# ProbSet2s - ME 416 Computer Assisted Design of Thermal...

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1 ME 416 Computer Assisted Design of Thermal Systems Solutions to Practice Problem Set #2 Turbomachinery 1. We begin by determining the actual power. The actual power will be determined from the generic centrifugal compressor graph. The inlet volume flow rate must be calculated from m v = V 1 1 where v = RT P = (0.287)(295) 98 = 0.864 m / kg 1 1 1 3 Calculating V = v m = (0.864)(16.4) = 14.2 m / s = (14.2 m / s)(60 s / min)(3.281 ft / m) = 30.1 x 10 cfm 1 1 3 3 3 3 Reading from the graph for a pressure ratio of 7, we find Basic BHP = 7,000 hp Correcting for the inlet pressure, since it’s not 14.5 psia gives [ ] / . ) . W = (P Basic BHP = (98 / 6.895) 14.5 (7000) = 6862 hp = (6862 hp)(0.7455 kW / hp) = 5,115 kW act 1 6895 145 The isentropic efficiency is given by m / W w = w w = act ideal act ideal s η This requires the ideal work. Assuming a constant specific heat or 1.005 J/kg, we can write w ideal = c P (T 2i - T 1 ) where T 2i comes from our isentropic ratio K 514 ) 7 )( 295 ( P P T T 4 . 1 / ) 1 4 . 1 ( / ) 1 ( 1 2 1 i 2 = = = - γ - γ so that w ideal = (1.005)(514-295) = 220 kJ/kg

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ME 416 CAD of Thermal Systems 2 η s ideal act ideal act = w w = w W m = 220 5115/ 16.4 = 0.705 / 2. We begin with the single stage compression system. With a compression ratio of 12:1, we have P 2 = 12P 1 = 1200 kPa The ideal exit temperature is calculated from K 610 ) 12 )( 300 ( P P T T 4 . 1 / ) 1 4 . 1 ( / ) 1 ( 1 2 1 i 2 = = = - γ - γ The ideal work is calculated from w ideal = c P (T 2i - T 1 ) = (1.005)(610-300) = 312 kJ/kg Then using the adiabatic efficiency, we calculate the actual work w act = w ideal / η s = 312/0.85 = 367 kJ/kg For the two stage compression with intercooling system, we have for stage #1 P 2 = 6.5P 1 = 650 kPa The ideal exit temperature is calculated from K 512 ) 5 . 6 )( 300 ( P P T T 4 . 1 / ) 1 4 . 1 ( / ) 1 ( 1 2 1 i 2 = = = - γ - γ The ideal work is calculated from w ideal = c P (T 2i - T 1 ) = (1.005)(512-300) = 213 kJ/kg Then using the adiabatic efficiency, we calculate the actual work w act,1 = w ideal / η s = 213/0.85 = 251 kJ/kg For the intercooler analysis we will need to calculate the actual exit temperature for the first stage, which comes from T 2a = T 1 + w actl /c P = 300 + (251)/1.005 = 550 K The intercooler is assumed to be a constant pressure device, so that
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## This note was uploaded on 07/25/2008 for the course ME 416 taught by Professor Somerton during the Fall '07 term at Michigan State University.

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ProbSet2s - ME 416 Computer Assisted Design of Thermal...

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