = P2 / Pi, are same for both stages.
So I.P. = (n/n-1)mo
RT{(P2/P1)n-1/n
– 1} = 15.3 KW
ηiso. = Isothermal Work / Indicated Work ------------- (1)
Isothermal work = mo
RT1ln(P2/P1) = 137 KJ
Indicated Work = (n/n-1)mo
RT1{(P2/P1)n-1/n
–1} = 15.3 KJ
ηiso. = 89.8 %
PROBLEM 12.12:
Air at 1.013 bar and 15 C is to be compressed at the rate of
5.6 m3
/min to 1.75 bar. Two machines are considered: (a)
the Roots Blower; and (b) a sliding Vane Rotary
Compressor. Compare the powers required, assuming for
the vane type that internal compression takes place through
75% of the pressure rise before delivery takes place, and
that the compressor is an ideal uncooled machine.
GIVEN:
P1 = 1.013 bar P2 = 1.75 bar
T1 = 15C = 288K
Vo
= 5.6 m3
/min = 5.6/60 m3
/sec
REQUIRED:
Power Required;
(1) Root Blower (2) Vane Type
ASSUMPTION:
(1) For Vane Type, Internal compression takes place
through 75 % of pressure rise before delivery takes
place.
(2) Compressor is an ideal uncooled machine.
SOLUTION:
(1) For Root Blower
Power Required = (P2 – P1)xVo
= 6.88 KW
(2) For Vane Type
Power = (r/r-1) P1Vo1 {(Pi/P1)r-1/r
–1}+(P2 – P1)xVo
2
Here Pi = 0.75 (P2 – P1) + P1 = 1.566 bar
Vo
1 = 5.6 m3
/min
Vo
i = Vo
1 (Pi/P1)1/r
= 4.1 bar
Now Power = 5.64 KW

8. Positive Displacement Machines
Designed by Sir Engr. Masood Khan