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[( ) 1 1. An air compressor takes air at 100 Kpa and discharges to 600 Kpa. If the volume flow of discharge is 1.2 𝑚 3 ⁄ 𝑠𝑒𝑐 , determine the capacity of air compressor. 𝑃 1 𝑉 1 𝑛 = 𝑃 1 𝑉 1 𝑛 𝑛 = 1.4 (𝑓𝑜𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑎𝑖𝑟) 100(𝑉 1 1.4 ) = (600)(1.2) 1.4 𝑉 = 4.315 𝑚 3 ⁄ 1 𝑠𝑒𝑐 2. The discharge pressure of an air compressor is 5 times the suction pressure. If volume flow at suction is 0.1 𝑚 3 ⁄ 𝑠𝑒𝑐 , what is the compressor power assuming n= 1.35 and suction pressure is 98 Kpa? 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 1.35(98)(0.1) 5𝑃 1 1.35−1 1.35 𝑊 = 1.35 − 1 [( 𝑃 ) − 1] 𝑊 = 19.57 𝐾𝑤 3. A 10 Hp motor is use to drive an air compressor. The compressor efficiency is 75%. Determine the compressor work. 𝑊 ƞ 𝑐 = 𝐵𝑟𝑎𝑘𝑒 𝑃𝑜𝑤𝑒𝑟 0.75 = 𝑊 10𝑥0.746 𝑊 = 5.59 𝐾𝑤

1 𝑉 = 619.75 1 𝑚 3 ⁄ ℎ𝑟 4. The initial condition if an air compressor is 98 Kpa and 27°C and discharges air at 400 Kpa. The bore and stroke are 355 mm and 381 mm, respectively and percent clearance od 5% running at 300 rpm. Find the volume of air at suction. ƞ 𝑣 = 1 + 𝑐 − 𝑐 1 𝑃 2 𝑛 ( 𝑃 ) 400 1 1.4 ƞ 𝑣 = 1 + 0.05 − 0.05 ( 98 ) ƞ 𝑣 = 0.913 𝑉 𝐷 𝑉 = 𝜋 𝐷 2 𝐿𝑁 4 = 𝜋 (0.355)2(0.381) ( 300 ) 𝐷 𝑉 𝐷 4 60 = 0.1885 𝑚 3 ⁄ 𝑠𝑒𝑐 𝑉 1 = 0.1885(0.913) 𝑉 = 0.17215 𝑚 3 ⁄ 𝑥 ( 3600 1 𝑠𝑒𝑐 ℎ𝑟 )

[( ) [( ) 5. An air compressor has a suction volume of 0.25 𝑚 3 ⁄ 𝑠𝑒𝑐 at 97 Kpa and discharges 650 Kpa. How much power saved by the compressor if there are two steps? For single stage: 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 1.4(97)(0.25) 650 1.4−1 1.4 𝑊 = 1.4 − 1 [( 97 ) − 1] 𝑊 = 61.28 𝐾𝑤 For two stages: 𝑃𝑥 = √𝑃1𝑃2 𝑃 𝑥 = √(97)(650) 𝑃 𝑥 = 251.097 𝐾𝑝𝑎 𝑊 = 2𝑛𝑃 1 𝑉 1 𝑃 𝑥 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 𝑊 = (2)(1.4)(97)(0.25) 1.4 − 1 [( 251.097 97 ) 1.4−1 1.4 − 1] 𝑊 = 53 𝐾𝑤 𝑃𝑜𝑤𝑒𝑟 𝑆𝑎𝑣𝑒𝑑 = 61.28 − 53 𝑃𝑜𝑤𝑒𝑟 𝑆𝑎𝑣𝑒𝑑 = 8.27 𝐾𝑤

