AIR
COMPRESSOR
1.) An air compressor has a suction volume of 0.25 m3 / sec at 97 Kpa and discharges to 650 Kpa.
How much power saved by the compressor if there are two stages?
Solution:
For single stage:
W = (nP1V1 / n – 1)((P2 / P1 )n-1 / n -1)
W = (1.4)(97)(0.25) / 1.4 – 1 ((650 / 97)1.4 – 1 /1.4 – 1)
W = 61.28 Kw
For two stage:
𝑃𝑥 = √𝑃1 𝑃2
𝑃𝑥 = √(97)(650)
𝑃𝑥 = 251.097 Kpa
2𝑛𝑃1 𝑉1
W=
2(1.4)(97)(0.25)
𝑛−1
𝑃
𝑛−1
𝑛
W=
( (𝑃𝑥 )
1
1.4−1
- 1)
((
251.097 1.4−1
97
)
1.4
- 1)
W = 53 Kw
Power saved = 61.28 – 53
Power saved = 8.27 Kw
2.) The suction condition of an air compressor is 98 Kpa, 27 ℃ and 0.2 m3 /sec. if surrounding air
is 100 Kpa, 20 ℃, determine the free air capacity in m3/sec.
Solution:
𝑃𝐹 𝑉𝐹
𝑇𝐹
=
(100)𝑉𝐹
(20+273)
𝑃𝑠 𝑉𝑠
𝑇𝑠
=
98(0.2)
(27+273)
VF = 0.1914 m3 / sec
3.) A 355 mm x 381 mm air compressor has a piston displacement of 0.1885 m3 / sec. determine
the operating speed of the compressor.
Solution:
VD = 𝜋/4 D2 L N
0.1885 = 𝜋/4 (0.355)2 (0.381) N
N = 300 rpm
4.) Determine the percent clearance of an air compressor having 87% volumetric efficiency and
compressor air pressure to be thrice the suction pressure.
Solution:
= 1 + c – c (P2/P1)1/n
V
c – c (3P1 / P1)1/1.4
C = 10.91%
5.) The compressor work of an air compressor is 100 Kw. If the piston speed is 15 m3/min,
determine the mean effective pressure.
Solution:
W = PM x VD
100 = PM (15/60)
PM = 400 Kpa
6.) An air compressor takes air 100 Kpa and discharged to 600 Kpa. If the volume flow of
discharge is 1.2 m3/sec, determine the capacity of the compressor.
Solution:
P1v1n = p2v2n
N = 1.4 (for standard air)
100 (v1)1.4 = (600)(1.2)1.4
V1 = 4.315 m3/sec
7.) The discharge pressure of an air compressor is 5 times the suction pressure. If volume flow at
suction is 0.1 m3/sec, what is the compressor power assuming n=1.35 and suction pressure is 98
Kpa?
Solution:
W = (nP1V1 / n – 1) ((P2/P1) n-1/n-1)
W = (1.35)(98)(0.1) / 1.35 – 1 ((5P1/P1)1.35 – 1 / 1.35 – 1)
W = 19.57 KW
8.) A 10 Hp motor is use to drive an air compressor. The compressor efficiency is 75%. Determine
the compressor work.
Solution:
c
= w / Brake Power
0.75 = W / 10 x 0.746
W = 5.59 KW
9.) The initial condition of air in an air compressor is 98 Kpa and 27 ℃ and discharged air at 400
Kpa. The bore and stroke are 355 mm and 381 mm respectively with percent clearance of 5%
running at 300 rpm. Find the volume of air at suction.
Solution:
v
= 1 + c – c (P2 / P1)1/n
v
= 1 + 0.05 – 0.05 (400/98)1/1.4
v
=0.913
VD = 𝜋 / 4 D2 L N
VD = 𝜋 / 4 (0.355)2 (0.381) (300/60)
VD = 0.1885 m3 / sec
V1 = 0.1885 (0.913)
V1 = 0.17215 m3 / sec
V1 = 619.75 m3 / hr
10.) A double acting air compressor has 16 x 7 in, 600 rpm has what volume displacement?
