This example presents a gas with a temperature limit and is typically found in a halogen mixture. A multi-section compressor is required to accommodate the limit. This example illustrates one approach for the division of work between the sections to achieve a discharge temperature within the specified bound.

mw = 69 kj = 1.35 k; = I 33 Zi = .98 Z2 = .96 t| = 80°F Pi = 24 psia P2 - 105 psia w = 3,200 lbs/min

Step 1. Use Equation 2.5 to calculate the specific gas constant.

Step 2. Convert the inlet temperature to absolute. Substitute into Equation 2.10 and using the conversion constant of 144

Step 3. Calculate the overall poltropic exponent using Equation 2.71 The average was used in evaluating k because the values were not all that


Step 5. Calculate the discharge temperature for the total pressure ratio to check against the stated temperature limit, using the assumed efficiency, r|p = 75 and the polytropic exponent. Apply Equation 5.14,

Since the limit is 265°F and the overall temperature is obviously in

Intercooler outlet temperature must be determined. If cooling water at 90°F and an approach temperature of 15°F are assumed, the gas outlet from the cooler returning to the compressor will be 105°F,

If Equation 3.12 is borrowed from the reciprocating compressor chapter and used for an uncooled section, the pressure ratio per section may be calculated assuming an approximate equal-work division. For the first trial, assume the limit of temperature may be achieved in two sections.

From the rule of thumb given for estimating intercooler pressure drop, a value of 2 psi is used because it is larger than 2% of the absolute pressure at the cooler. The pressure drop must be made up by the compressor by additional head, and can be added to the first or second section pressure ratio. By applying a little experience, the guessing can be improved. The front section has a lower inlet temperature and is generally more efficient, so the best location for additional pressure would be in the first section. The first section discharge pressure is 50.2 + 2 = 52.2 psia. A new pressure ratio for the first section must be evaluated.

Step 6. Evaluate the discharge temperature, continuing the use of the previously calculated polytropic exponent.

t2 = 242.2°F first section discharge temperature

This temperature is within the limit.

intercooler outlet pressure is 50.2 psia. Calculate the second section pressure ratio.

Evaluate the Section 2 discharge temperature.

t2 = 265 °F discharge temperature

Because the temperature just calculated is right on the temperature limit and there is margin in the Section 1 temperature, the pressure may be arbitrarily adjusted to the first section to better balance the temperatures. A Section 1 discharge pressure of 54.5 psia is selected, which results in a new pressure ratio.

Now calculate a new Section 1 discharge temperature for the pressure just assumed.

The temperature is still within the required limit. Correct the cooler outlet pressure and evaluate a new ratio for Section 2, The corrected cooler outlet pressure is 52.5 psia.

Recalculate the discharge temperature for Section 2, using the previous cooler outlet temperature.

The temperature is now below the 265°F limit and consistent with the Section 1 temperature. At this point, the initial assumption for 2 sections can be considered a firm value.

Step 7. Calculate the poly tropic head for each section, using the overall average compressibility of Z2aVg = .97.

Section 1

Hp = .97 x 22.39 x 540 x2.956(2.271)-338 Hp = 11,074 ft-lb/lb

Section 2

Hp = .97 x 22.39 x 565 x 2.956(2.0)-338 Hp = 9,576 ft-lb/lb

Step 8. Develop the allowable head per stage by the use of one of (ho rules of thumb.

Divide the total head per section by the allowable head per stage to develop the number of stages required in each section.

Section 1

Section 2

Step 9. Calculate a head per stage for each section based on two stages each.

Section 1

HP = 5,537 ft-lb/lb head per stage, Section 1

Section 2

Hp = 4,788 ft-lb/lb head per stage, Section 2

Use the geometric portion of Equation 5.12 to calculate the tip speed. Assume [Xp = .48 for the pressure coefficient.

Section 1

u2 = (5,537 X32.2/.48)-5 u2 = 609.5 fps tip speed first two stages

Section 2

u2 = 566.7 fps tip speed last two stages

Step 10. From Figure 5-26 and the inlet volume to the first section, select an initial impeller diameter.