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1 6. The suction condition of an air compressor is 98 Kpa, 27°C and 0.2 𝑚 3 ⁄ 𝑠𝑒𝑐 . If surrounding air is 100 Kpa and 20°C, determine the free air capacity in 𝑚 3 ⁄ 𝑠𝑒𝑐 . 𝑃 𝐹 𝑉 𝐹 = 𝑃 𝑆 𝑉 𝑆 𝑇 𝐹 𝑇 𝑠 (100)𝑉 𝐹 (20 + 273) 98(0.2) = (20 + 273) 𝑉 = 0.1914 𝑚 3 ⁄ 𝐹 𝑠𝑒𝑐 7. A 355 mm x 381 mm air compressor has a piston displacement of 0.1885 𝑚 3 ⁄ 𝑠𝑒𝑐 . Determine the operating speed of the compressor. 𝑉 𝐷 = 𝜋 𝐷 2 𝐿𝑁 4 0.1885 = 𝜋 (0.355)2(0.381)(𝑁) 4 𝑟𝑒𝑣 𝑁 = 5 ( sec ) 𝑥 60 ( 𝑠𝑒𝑐 ) 𝑁 = 300 𝑟𝑝𝑚 8. Determine the percent clearance of an air compressor having 87% volumetric efficiency and compressor air compressor air pressure to be thrice the suction pressure. 1 ƞ 𝑣 = 1 + 𝑐 − 𝑐 𝑃 2 𝑛 ( 𝑃 ) 1 0.87 = 1 + 𝑐 − 𝑐 ( 3𝑃 1 ) 1.4 𝑃 1 𝑐 = 10.91% min

9. The compressor work of an air compressor is 100 KW. If the piston speed is 15 𝑚 3 ⁄ 𝑠𝑒𝑐 , determine the mean effective pressure. 𝑊 = 𝑃 𝑚 𝑉 𝐷 15 100 = 𝑃 𝑚 ( 60 ) 𝑃 𝑚 = 400 𝐾𝑝𝑎 10.A double acting air compressor has 16 in x 7 in, 600 rpm has what volume displacement? 𝑉 𝐷 = 2 ( 𝜋 𝐷 2 𝐿𝑁) 4 𝜋 𝑉 𝐷 = 2 ( 4 16 2 ( 12 ) 7 ( 12 ) (600)) 𝑉 𝐷 = 977.38 𝑓𝑡 3 𝑚𝑖𝑛 11.A two-stage air compressor has a suction pressure of 14 psi and discharge pressure of 130 psig. What is the intercooler pressure in Kpag. 𝑃 2 = 130 + 14.7 𝑃 2 = 144.7 𝑝𝑠𝑖𝑎 𝑃𝑥 = √𝑃1𝑃2 𝑃 𝑥 = √(14)(144.7) 101.325 𝐾𝑝𝑎 𝑃 𝑥 = 45 𝑝𝑠𝑖 𝑥 ( 14.7 𝑝𝑠𝑖 ) 𝑃 𝑥 = 310.24 𝐾𝑝𝑎𝑎 − 101.325 𝑃 𝑥 = 208.91 𝐾𝑝𝑎𝑔 ⁄

2 2 ( 12. A two stage air compressor has an intercooler pressure of 3 𝑘𝑔 ⁄ 2 . What is the discharge pressure if suction is 1 𝑘𝑔 ⁄ 2 ? cm 𝑃𝑥 = √𝑃1𝑃2 𝑃 𝑥 2 = 𝑃 1 𝑃 2 3 2 = (1)𝑃 2 𝑃 = 9 𝑘𝑔 ⁄ cm 13. The piston speed of an air compressor is 140 𝑚 ⁄ 𝑚𝑖𝑛 and has a volume displacement of 0.2 𝑚 3 ⁄ 𝑠𝑒𝑐 cylinder. . determine the diameter of compressor 𝑉 𝐷 = ( 𝜋 𝐷 2 𝐿𝑁) 4 𝑃𝑖𝑠𝑡𝑜𝑛 𝑆𝑝𝑒𝑒𝑑 = 2 𝐿𝑁 140 = 2 𝐿𝑁 2 𝐿𝑁 = 70 𝑚 ⁄ 𝑚𝑖𝑛 0.2 = ( 𝜋 ) 4 𝐷 2 70 )) 60 𝐷 = 467.19 𝑚𝑚 cm (