Solution:
VD = 2 (𝜋/4 D2 L N)
VD = 2 (𝜋/4 (16/12)2 (7/12) (600))
VD = 977.38 ft3 / min.
11.) A two stage compressor has a suction pressure of 14 psi and discharge pressure of 130 psig.
What is the intercooler pressure in Kpag.
Solution:
P2 = 130 + 14.7
P2 = 147.5 psia
PX = √𝑃1 𝑃2
PX = √14(144.7)
PX = 45 psi x 101.325
PX = 208.91 Kpag
12.) A two stage air compressor has an intercooler pressure of 3 kg / cm3. What is the discharge
pressure if suction pressure is 1 kg/cm2?
Solution:
PX = √𝑃1 𝑃2
(PX)2 = P1P2
32 = 1(P2)
P2 = 9 kg/cm2
13.) The piston speed of an air compressor is 140 m/min and has a volume displacement of 0.2 m3
/sec. Determine the diameter of compressor cylinder.
Solution:
VD = 𝜋/4 D2 L N
Piston speed = 2 L N
140 = 2 LN
LN = 70 m /min
0.2 = 𝜋/4 D2 (70/60)
D = 467.19 mm
14.) A single acting reciprocating air compressor has a clearance volume of 10%. Air is received at
90 Kpa and 29.3 ℃ and is discharged at 600 Kpa. The compression and expansion are polytropic
with n = 1.28. The pressure drop is 5 Kpa at suction port and 10 Kpa at the discharged port. The
compressor piston displacement is 500 cm3 when operating at 900 rpm. Determine the mass of
compressed air in kg/hr.
Solution:
VD = (𝜋/4 D2 L)N
VD = (500)(900)
VD = 450,000 cm3 / min
VD = 0.45 m3/min
P1 = 90 – 5
P1 = 85 Kpa
P2 = 600 + 10
P2 = 610 Kpa
V
= 1 + c – c (P2/P1)1/n
V
= 1 + 0.10 – 0.10 (610/85)1/1.28
V
= 0.633684
V1 = 0.45 (0.633684)
V1 = 0.285 m3 / min
M = PV/RT
M = 85(0.285)/(0287)(29.3+273)
M = 0.2792 kg/min
M = 16.76 kg/hr
15.) A two stage compressor has a suction pressure of 14 psi and discharge pressure of 130 psig.
What is the intercooler pressure in Kpag.
Solution:
P2 = 130 + 14.7
P2 = 147.5 psia
PX = √𝑃1 𝑃2
PX = √14(144.7)
PX = 45 psi x 101.325
PX = 208.91 Kpag
16.) The piston speed of an air compressor is 140 m/min and has a volume displacement of 0.2 m3
/sec. Determine the diameter of compressor cylinder.
Solution:
VD = 𝜋/4 D2 L N
Piston speed = 2 L N
140 = 2 LN
LN = 70 m /min
0.2 = 𝜋/4 D2 (70/60)
D = 467.19 mm
17.) A 10 Hp motor is use to drive an air compressor. The compressor efficiency is 75%.
Determine the compressor work.
Solution:
c
= w / Brake Power
0.75 = W / 10 x 0.746
W = 5.59 KW
18.) An air compressor piston displacement is 5000 cm3 when operates at 900 rpm and volumetric
efficiency of 75%. Determine the mass flow of air at standard density.
Solution:
V
= V 1 / VD
0.75 = V1 / 5000
V1 = 3750 cm3 (900)
V1 = 337500 cm3/min
W = 1.2 kg/m3 (at standard)
M = 1.2 (3375000/1003)
M = 4.05 kg/min
M = 243 kg/hr
19.) A two stage compressor air at 100 Kpa and 22 ℃ discharges to 690 Kpa. If intercooler intake
is 105℃, determine the value of n.