Because the second section shares a common shaft with the first section, it is not necessary to look up a new impeller size. Apply the Section 1 impeller diameter, Equation 5.15, and the conversion constants of 12 in./ft and 60 sec/min. to calculate a shaft speed.

12x60x566.7 rex 5,588

With the shaft speed and the tip speed calculated in Step 9 for the Section 2 stages, calculate an impeller diameter using Equation 5.15.



d2 ~ 23.24 in. Section 2 impeller diameter

Step 11. Calculate the inlet volume into Section 2. Use Zavg = 97, Pj = 52.5 psia, and tj = 105°F. Substitute into Equation 2.10 as was done in

Step 2.

1 52.5x144

Q, = 5,194 cfm inlet volume into Section 2

Calculate the last impeller volume for each section using Equation 5.18.

Section 1


Q]s = 8,363.4 cfm last stage volume, Section 1

Section 2



Qls = 4,129.4 cfm last stage volume, Section 2

Use Equation 5.19 to evaluate the flow coefficient for the first and last impeller of each section.

Section 1 S = (700 x 10,971)/(5,588 x 253) 8 = .088 flow coefficient, first stage 8 = (700 x 8,364.4)/(5,588 x 253) 8 - .067 flow coefficient, last stage

Section 2 8 = (700 x 5,194)/(5,588 x 23.243) 5 = .052 flow coefficient, first stage 8 = (700 x 4,129.4)/(5,588 x 23.243) 8 = .041 flow coefficient, last stage

Step 12. Use Figure 5-27 and the flow coefficients to determine the efficiencies for the stages.

Section 1

8 =.088, rip =.79 8 = .067, Tip = .80 The average is Tip = .795

Section 2

The average is rip--787

Step 13. Recalculate the poly tropic exponent

Section 1

Use kavg= 1.345

Section 2

Use kavg= 1.335

Step 14. Use the poly tropic exponents calculated in the previous step and recalculate the discharge temperature of each section to correct for the average stage efficiency.

Section 1

T2 = 540(2.271>323 T2 = 703.8 °R t3 = 703.8 -460

t2 = 243.8°F final Section 1 discharge temperature.

Section 2

t2 ~ 244.8°F final Section 2 discharge temperature.

The temperature is approximately 20°F below the 265°F temperature limit. The sections differ by less than 1°F. This is probably just luck because that good a balance is not really necessary. Also, it should be noted that to maintain simplicity the additional factors were ignored, such as the 10°F temperature pickup in the return stream due to internal wall heat transfer. Also, nozzle pressure drops for the exit and return were not used. Balance piston leakage was not used as it was in Example 5-3. When all the factors are used, the pressures for each section would undoubtedly need additional adjustment as would the efficiency. Howev er, for the actual compression process, the values are quite realistic, and for doing an estimate, this simpler approach may be quite adequate.

Step 15. To complete the estimate, calculate the shaft power, using the conversion of 33,000 ft-lb/min/hp.

Section 1


Wp = 1,350.8 hp gas horsepower, Section 1

Section 2

Wp = 1,179.9 hp gas horsepower, Section 2

Combine the two gas horsepower values and add 1 % for the mechanical losses.

Wp = 2,556.0 hp compressor shaft power

Fan Laws

These relationships were actually developed for pumps instead of compressors, but they are very useful in rating compressors that are being considered for reapplication. The equations used to this point are adequate to perform any rerate calculation; however, looking at the fan laws may help establish another perspective. The following relationships are a statement of the fan laws.

The equations have been expressed as proportionals; however, they can be used by simply "ratioing" an old to a new value. To add credibility to fan law adaptation, recall the flow coefficient, Equation 5.19. The term Qj/N is used which shows a direct proportion between volume Qj and speed N. Equation 5.12 indicates the head, Hp, to be a function of the tip speed, u2 squared. The tip speed is, in turn, a direct function of speed making head proportional to speed. Finally, the power, Wp, is a function of head multiplied by flow, from which the deduction of power, proportional to the speed cubed, may be made.

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