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COMPRESSORS
Centrifugal compressors are best discussed by going through a calculation procedure and discussing each part.
First, the required head is calculated. Either the polytropic or adiabatic head can be used to calculate horsepower as long as the polytropic or adiabatic efficiency is used with the companion head.
Polytropic Head
Hpoly = Z R T1 [ (P2/P1)^{(K-1)/(K Ep)} -1 ] / {(K - 1)/(K Ep)}
Adiabatic Head
HAD = Z R T1 [ (P2/P1)^{(K-1)/K} -1 ] / {(K - K)/K}
where,
Z = average compressibility factor; using 1.0 will yield conservative results
R = 1,544/mol. wt.
T1 = suction temperature, °R
P1, P2 = suction, discharge pressures, psia
K = adiabatic exponent, Cp/Cv
Ep = polytropic efficiency, use 75% for preliminary work
T = Gas Temperature °R (°F + 460)
From Polytropic Head
HP = W Hpoly / [ Ep 33000 ]
From Adiabatic Head
HP = W HAD / [ EA 33000 ]
where,
HP = gas horsepower
W = flow, lb/min
To the gas horsepower is added bearing and oil seal losses.
Use 50 hp in lieu of manufacturer's data for large machines.
The discharge temperature is calculated as follows:
t2 = t1 + {Hpoly / [ Z R [K / (K-1) ] Ep ] }
Often the temperature of the gas must be limited. High temperature requires a special and more cosily machine. Most multistage applications are designed to stay below 250-300 °F. At temperatures greater than 450-500 °F the approximate mechanical limit, problems of sealing and casing growth start to occur.
Intercooling can be used to hold desired temperatures for high overall compression ratio applications. This can be done between stages in a single compressor frame or between series frames. Sometimes economics rather than a temperature limit dictate intercooling.
Sometimes for high compression ratio applications, the job cannot be done in a single compressor frame. Usually a frame will not contain more than about eight stages (wheels). There is a maximum head that one stage can handle.
This depends upon the gas properties and inlet temperature. Usually this will run 7,000 to 11,000 feet for a single stage. In lieu of manufacturer's data use eight maximum stages per frame. Then subtract one stage for every side nozzle, such as to and from an intercooler, side gas injection, etc. For many applications the compression ratio across a frame will run 2.5-4.0.
Centrifugal Compressor Power Calculation
Natural gas composition rich in Methane, usually between 76 and 90 mole %, the remaining are the concentrations of hydrocarbons C4, C3 and C2.
Case 1: Methane (CH4) 90%, Ethane (C2H6) 10%
Case 2: Methane (CH4) 90%, Butane (C4H10) 10%
Case 3: Methane (CH4) 76%, Ethane (C2H6) 24%
Case 4: Methane (CH4) 76%, Butane (C4H10) 24%
Case 1: Methane (CH4) 76%, Butane (C4H10) 24%
.
.
.
Case 2: Methane (CH4) 90%, Butane (C4H10) 10%
.
Case 3: Methane (CH4) 76%, Ethane (C2H6) 24%
.
Case 4: Methane (CH4) 90%, Ethane (C2H6) 10%
.
Pressure of a gas stream (30 degC, 200 MMscfd, 90% C1, 10% C2) must increase from 600 psia to 1100 psia. Design a Centrifugal compressor.
Calculation:
T1 = 546 °R
r = 1.833
Mg = 17.45
ɣg = 0.602
Ppc = 674.3 psia
Tpc = 355.5 °R
Tpr = 1.54
Ppr = 0.89
z1 = 0.93
q1 = 3.33 Mcfd
k = 1.271
ηp = 0.715
Hp = 29848
ns = 3
T2 = 237 °F < 350
GHP = 8085 (7935)
M.L = 27
BHP = 8112
S = 11000 rpm
Number of stages of compression
Using the specified overall pressure ratio and suction temperature (and an assumed efficiency), the discharge temperature for compression of gas with a known k value in a single stage can be estimated by rewriting Eq. 7.
................(14)
where
T2 = estimated absolute discharge temperature, °R,
T1 = specified absolute suction temperature, °R,
P1 = specified absolute suction pressure, psia,
P2 = specified absolute discharge pressure, psia,
k = ratio of specific heats,
ηp = assumed polytropic efficiency,
≈ 0.72 to 0.85 for centrifugal compressors,
If the single-stage discharge temperature is too high (typical limit is 300 to 350 °F), it is necessary to configure the compression equipment in more than one stage. Calculating the compression ratio per stage with Eq. 15 does the evaluation of a multistage design.
................(15)
where
Rsect = compression ratio per section,
and
n = number of sections.
Using the previous equations and prudent assumptions, it is possible to determine the minimum number of stages required to accomplish a given overall compression ratio without exceeding temperature limits.