Solution:
PX = √𝑃1 𝑃2
PX = √100(690)
PX = 262.68 Kpa
TX / T1 = (PX /P1) n-1/n
(105+273) / (22 + 273) = (262.68 / 100) n-1/n
1.281 = (2.6268) n-1/n
n- 1 / n = ln(1.281) / ln(2.6268)
n-1 = 0.2564n
n = 1.345
20.) The piston displacement of a double acting compressor running at 300 rpm is 0.4 m3 / sec. if
the bore and stroke are unity, determine the length of stroke.
Solution:
VD = 2 (𝜋/4 D2 L N)
L = D (for unity)
0.4 = 2 ((𝜋/4 (D)2 (D) (300/60))
D = 0.37067 m
L=D
L = 370.67 mm
21.) an air compressor takes air at 97 Kpa at the rate of 0.5 m 3/sec and discharge 500 Kpa. If
power input to the compressor is 120 Kw, determine the heat loss in the compressor
.
Solution:
W=
2𝑛𝑃1 𝑉1
𝑛−1
𝑃
( (𝑃𝑥 )
1
𝑛−1
𝑛
- 1)
W=
1.4(97)(0.5)
1.4−1
((
500 1.4−1
97
)
1.4
- 1)
W = 101.45 KW
Heat loss = 120 – 101.45
Heat loss = 18.55 KW
22.) A single acting reciprocating air compressor has a clearance volume of 10%. Air is received at
90 Kpa and 29.3 ℃ and is discharged at 600 Kpa. The compression and expansion are polytropic
with n = 1.28. The pressure drop is 5 Kpa at suction port and 10 Kpa at the discharged port. The
compressor piston displacement is 500 cm3 when operating at 900 rpm. Determine the mass of
compressed air in kg/hr.
Solution:
VD = (𝜋/4 D2 L)N
VD = (500)(900)
VD = 450,000 cm3 / min
VD = 0.45 m3/min
P1 = 90 – 5
P1 = 85 Kpa
P2 = 600 + 10
P2 = 610 Kpa
V
= 1 + c – c (P2/P1)1/n
V
= 1 + 0.10 – 0.10 (610/85)1/1.28
V
= 0.633684
V1 = 0.45 (0.633684)
V1 = 0.285 m3 / min
M = PV/RT
M = 85(0.285)/(0287)(29.3+273)
M = 0.2792 kg/min
M = 16.76 kg/hr
23.) A single acting air compressor operates at 150 rpm with an initial condition of air at 97.9 Kpa
and 27℃ and discharges the air at 379 Kpa to a cylindrical tank. The bore and stroke are 355 mm
and 381 mm, respectively, with 5% clearance. If the surrounding air is at 100 Kpa and 20℃ while
the compression and expansion process are PV1.3 = C, determine free air capacity, m3/sec.
Solution:
𝜋
VD = 4 D2 L N
𝜋
VD = (0.335)2 (0.381) (150/60)
4
VD = 0.094278 m3/sec
V
= 1 + c – c (P2/P1)1/n
V
= 1 + 0.05 – 0.05 (379/97.9)1/1.3
V
= 0.908
V1 = 0.908 (0.094278)
V1 = 0.085604 m3 / sec
Solving for free air capacity:
PFVF/TF = P1V1/T1
100(VF) / (20+273) = 97.9(0.085604) / (27+273)
VF = 0.081851 m3 / sec
24.) The piston displacement of a double acting compressor is 0.358 m3/sec, delivers gas from
101.325 Kpa to 675 Kpa at the rate of 0.166 m3/sec at 150 rpm. Value of n for compression and
expansion is 1.33. find the bore and stroke assuming bore = stroke.
Solution:
𝜋
VD = 2 (4 D2 L N)
𝜋
0.358 = 2 (4 D2 (D) (150/60))
D = 0.45m
D = 450 mm
D = L = 450 mm
25.) An air compressor takes air 100 Kpa and discharged to 600 Kpa. If the volume flow of
discharge is 1.2 m3/sec, determine the capacity of the compressor.
Solution:
P1v1n = p2v2n
N = 1.4 (for standard air)
100 (v1)1.4 = (600)(1.2)1.4
V1 = 4.315 m3/sec