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ℎ𝑟 14.An air compressor piston displacement is 5000 cm 3 when operates at 900 rpm and volumetric efficiency of 75%. Determine the mass flow of air at standard density. ƞ 𝑣 = 𝑉 1 𝑉 𝐷 0.75 = 𝑉 1 5000 𝑉 1 = 3750 cm 3 (900) cm 3 𝑉 1 = 3375000 𝑤 = 1.2 𝑘𝑔 ⁄ 𝑚 ⁄ 𝑚𝑖𝑛 3 (𝑎𝑡 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑) 𝑚 = 1.2 ( 3375000 100 3 ) 𝑚 = 4.05 𝑘𝑔 ⁄ 𝑚𝑖𝑛 𝑚 = 243 𝑘𝑔 ⁄ 𝑥 3600𝑠 ⁄ ℎ𝑟

1 15.A two-stage compressor air at 100 Kpa and 22°C discharges to 690 Kpa. 1 intercooler intake is 105°C, determine the value of n. 𝑃𝑥 = √𝑃1𝑃2 𝑃 𝑥 = √(100)(690) 𝑃 𝑥 = 262.68 𝐾𝑝𝑎 𝑇 𝑥 = 𝑇 1 𝑃 𝑥 ( 𝑃 ) 𝑛−1 𝑛 (105 + 273) (22 + 273) = ( 262.68 100 ) 𝑛−1 𝑛 𝑛−1 1.281 = (2.6368) 𝑛 𝑛 − 1 𝑛 = ln 1.281 ln 2.6268 𝑛 − 1 = 0.2564𝑛 𝑛 = 1.345

[( ) 16.The piston displacement of a double acting compressor running at 300 rpm is 0.4 𝑚 3 ⁄ 𝑠𝑒𝑐 . If bore and stroke are unity, determine the length of stroke. 𝑉 𝐷 = 2 ( 𝜋 𝐷 2 𝐿𝑁) 4 𝐿 = 𝐷 (𝑓𝑜𝑟 𝑢𝑛𝑖𝑡𝑦) 0.4 = (2 ( 𝜋 ) 𝐷 2 (𝐷) ( 300 4 60 )) 𝐷 = 0.37067 𝑚 𝐿 = 𝐷 𝐿 = 370.67 𝑚𝑚 17. 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. 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 (1.4)(97)(0.5) 500 1.4−1 1.4 𝑊 = 1.4 − 1 [( 97 ) − 1] 𝑊 = 101.45 𝐾𝑤 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 = 120 − 101.45 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 = 18.55 𝐾𝑤

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[( ) 18.A single acting air compressor has a volumetric efficiency of 87%, operates at 500 rpm. It takes in air at 100 Kpa and 30°C and discharges it at 600 Kpa. 𝑚 3 The air handled is 6 ⁄ 𝑚𝑖𝑛 measured at discharge condition. If compression is isentropic, find mean effective pressure in Kpa. 𝑃 1 𝑉 1 𝑘 = 𝑃 1 𝑉 1 𝑘 (100)(𝑉 1 1.4 ) = (600)(6) 1.4 𝑚 3 𝑉 1 = 21.58 21.58 ⁄ 𝑚𝑖𝑛 𝑉 𝐷 = 0.87 𝑚 3 𝑉 𝐷 = 24.8 ⁄ 𝑚𝑖𝑛 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 (1.4)(100)(21.58) 600 1.4−1 1.4 𝑊 = 1.4 − 1 [( 100 ) − 1] 𝑊 = 5,049.26 𝐾𝐽 ⁄ 𝑚𝑖𝑛 𝑊 = 𝑃 𝑚 𝑉 𝐷 5,049.26 = 𝑃 𝑚 (24.8) 𝑃 𝑚 = 203.6 𝐾𝑝𝑎

