## Example

Calculate the performance of a compressor using air for the following conditions:

d = 10.5 in. rotor diameter L/d - 3.5 length-diameter ratio mw = 23

O, = 2,500 acfm inlet volume i. = I00°F inlet temperature Pi. = 14.5 psia inlet pressure P2 --■ 43.5 psia discharge pressure rp = 3.0 pressure ratio k= l .23

w = 138.8 lbs/min weight flow

Step 1. Using Equation 4.2, solve for the displaced volume per revolution. Convert the units using 1728 in.3/ft3,

From Figure 4-6, read a volumetric efficiency for pressure ratio, 3, where Ev -- 89%. Use Equation 4.4 and solve for the total displaced volume.

Qd = 2,809 cfm total displacement volume

Now calculate the required speed by substituting into Equation 4.3. N = 2.809/.450 N = 6,242 rpm rotor speed

Step 2. Find the velocity of sound for air at the inlet conditions given, using Equation 2.32 from Chapter 2.

R = 67 17 specific gas constant a = (1.23 x 67.17 x 32.2 x 560)1/2 a = 1,220.6 fps velocity of sound

Compute the rotor tip velocity using Equation 4.5 and the unit conversions of 12 in./ft and 60 sec/min.

kx 10.5x6242

60 x 12

u = 286.0 fps rotor tip velocity

Refer to Figure 4-4 with pressure ratio = 3.0 and read the rotor Mach number uc/a = .25. Calculate uD, the optimum tip velocity.

u„ = 305 fps optimum tip velocity

Then calculate the tip speed ratio.

Step 3. Refer to Figure 4-7 and select an efficiency at a pressure ratio of 3 and a volume ratio, rv of 2.44. The adiabatic efficiency is 74%. Now, from Figure 4-5, select a value of efficiency ratio using the tip speed ratio just calculated. Because the value is .99+, round off to an even 1.0. With a multiplier of 1.0, the final adiabatic efficiency is the value read directly off the curve or T|a = 74. The molecular weight correction for efficiency, per rule of thumb, is 0.6 for a final efficiency of 73.4.

Step 4. The adiabatic power can be solved by substituting into Equation 3.18.

k/(k - 1) = 1.23/.23 k/(k - I) = 5.34 (k- 1 )/k = . 187

Calculate the power using the conversions of 144 in.2/ft2 and 33,000 ft Ibs/min/hp for a net value of 229.

Substitute into Equation 4.6 for the discharge temperature.

.734

12 = 256.6°F discharge temperature

Step 5. Solve for the shaft power substituting into Equation 4.7.

Ps — 281.0 hp shaft horsepower ## Living Off The Grid

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