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• 120 Rules of Thumb for Mechanical Engineers

Centrifugal Compressor Performance Calculations

Centrifugal compressors are versatile, compact, and generally used in the range of 1,000 to 100,000 inlet cubic ft per minute (ICFM) for process and pipe line compression applications.

Centrifugal compressors can use either a horizontal or a vertical split case. The type of case used will depend on the pressure rating with vertical split casings generally being used for the higher pressure applications. Flow arrange- ments include straight through, double flow, and side flow configurations.

Centrifugal compressors may be evaluated using either the adiabatic or polytropic process method. An adiabatic process is one in which no heat transfer occurs. This doesn't imply a constant temperature, only that no heat is trans- ferred into or out of the process system. Adiabatic is nor- mally intended to mean adiabatic isentropic. A polytropic process is a variable-entropy process in which heat transfer can take place.

When the compressor is installed in the field, the power required from the driver will be the same whether the pro- cess is called adiabatic or polytropic during design. There- fore, the work input will be the same value for either pro- cess. It will be necessary to use corresponding values when making the calculations. When using adiabatic head, use adiabatic efficiency and when using polytropic head, use polytropic efficiency. Polytropic calculations are easier to make even though the adiabatic approach appears to be simpler and quicker.

The polytropic approach offers two advantages over the adiabatic approach. The polytropic approach is indepen- dent of the thermodynamic state of the gas being com- pressed, whereas the adiabatic efficiency is a function of the pressure ratio and therefore is dependent upon the ther- modynamic state of the gas.

If the design considers all processes to be polytropic, an impeller may be designed, its efficiency curve determined, and it can be applied without correction regardless of pres- sure, temperature, or molecular weight of the gas being compressed. Another advantage of the polytropic approach is that the sum of the polytropic heads for each stage of compression equals the total polytropic head required to get from state point 1 to state point 2. This is not true for adiabatic heads.

Sample Performance Calculations

Determine the compressor frame size, number of stages, rotational speed, power requirement, and discharge tem- perature required to compress 5,000 lbm/min of gas from 30 psia at 60F to 100 psia. The gas mixture molar compo- sition is as follows:

Ethane 5%

n-Butane 15 % Propane 80 %

The properties of this mixture are as follows:

MW = 45.5 P, = 611 psia T, = 676"R cp = 17.76 kl = 1.126 Z1 = 0.955

Before proceeding with the compressor calculations, let's review the merits of using average values of Z and k in cal- culating the polytropic head.

The inlet compressibility must be used to determine the actual volume entering the compressor to approximate the size of the compressor and to communicate with the vendor via the data sheets. The maximum value of 8 is of interest and will be at its maximum at the inlet to the compressor where the inlet compressibility occurs (although using the average compressibility will result in a conservative esti- mate of e).

Compressibility will decrease as the gas is compressed. This would imply that using the inlet compressibility would be conservative since as the compressibility de- creases, the head requirement also decreases. If the varia- tion in compressibility is drastic, the polytropic head re-

• Pumps and Compressors 121

quirement calculated by using the inlet compressibility would be practically useless. Compl.essor manufacturers calculate the performance for each stage and use the inlet compressibility for each stage. An accurate approximation may be substituted for the stageby-stage calculation by calculating the polytropic head for the overall section using the average compressibility. This technique d t s in over- estimating the first half of the impellers and Underestimat- ing the last half of the impellers, thmby calculating a polytropic head very near that calculated by the stapby- stage technique.

Determine the inlet flow volume, Q1:

where m = mass flow Z1= inlet compdbility factor

Pi = inletpresfllre

R = gas constant = 1,545/MW TI .- inlet temperature "R

Q1 = 5,OOO[(O.955)(1,545)(80 + ~ 8 0 ) / ( ~ 5 . 5 ) ( 1 ~ ) ~ ~ ) ] = 19,517 ICFM

Refer to Bible 3 and select a compressor frame that will handle a flow rate of 19,517 ICFM. A name C Compressor will handle a range of 13,000 to 31,000 ICFM and would have the following nominal dak

!&,- = 10,OOO ft-lb/lbm (nominal polytropic head)

np = 77% (polytropic &iency)

N,, = 5,900 rpm

Determine the pl.iessure ratio, rp.

rp = PB/P1 = 100/30 = 3.33

Determine the approximate discharge temperature, Tg.

nh - 1 = [Wk - 11% = [1.126/(1.126 - 1.000)](0.77) = 6.88

T2 = Tl(rp)(n-l)/n = (60 + 460)(3.33)"f3.88 = 619"R = 159F

Determine the average compressibility, Z,.