19.A single reciprocating air compressor has a clearance volume of 10%. Air is received at 90 Kpa and 29.3⁰C and is discharge 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 discharge port. The compressor piston displacement is 500 cm³ when operating at 900 rpm. Determine the mass of compressed air in kg/hr. 𝜋 𝑉 𝐷 = ( 4 𝐷²𝐿)𝑁 𝑉 𝐷 = (500)(900) 𝑉 𝐷 = 450,000 𝑐𝑚³/𝑚𝑖𝑛 𝑉 𝐷 = 90 – 5 𝑃 1 = 85 𝐾𝑝𝑎 𝑃 2 = 600 + 10 𝑃 2 = 610 𝑘𝑝𝑎 ƞ 𝑉 1 = 1 + 𝑐 – 𝑐 ( 𝑃 2 ) 𝑛 𝑃 1 610 1 1.28 ƞ 𝑉 = 1 + 0.10 – 0.10 ( 85 ) ƞ 𝑉 = 0.6333684 𝑉 1 = 0.45(0.633684) 𝑚 3 𝑉 1 = 0.285 𝑃𝑉 ⁄ 𝑚𝑖𝑛 𝑚 = 𝑅𝑇 85(0.285) 𝑚 = (0.287)(29.3 + 273) 𝑚 = 0.2792 𝐾𝑔 ⁄ 𝑚𝑖𝑛 𝑚 = 16.76 𝐾𝑔 ⁄ 𝑚𝑖𝑛 𝑥 3600𝑠 ⁄ 𝑚𝑖𝑛

19.A single acting air compressor operates at 150 rpm with an initial condition of air at 97.9 kpa and 27⁰C and discharges the air at 379 kpa to a cylindrical tank. The bore and stroke are 355mm and 381 mm, respectively, with 5% clearance. If the surrounding air is at 100kpa and 20⁰C while the compression and expansion process are PV 1.3 = C, determine free air capacity, 𝑚 3 ⁄ 𝑠𝑒𝑐 . 𝑉 𝐷 = 𝜋 4 𝐷²𝐿𝑁 𝑉 = 𝜋 (0.355)2(0.381) ( 150 ) 𝐷 4 60 𝑉 𝐷 = 0.094278 𝑚³/𝑠𝑒𝑐 ƞ 𝑉 1 = 1 + 𝑐 – 𝑐 ( 𝑃 2 ) 𝑛 𝑃 1 1 379 1.3 ƞ 𝑉 = 1 + 0.01 – 0.05 ( 97.9 ) ƞ 𝑉 = 0.908 𝑉 1 = 0.908(0.094278) 𝑉 1 = 0.085604 𝑚 3 ⁄ 𝑠𝑒𝑐 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑓𝑟𝑒𝑒 𝑎𝑖𝑟 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑃 𝐹 𝑉 𝐹 = 𝑃 𝑆 𝑉 𝑆 𝑇 𝐹 𝑇 𝑠 100(𝑉 𝐹 ) (20 + 273) 97.9(0.085604) = (27 + 273) 𝑉 = 0.081851 𝑚 3 ⁄ 𝐹 𝑠𝑒𝑐

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20.The piston displacement of a double acting compressor is 0.358 𝑚 3 ⁄ 𝑠𝑒𝑐 , delivers gas from 101.325 Kpa and 300 ⁰K to 675 Kpa at the rate of 0.166 𝑚 3 ⁄ 𝑠𝑒𝑐 at 150 rpm value of n for compression and expansion is 1.33. Find the compressor percent clearance. ƞ 𝑉 = 𝑉₁ 𝑉ᴅ 0.166 ƞ 𝑉 = = 0.358 ƞ 𝑉 = 0.4637 1 ƞ 𝑉 = 1 + 𝑐 – 𝑐 ( 𝑃 2 ) 𝑛 𝑃 1 1 ƞ 𝑉 = 1 + 𝑐 − 𝑐 ( 675 101.325 ) 1.33 𝐶 = 0.1696 𝑥100 𝐶 = 16.96 % 22. The piston displacement of a double acting compressor is 0.358 𝑚 3 ⁄ 𝑠𝑒𝑐 , delivers gas from 101.325 Kpa and 300⁰K to 675 Kpa at the rate of 0.166 𝑚 3 ⁄ 𝑠𝑒𝑐 at 150rpm. Value of n for compression and expansion is 1.33. Find the bore and stroke assuming bore = stroke 𝑉 𝐷 = 2 ( 𝜋 𝐷 2 𝐿𝑁) 4 0.358 = 2 ( 𝜋 (𝐷)2 ( 150 )) 4 60 𝐷 = 0.45 𝑚 𝐷 = 450𝑚𝑚 𝐷 = 𝐿 = 450𝑚𝑚

