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Given Data: We Know That Power 2 (W) (N) For Double Acting Polytropic Process W

1. The size of the cylinder for a 150KW, 160rpm air compressor with a piston speed of 150 m/min and operating between 16 bar and 6 bar is determined to be 0.58 meters based on the given polytropic compression law and power equation. 2. For an air compressor cylinder with a 150mm bore, 150mm stroke, and 15% clearance operating between 1 bar and 5 bar with a polytropic exponent of 1.3: i) The cylinder volumes at various points are calculated to be 2.1948 m3, 0.6366 m3, 0.2862 m3, and 0.98674 m3. ii) The flow rate at 720 rpm is

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559 views4 pages

Given Data: We Know That Power 2 (W) (N) For Double Acting Polytropic Process W

1. The size of the cylinder for a 150KW, 160rpm air compressor with a piston speed of 150 m/min and operating between 16 bar and 6 bar is determined to be 0.58 meters based on the given polytropic compression law and power equation. 2. For an air compressor cylinder with a 150mm bore, 150mm stroke, and 15% clearance operating between 1 bar and 5 bar with a polytropic exponent of 1.3: i) The cylinder volumes at various points are calculated to be 2.1948 m3, 0.6366 m3, 0.2862 m3, and 0.98674 m3. ii) The flow rate at 720 rpm is

Uploaded by

Alex Anders
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© © All Rights Reserved
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1.

A single stage double acting air compressor of 150KW power takes air in at 16 bar &
delivers at 6 bar. The compression follows the law PV1.35 = C. the compressor runs
at 160rpm with average piston speed of 150 m/min. Determine the size of the
cylinder.

Given data

Power (P) = 150KW


Piston speed (2lN) = 150m /min
1𝑚𝑖𝑛 2𝜋 𝑟𝑎𝑑 𝑟𝑎𝑑
Speed (N) = 160rpm 60𝑠 × 𝑟𝑒𝑣 = 16.75 𝑠
Pressure (P1) = 1bar = 100 KN/m2
Pressure (P2) = 6bar = 600 KN/m2
PV1.35 = C, n = 1.35
Hence it is a polytropic process.

To find
Size of the cylinder (d) ?
Solution

𝑚
2𝑙𝑁 = 2.5
𝑠
𝑟𝑎𝑑 𝑚
2(𝑙) (16.75 ) = 2.5
𝑠 𝑠
2.5
𝑙= = 0.07463𝑚
2(16.75)
𝜋
𝑠𝑖𝑛𝑐𝑒 𝑉1 = 𝑉𝑠 = 𝑑2 𝑙
4
𝜋 2
𝑉1 = 𝑉𝑠 = 𝑑 (0.07463)
4
𝑉1 = 0.0586𝑑2
We know that
Power = 2(W)(N) for double acting polytropic process
W=
𝑛−1
𝑛 𝑃2 𝑛
𝑊= (𝑃1 𝑉1 ) [( ) − 1]
𝑛−1 𝑃1
1.35 1.35−1
𝑊= (100 × 0.0586𝑑2 ) [(6) 1.35 − 1]
1.35 − 1
𝑊 = 13.36𝑑2
𝑃𝑜𝑤𝑒𝑟 (𝑝) = 2 × 𝑊 × 𝑁
150 = 2 × 13.36𝑑2 × 16.75
𝑑2 = 0.3352
𝑑 = 0.58𝑀
2. An air compressor cylinder has 150mm bore and 150mm stroke and the
clearance is 15%. It operates between 1 bar, 2℃and 5 bar. Take polytrophic
exponent n=1.3 for compression and expansion processes find?
i. Cylinder volume at the various salient points of in cycle.
ii. Flow rate in m3/min at 720 rpm and.
iii. The deal volumetric efficiency.
Given
𝐷 = 150 × 10−3 𝑚
𝐿 = 150 × 10−3 𝑚
𝑉𝑐 = 0.15𝑉𝑠
𝑁
𝑃1 = 1 × 105
𝑚2
𝑁
𝑃2 = 5 × 105
𝑚2
𝑇1 = 27 + 273 = 300𝑘
𝑁 = 720𝑟𝑝𝑚
𝑝𝑣 𝑛 = 𝐶𝑛 = 1.3
Find
V1, V2, V3
FAD (Va)
𝜂𝑣
Solution
𝑉1 = 𝑉𝑐 + 𝑉𝑠
𝜋 4 𝜋 𝑚3
𝑉𝑠 = 𝐷 𝐿. 𝑁 = (0.15)2 (0.15)(720) = 1.9085
4 4 𝑚𝑖𝑛
𝑉𝑐 = 0.15𝑉𝑠
𝑉𝑐 = 0.15(1.9085)
𝑚3
𝑉𝑐 = 0.2826
𝑚𝑖𝑛
𝑉1 = 𝑉𝑐 + 𝑉𝑠
𝑉1 = 0.2862 + 1.9085
𝑚3
𝑉1 = 2.1948
𝑚𝑖𝑛
𝑃1 𝑉1 𝑛 = 𝑃2 𝑉2 𝑛
1⁄
𝑃1 𝑛
𝑉2 = 𝑉1 ( )
𝑃2
1⁄
1 × 105 1.3
= 2.1948 ( )
5 × 105

𝑚3
𝑉2 = 0.6366
𝑚𝑖𝑛
𝑉𝑐 = 𝑉3
𝑚3
𝑉3 = 0.2862
𝑚𝑖𝑛
𝑃3 𝑉3 𝑛 = 𝑃4 𝑉4 𝑛
1⁄
𝑃3 𝑛
𝑉4 = 𝑉3 ( )
𝑃4
𝑃2 = 𝑃3
𝑃1 = 𝑃4
1⁄
𝑃2 𝑛
𝑉4 = 𝑉3 ( )
𝑃1
1⁄
5 × 105 1.3
= 0.2862 ( )

𝑚3
𝑉4 = 0.98674
𝑚𝑖𝑛
1⁄
𝑃2 𝑛
𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑓𝑓𝑖𝑒𝑐𝑖𝑒𝑛𝑐𝑦 (𝜂𝑣 ) = 1 + 𝑘 − 𝑘 ( )
𝑃1
𝑉𝑐 0.2862
𝐾 = 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑖𝑜 = =
𝑉𝑠 1.9085
𝐾 = 0.1499
1⁄
5 × 105 1.3
(𝜂𝑣 ) = 1 + 0.1499 − 0.1499 ( )
1 × 105

𝜂𝑣 = 0.633 𝑜𝑟 63.3%
𝐹𝐴𝐷
𝜂𝑣 =
𝑉𝑠
𝐹𝐴𝐷 = 𝜂𝑣 (𝑉𝑠 )
= 0.633(1.9085)
𝐹𝐴𝐷 = 1.2083

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