Z1= 0.955 (from gas properties calculation)

where Z1= inlet compressibility

(PJ2 = pzlp, = 100/611 = 0.164

( T J B = TdTC = 619/676 = 0.916

nble 3

Typical Centrifugal Compressor Frame Data* Nominal

Impeller Diameter Nominal Nominal Polytropic Rotational

Nominal Polytropic Head Nominal Inlet Volume Flow

English Metric Engllsh Metric Efficiency Speed Engllrh MeMc Frame (ICFM) Im'/h) Ift-lbf/lbml (k-Nm/kg) 1%) IRPM) (in) Imm) A 1 ,ONk7,000 1.7oO-12.OOO 1o.ooo 30 76 11,000 16 406 B 6,000-1 8 ,OW 10.000-31 ,000 10,000 30 76 7.700 23 584 C 13,ooO-31 .OOO 22,000-53.000 10,ooo 30 n 5.900 30 762 D 23,000-44.000 39.000-75.000 10,ooo 30 n 4.900 36 914 E 33 ,ooo65 .OOO 56.000-1 10,000 10,000 30 78 4,000 44 i.im F 48.000-100.000 82.OCS170.000 10.000 30 78 3.300 54 1.370 *While this table is based on 8 survey of currently available equipment, the instance of any machinery duplicating this table would be purely

coincidental.

• 122 Rules of Thumb for Mechanical Engineers

e

Figure 10. Maximum polytropic head per stageEnglish system.

Refer to Figure 5 to find Zz, discharge compressibility. temperature but also at the estimated discharge tempera- ture.

Zz = 0.925 The suggested approach is as follows:

z, = (Z, + Z2)/2 = 0.94

Determine average k-value. For simplicity, the inlet value of k will be used for this calculation. The polytropic head equation is insensitive to k-value (and therefore n- value) within the limits that k normally varies during com- pression. This is because any errors in the n/(n - 1) multi- plier in the polytropic head equation tend to balance corresponding errors in the (n - l)/n exponent. Discharge temperature is very sensitive to k-value. Since the k-value normally decreases during compression, a discharge tem- perature calculated by using the inlet k-value will be con- servative and the actual temperature may be several de- grees higher-possibly as much as 2540F. Calculating the average k-value can be time-consuming, especially for mixtures containing several gases, since not only must the mol-weighted cp of the mixture be determined at the inlet

1. If the k-value is felt to be highly variable, one pass should be made at estimating discharge temperature based on the inlet k-value; the average k-value should then be calculated using the estimated discharge tem- perature.

2. If the k-value is felt to be fairly constant, the inlet k- value can be used in the calculations.

3. If the k-value is felt to be highly variable, but suffi- cient time to calculate the average value is not avail- able, the inlet k-value can be used (but be aware of the potential discrepancy in the calculated discharge temperature).

kl = k, = 1.126

Determine average n/(n - 1) value from the average k- value. For the same reasons discussed above, use n/ (n - 1) = 6.88.

Table 4

Approximate Mechanical Losses as a Percentage of Gas Power Requirement.

~

Gas Power Requirement Mechanical English Metric Losses, L,

(hp) IkWI (%I 0-3.000

3.000-6.000 6,000-10.000

10,000+

0-2,500 2,500-5,000 5,000-7.500

7,500+

3 2.5 2 1.5

*There is no way to estimate mechanical losses from gas power requirements. This table will. however, ensure that mechanical losses are considered and yield useful values for es- timating purposes.

• Pumps and Compressors 123

Hp = Z,RTl(n/n - l)[rp(n-L)'n - 11 = (0.94) (1,545/45.5) (520) (6.88) (3 .33)'/".88 - 11 = 21,800 ft-lbf/lbm

Determine the required number of compressor stages, 8:

8 = [(26.1MW)/(kiZ1T1)]0.5 = [ (26.1) (45.5)/ (1.126) (0.955) (520)]0.5 = 1.46

max Hp/stage from Figure 10 using 8 = 1.46

Number of stages = Hp/max. H,/stage = 21,800/9,700 = 2.25 = 3 stages

Determine the required rotational speed:

Mechanical losses (L,) = 2.5% (from Table 4) L, = (0.025)(4,290)

= 107 hp

PWR, = PWR, + L,,, = 4,290 + 107 = 4,397 hp

Determine the actual discharge temperature:

TZ = Tl(rp)(n-l)/n = 520(3.33)"6.88 = 619"R = 159F

The discharge temperature calculated in the last step is the same as that calculated earlier only because of the deci- sion to use the inlet k-value instead of the average k-value. Had the average k-value been used, the actual discharge temperature would have been lower.

N = N,,[HP/Hpn,, x no. = 5,900[21,800/(10,000) (3)]0.5 = 5,030 rpm

Determine the required shaft power: Source

PWR, = m