23. A stage, double acting L-type air compressor 16” x 10” x 7”, 600 rpm, has a free air unloader at each end for capacity control. It is driven through V- belts by a 150 Hp electric motor, 460 V, 3 phase, 60 Hz, 1200 rpm. Barometric 𝑚 3 pressure is 125 psi gage. Calculate piston displacement in ⁄ ℎ𝑟. 𝑉 𝐷 = 2 ( 𝜋 𝐷 2 𝐿𝑁) 4 𝜋 𝑉 𝐷 = 2 ( 4 16 2 ( 12 ) 7 ( 12 ) (600)) 𝑉 𝐷 = 977.384 𝑐𝑓𝑚 60 𝑉 𝐷 = 977.384 ( 35.31 ) 𝑚 3 𝑉 𝐷 = 1661 ⁄ ℎ𝑟. 𝐹𝑜𝑟 𝑝𝑖𝑠𝑡𝑜𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝜋 𝑉 𝐷 = 2 ( 4 10 2 ( 12 ) 7 ( 12 ) (600)) 𝑉 𝐷 = 381.791 𝑐𝑓𝑚 60 𝑉 𝐷 = 381.791 ( 35.31 ) 𝑚 3 𝑉 𝐷 = 648.75 ⁄ ℎ𝑟.

𝑉 = 5.0.13 𝑚 3 𝑠 ⁄ 𝑚𝑖𝑛 24. A two-cylinder single-acting compressor is directly couple to an electric motor running at 1000 rpm. Other date are as follow. a. size of each cylinder = 150 mm x 200 mm b. clearance volume = 10% of displacement c. exponent (n) for both compression and re-expansion process = 1.6 d. air constant = 1.4 e. air molecular mass M = 29 Calculate the volume rate of air delivery in terms of standard air for a delivery pressure 8 times ambient pressure under ambient air conditions of 300⁰K and 1 bar. 𝜋 𝑉 𝐷 = 4 𝐷²𝐿𝑁 𝑉 𝐷 = 𝜋 (0.15)2(0.2)(1000) 4 𝑉 𝐷 = 7.068 𝑚³/𝑚𝑖𝑛 1 ƞ 𝑉 ƞ = 1 + 𝑐 – 𝑐 ( 𝑃 2 ) 𝑛 𝑃 1 = 1 + 𝑐 – 𝑐 ( 8𝑃 1 1 1.6 𝑉 ƞ 𝑉 = 0.733 ) 𝑃 1 𝑉₁ = 0.733(7.068) 𝑚 3 𝑉₁ = 5.181 ⁄ 𝑚𝑖𝑛 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑎𝑖𝑟 𝑖𝑠 𝑎𝑡 70⁰𝐹 (21.11⁰𝐶) 𝐴𝑁𝐷 14.7 𝑝𝑠𝑖(101.325𝐾𝑝𝑎) 𝑃 𝑠 𝑉 𝑠 = 𝑃₁𝑉₁ 𝑇 𝑠 𝑇₁ 101.325(𝑉 𝑠 ) = 100(2.181) 21 + 273 (300)

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25. A two stage compressor with first stage piston displacement of 94390 𝑐𝑚 3 ⁄ 𝑠𝑒𝑐 is driven by a motor. Motor output is 35 Hp, suction temperature 22⁰C, volumetric efficiency is 85%. Mechanical efficiency is 95%, the intercooler are 105⁰C and 44⁰C. Find discharge is 100 psi gage, suction gage estimated 14.5 psi. find the compression efficiency 𝑉 𝐷1 𝑉 𝐷1 = 94390 𝑐𝑚 3 ⁄ 𝑠𝑒𝑐 = 0.094390 𝑐𝑚 3 ⁄ 𝑠𝑒𝑐 𝑉 1 = 𝑉 𝐷1 (ƞ 𝑉 ) = 0.009439(0.85) 𝑉 1 = 0.08020315 𝑚³/𝑠𝑒𝑐 101.325 𝑃 1 = 14.5 𝑝𝑠𝑖 ( 𝑃 1 = 99.946 𝐾𝑝𝑎 14.7 ) 101.325 𝑃 𝑋 = (30 + 14.7) ( 𝑃 𝑋 = 308.11 𝐾𝑝𝑎 𝑃 2 = (100 + 14.7) ( 𝑃 2 = 760.61 𝐾𝑝𝑎 14.7 ) 101.325 14.7 ) 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒: 𝑃₁𝑉₁ = 𝑚𝑅𝑇 99.946(0.0802315) = 𝑚(0.287)(22 + 273) 𝑚 = 0.09471 𝐾𝑔 ⁄ 𝑠𝑒𝑐

𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑝𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡 𝑛: 𝑛−1 𝑇 𝑥 ( 𝑇 1 ) = ( 𝑃 𝑥 ) 𝑛 𝑃 1 𝑛−1 101 + 273 ( 22 + 273 ) = ( 308.11 𝑛 ) 99.946 𝑛−1 1.2812 = (3.08277) 𝑛 𝑛 − 1 = 𝑛 𝑙𝑛1.2813 𝑙𝑛3.08277 𝑛 = 1.2824 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑠𝑡𝑎𝑔𝑒 𝑊 = 𝑛(𝑚𝑅𝑇) (( 𝑛 − 1 𝑃 2 𝑃 1 𝑛−1 𝑛 ) − 1) 𝑊 = 1.2824((0.0947)(0.287)(22 + 273)) 1.2824 − 1 308.11 (( 99.946 1.2824−1 1.2824 ) − 1) 𝑊 = 10.245 𝑘𝑤 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑡𝑎𝑔𝑒: 𝑊 = 1.2824((0.0947)(0.287)(44 + 273)) 1.2824 − 1 790.611 (( 308.11 1.2824−1 1.2824 ) − 1) 𝑊 = 8.956 𝑘𝑤 𝑊𝑇 = 10.245 + 8.956 𝑊𝑇 = 19.20 𝑘𝑤 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 19.20 (35)(0.746)(0.95) 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 77.41%

[( ) [( ) 60 26. An air compressor is to compress 8.5 3 ⁄ 𝑚𝑖𝑛 from 98.56 Kpa to 985.6 Kpa. Assuming condition ideal, and with n = 1.3, what will be the saving in the work due to two staging? 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 𝑤 = 1.3(98.56) ( 8.5 ) 60 (( 1.3 − 1 985.6 ) 98.56 1.3−1 1.3 − 1) 𝑤 = 42.43 𝐾𝑤 𝐹𝑜𝑟 𝑡𝑤𝑜 𝑠𝑡𝑎𝑔𝑒 − 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑃 𝑋 = √ 98.56(985.6) 𝑃 𝑋 = 311.67 𝐾𝑝𝑎 𝑊 = 2𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 𝑊 = 2(1.3)(98.56) ( 8.5 ) 311.67 (( ) 1.3−1 1.3 − 1) 1.3 − 1 98.56 𝑊 = 36.83 𝑘𝑤 𝑃𝑜𝑤𝑒𝑟 𝑠𝑎𝑣𝑒𝑑 = 42.43 – 36.83 𝑃𝑜𝑤𝑒𝑟 𝑠𝑎𝑣𝑒𝑑 = 5.6 𝐾𝑊 𝑚

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27. A single air compressor handles 0.454 𝑚 3 ⁄ 𝑠𝑒𝑐 of atmospheric pressure, 27⁰C air, and delivers it to a receiver at 625.75 Kpa. Its volumetric efficiency on an isothermal basis is 0.85 and its mechanical efficiency 0.90. if it operates at 350 rpm, what power in KW required to drive it? 𝑊 = 𝑃 𝑉 𝑙𝑛 𝑃 2 1 1 ( 𝑃 ) 𝑊 = 101.30(0.454)𝑙𝑛 ( 𝑊 = 85.685 𝑘𝑤 652.75 101.3 ) 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 = 85.685 0.85(0.90) 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 = 112𝑘𝑤 28. A two stage compressor receives 0.50 𝐾𝑔 ⁄ 𝑠𝑒𝑐 of air at 120 Kpa and 300⁰K and deliovers it at 7mpa. Find the heat transferred in the intercooler. 𝑃₂ = 7 𝑀𝑝𝑎 𝑃₂ = 7000 𝐾𝑝𝑎 𝑃 𝑋 = 𝑖𝑛𝑡𝑒𝑟𝑐𝑜𝑜𝑙𝑒𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃 𝑋 = √( 120 ) (7000) 𝑃 𝑋 = 916.515 𝐾𝑝𝑎 𝑇 𝑋 916.515 1.4−1 1.4 (300) = ( 120 ) 𝑇𝑥 = 536.28⁰𝐾 𝑄 = 𝑚 𝐶 𝑃 ( 𝑇 𝑋 – 𝑇 1 ) 𝑄 = 0.5(1)(536.28 – 300) 𝑄 = 118.14 𝐾𝑤 1

