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3. Air enters a compressor at ambient conditions of 15 psia and 77°F with a low velocity and exits at 180 psia, 740°F, and 480 ft/s. The compressor is cooled by the ambient air at 77°F at a rate of 1800 Btu/min. The power input to the compressor is 450 hp. Determine (a) the mass flow rate of air, (b) the portion of the power input that is used just to over-come the irreversibilities, (c) the maximum amount of power input, and (d) the second-law efficiency o this compressor.

3. Air enters a compressor at ambient conditions of 15 psia and 77°F with a low velocity and exits at 180 psia, 740°F, and 480 ft/s. The compressor is cooled by the ambient air at 77°F at a rate of 1800 Btu/min. The power input to the compressor is 450 hp. Determine (a) the mass flow rate of air, (b) the portion of the power input that is used just to over-come the irreversibilities, (c) the maximum amount of power input, and (d) the second-law efficiency o this compressor.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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› → → Step 1 Given Data • Work inlet is: Win 400 hp 282.72 Btu/st • Heat output is: - 1500 Btu/min=25 btu/s Step 2 (a) By using the energy balance equation W + m (h₁ + 2) = + m (h₂ + ² ) W-Qm (h₂-h₁+¹²)............. ·(1) Enthalpy and entropy at 15 psia and 60°F h = 124.27 Btu/lbm $10.59173 Btu/lbm R. Enthalpy and entropy at h₂260.97 Btu/lbm $20.76964 Btu/lbm R Step 3 Substitute values in expression (1) 282.72-25-m (260.97-124.27+350-0² (2007)) m 1.852 lbm/s Thus the mass flow rate of the air is 1.852 lbm/s 150 psia and 620°F (b) By using the entropy balance equation Sgen = m (82-81) + (²) The change in entropy is 52-51-0.76964-0.59173-R In () -0.76964 -0.59173 -0.06855 In (1) -0.02007 Btu/lbm. R. Step 4 Substitute values in expression (2) Spew (1.852) (0.02007) +25 (60+460) -0,0853 Btu/s R The rate of exergy destruction is x=Topen =400 (60+460) (0,0853) -44.656 Btu/sx 768 B =63 hp The portion of the power input is -0.15 Thus the portion of the power input is 0.15 is is
Transcribed Image Text:› → → Step 1 Given Data • Work inlet is: Win 400 hp 282.72 Btu/st • Heat output is: - 1500 Btu/min=25 btu/s Step 2 (a) By using the energy balance equation W + m (h₁ + 2) = + m (h₂ + ² ) W-Qm (h₂-h₁+¹²)............. ·(1) Enthalpy and entropy at 15 psia and 60°F h = 124.27 Btu/lbm $10.59173 Btu/lbm R. Enthalpy and entropy at h₂260.97 Btu/lbm $20.76964 Btu/lbm R Step 3 Substitute values in expression (1) 282.72-25-m (260.97-124.27+350-0² (2007)) m 1.852 lbm/s Thus the mass flow rate of the air is 1.852 lbm/s 150 psia and 620°F (b) By using the entropy balance equation Sgen = m (82-81) + (²) The change in entropy is 52-51-0.76964-0.59173-R In () -0.76964 -0.59173 -0.06855 In (1) -0.02007 Btu/lbm. R. Step 4 Substitute values in expression (2) Spew (1.852) (0.02007) +25 (60+460) -0,0853 Btu/s R The rate of exergy destruction is x=Topen =400 (60+460) (0,0853) -44.656 Btu/sx 768 B =63 hp The portion of the power input is -0.15 Thus the portion of the power input is 0.15 is is
3. Air enters a compressor at ambient conditions of 15 psia and 77°F with a low velocity and exits at 180 psia, 740°F, and 480 ft/s. The compressor is cooled by the ambient air at 77°F at a rate of 1800 Btu/min. The power input to the compressor is 450 hp. Determine (a) the mass flow rate of air, (b) the portion of the power input that is used just to over-come the irreversibilities, (c) the maximum amount of power input, and (d) the second-law efficiency o its this compressor.
Transcribed Image Text:3. Air enters a compressor at ambient conditions of 15 psia and 77°F with a low velocity and exits at 180 psia, 740°F, and 480 ft/s. The compressor is cooled by the ambient air at 77°F at a rate of 1800 Btu/min. The power input to the compressor is 450 hp. Determine (a) the mass flow rate of air, (b) the portion of the power input that is used just to over-come the irreversibilities, (c) the maximum amount of power input, and (d) the second-law efficiency o its this compressor.
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