[( ) 29. An air compressor is tested and it is found that the electric motor used 37.3 Kw when the compressor handled 0.189 𝑚 3 ⁄ 𝑠𝑒𝑐 of air at 101.4 and 300K and discharge it at 377.1 Kpa. Determine the adiabatic efficiency. 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 𝑤 = 1.4(101.4)(0.189) 1.4 − 1 [( 377.1 ) 101.4 1.4−1 1.4 − 1] 𝑤 = 30.54 𝑘𝑤 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 = 30.54 37.3 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 = 67.48% 30. An air compressor is tedted and it is found that the electric motor used 37.3 kw when the compressor handled 0.089 𝑚 3 ⁄ 𝑠𝑒𝑐 of air at 101.4 Kpa and 300⁰K and discharge it at 377.1 kpa. Determine the isothermal efficiency. 𝑊 = 𝑃 𝑉 𝑙𝑛 𝑃 2 1 1 ( 𝑃 ) 𝑊 = 101.4(0.189)𝑙𝑛 ( 377.1 ) 101.4 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 25.17 37.3 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 67.48% 1

31. Calculate the volumetric efficiency of a single cylinder, double acting compressor with a bore and stroke of 0.42 x 0.45 m. The compressor is tested at 150 rpm and found to deliver gas from 101.3 Kpa and 300K TO 675Kpa at a rate of 0.166 𝑚 3 ⁄ 𝑠𝑒𝑐 process. when n = 1.33 for expansion and compression 𝑉 𝐷 𝑉 = 2 ( 𝜋 𝐷 2 𝐿𝑁) 4 = 2 [( 𝜋 ) (0.45)2(0.45) ( 150 )] 𝐷 𝑉 𝐷 4 60 = 0.3578 𝑚 3 ⁄ 𝑠𝑒𝑐 𝑉₁ ƞ 𝑉 = 𝑉 𝐷 ƞ 𝑉 = 0.166 0.3578 ƞ 𝑉 = 𝟒𝟔. 𝟑𝟗%

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1 32. A reciprocating compressor with 3% clearance receives air at 100 Kpa and 300K and discharge it at 1.0 Mpa. The expansion and compression are polytropic with n = 1.25. There is a 5% pressure drop trough the inlet and outlet valves. Find volumetric efficiency. Solution: 𝑃₁ = 100(1 – 0.05) 𝑃₁ = 95 𝐾𝑝𝑎 𝑃₂ = 1000 ( 1 + 0.05) 𝑃₂ = 1050 𝐾𝑝𝑎 ƞ 𝑉 = 1 + 𝑐 – 𝑐 1 𝑃 2 𝑛 ( 𝑃 ) 1050 1 1.25 ƞ 𝑉 = 1 + 0.03 – 0.03 ( ƞ 𝑽 = 𝟖𝟐. 𝟓𝟎% 95 )

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1 [( ) 33. A reciprocating compressor has a 5% clearance with a bore and stroke of 25 x 30 cm. The compressor operates at 500 rpm. The air enters the cylinder at 27°C and 95 Kpa and discharges at 2000 Kpa. If n = 1.3, determine the compressor power. ƞ 𝑣 = 1 + 𝑐 − 𝑐 1 𝑃 2 𝑛 ( 𝑃 ) 2000 1 1.3 ƞ 𝑣 = 1 + 0.05 − 0.05 ( ƞ 𝑣 = 52.89% 95 ) 𝑉 𝐷 𝑉 = 𝜋 𝐷 2 𝐿𝑁 4 = 𝜋 (0.25)2(0.3) ( 500 𝐷 𝑉 𝐷 4 60 ) = 0.1227 𝑚 3 ⁄ 𝑠𝑒𝑐 𝑉 1 = 0.1227(0.5289) 3 𝑉 1 = 0.0649 𝑚 ⁄ 𝑠𝑒𝑐 𝑛−1 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛 − 1] 𝑛 − 1 𝑃 1 𝑊 = 1.3(95)(0.0649) 1.3 − 1 [( 2000 95 ) 1.3−1 1.3 − 1] 𝑊 = 27.25 𝐾𝑤

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ℎ𝑟 1 [( ) 34. A water-jacketed air compressor handles 0.143 𝑚 3 ⁄ 𝑠𝑒𝑐 of air entering at 96.5 Kpa and 21°C and leaving at 480 Kpa and 132°C; 10.9 𝑘𝑔 ⁄ of cooling water enters the jacket at 15°C and leaves at 21°C. Determine the compressor power. 𝑇 2 = 𝑇 1 𝑃 2 ( 𝑃 ) 𝑛−1 𝑛 (132 + 273) 480 𝑛−1 𝑛 (21 + 273) = ( 96.5 ) 𝑛 − 1 𝑛 = ln 1.377 ln 4.974 𝑛 = 1.249 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 1.249(96.5)(0.143) 480 1.249−1 1.249 𝑊 = 1.3 − 1 [( 96.5 ) − 1] 𝑊 = 26.087 𝐾𝑤 𝑄 = 𝑚𝐶 𝑝 (𝑡 1 − 𝑡 2 ) 𝑄 = ( 10.9 ) (4.187)(21 − 15) 3600 𝑄 = 0.076𝐾𝑤 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑊𝑜𝑟𝑘 = 𝑊 + 𝑄 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑊𝑜𝑟𝑘 = 26.087 + 0.076 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑊𝑜𝑟𝑘 = 26.163 𝐾𝑤

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[( ) [( ) 35. A 2- stage compressor operates between constant pressure limits of 98.6Kpa and 1.103 Mpa. The swept volume of the low-pressure piston is 0.142 𝑚 3 . Due to failure of the cooling water supply to the intercooler, air is passed to the high-pressure cylinder without reduction in temperature. Using 𝑃𝑉 1.2 = 𝐶 ; determine the percentage increase in power. 𝑃𝑥 = √𝑃1𝑃2 𝑃 𝑥 = √(98.6)(1103) 𝑃 𝑥 = 329.8 𝐾𝑝𝑎 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑝𝑜𝑤𝑒𝑟 𝑑𝑢𝑒 𝑡𝑜 𝑡𝑤𝑜 𝑠𝑡𝑎𝑔𝑖𝑛𝑔: 𝑛−1 𝑊 = 2𝑛𝑃 1 𝑉 1 𝑃 𝑥 𝑛 − 1] 𝑛 − 1 𝑃 1 (2)(1.2)(98.6)(0.142) 329.8 1.2−1 1.2 𝑊 = 1.2 − 1 [( 98.6 ) − 1] 𝑊 = 37.45 𝐾𝐽 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑝𝑜𝑤𝑒𝑟 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑖𝑛𝑔𝑙𝑒 𝑠𝑡𝑎𝑔𝑖𝑛𝑔: 𝑊 = 𝑛𝑃 1 𝑉 1 𝑃 2 𝑛−1 𝑛 − 1] 𝑛 − 1 𝑃 1 1.2−1 (1.2)(98.6)(0.142) 1103 1.2 𝑊 = 1.2 − 1 [( 98.6 ) − 1] 𝑊 = 41.63 𝐾𝐽 % 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 = 41.63 − 37.45 37.45 % 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 = 11.16%

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⁄ [( ) 𝐾𝐽 𝑊 = −145 𝑘𝑔 ( 𝑛𝑜 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 ) 𝑐ℎ𝑜𝑖𝑐𝑒𝑠 36. An ideal-single stage air compressor without clearance takes in air at 100 Kpa with a temperature of 16°C and delivered it at 413 Kpa after isentropic compression. What is the discharge work done by the compressor in 𝐾𝐽 𝐾𝑔 ? 𝑊 = 𝑘𝑚𝑅𝑇 𝑃 2 𝑘−1 𝐾 − 1] 𝑛 − 1 𝑃 1 𝑊 (−1.4)(0.287)(16 + 273) 1103 1.4−1 1.4 𝑚 = 1 − 1.4 [( 98.6 ) − 1]

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