Nomenclature
k = Cp/Cv
Cp/Cv = ratio of specific heats, dimensionless
n = polytropic exponent
His = isentropic head, ft-lbf/lbm,
zavg = average compressibility factor, dimensionless,
Ts = suction temperature, °R,
S = gas specific-gravity (standard atmospheric air = 1.00),
Pd = discharge pressure, psia,
Ps = suction pressure, psia
Hp = polytropic head, ft-lbf/lbm,
ηp = polytropic efficiency
ηis = isentropic efficiency,
Ts = suction temperature, °R,
Td = discharge temperature (actual or predicted), °R,
k = ratio of specific heats, Cp/Cv
ηp = polytropic efficiency
P = pressure,
V = volume,
N = number of moles,
R = constant for a specific gas,
T = temperature
W = mass flow, lbm/min.,
R = universal gas constant = 1,545,
MW = molecular weight,
Ts = suction temperature, °R,
zs = compressibility at inlet,
Ps = absolute suction pressure, psia
Qg = standard volume flow, MMscf/D
GHP = gas power, horsepower,
W = mass flow, lbm/min.,
Hp = polytropic head, ft-lbf/lbm
P1 = inlet pressure, psia,
V1 = inlet volume, ACFM,
P2 = discharge pressure, psia,
CE = compression efficiency (assume 0.85 for estimating purposes)
T2 = estimated absolute discharge temperature, °R,
T1 = specified absolute suction temperature, °R,
P1 = specified absolute suction pressure, psia,
P2 = specified absolute discharge pressure, psia,
k = ratio of specific heats,
ηp = assumed polytropic efficiency,
≈ 0.72 to 0.85 for centrifugal compressors,
Rsect = compression ratio per section,
n = number of sections
Polytropic efficiency mathematical equation can be defined as follows:
npoly = n/(n-1) / k/(k-1)
where:
n= polytropic exponent, dimensionless
k = Cp/Cv =average (suction/discharge) ratio of specific heats, dimensionless
Calculate the polytropic exponent. For practical purpose as a first approximation an average polytropic efficiency of 75% (0.75) is considered per stage of compression & the polytropic exponent calculated based on the above equation knowing the average specific heat ratios.
Alternatively, if you know the discharge temperature (T2) of the compression stage based on field measured values then you can calculate the exponent value as follows:
exponent (n or gamma) = ln(P2/P1) / (ln(P2/P1) - ln(T2/T1))
where:
P1 = stage suction pressure, absolute units
P2 = stage discharge pressure, absolute units
T1 = stage suction temperature, absolute units
T2 = stage discharge temperature, absolute units
The discharge temperature from actual measurement then to calculate the discharge temperature you require the exponent value using the equation:
T2 = T1* (P2/P1)n-1/n
So if you are doing some sizing calculations it is always a practical approach to consider a 75% polytropic efficiency for starting the calculations.
The specific heat ratio (k = Cp/Cv) is considered constant for all the stages. Since Specific heat at constant pressure (Cp) is a function of temperature, you should recheck the Cp values at the outlet of each stage for your gas composition and then recalculate the specific heat ratio at the outlet of each stage as
k = Cp/Cv = Cp / (Cp-(8.314 / MW))
where:
Cp = specific heat at constant pressure, KJ/kg-K
MW = Molecular weight of gas or gas mixture, kg / kmole
__________
| __Component__ | _Formula_ | __M.W.___ | __mol %___ | _Pseudo_M.W.__ | ___Critical_Press.___ | ___Critical_Temp.___ | __MWcp___ | _MWcp x Mol %_ |
| __Methane__ | ___C1___ | __16.0___ | ___89%____ | _14.24_ | __668__ | ___343___ | __8.54__ | ___7.6___ |
| __Ethane__ | ___C2___ | __30.1___ | ___4%____ | _1.20__ | ___708___ | ___550___ | __12.60___ | ___0.504___ |
| __Propane__ | ___C3___ | __44.1___ | ___5%____ | _2.21__ | ___616___ | ___666___ | __17.6___ | ___0.88____ |
| __Carb.Dioxide__ | ___CO2___ | __44.0___ | ___2%____ | _0.88__ | ___1071___ | ___548___ | __8.89___ | ___0.1778___ |
| _______ | _____ | _______ | ||||||
| __Mixture_Gas__ | _ | _ | ___100% | _18.53__ | _ | _ | _ | ___9.162____ |
_____
Gas Flowrate at Standard Condition
Gas Flowrate at Standard Condition = 15 MMSCFD
Gas Flowrate at Standard Condition = 15,000,000 cubic foot per day (ft3/day)
Gas Flowrate at Standard Condition = 10,420 cubic foot per minute (ft3/min)
Standard volume flow
Standard volume flow is the most common term used by the industry to describe volumetric flow because it is independent of actual gas pressures or temperatures. It is the volume per unit of time using pressures and temperatures that have been corrected to "standard" conditions. These conditions apply to pressure, temperature, molecular weight, and compressibility. The standards must be known and held constant. For purposes of this text, the standard conditions used are
pressure = 14.7 psia,
temperature = 60 °F,
compressibility = 1.00,
and
molecular weight = MW of subject gas.
Standard volume flow is usually dry and expressed in millions of standard cubic feet per day (MMScf/D).
Weight Flow of the Gas at Standard Conditions
Standard reference conditions is at Temperature 60°F (520°R), Absolute pressure 14.73 psia as referred to Publishing or establishing entity per EGIA, OPEC, U.S. EIA
In thermodynamics, the reduced properties of a fluid are a set of state variables normalized by the fluid's state properties at its critical point.
Reduced pressure, Pr
The reduced pressure is defined as its actual pressure divided by its critical pressure, Pc
Pr = P / Pc
Reduced temperature, Tr
The reduced temperature of a fluid is its actual temperature, divided by its critical temperature, Tc
Tr = T / Tc
Development of Hartwick, W. (Chemical Engineering, Oct, 1956) indicates an approach to correcting the ideal gas horsepower for the effects of compressibility.
Determine gas specific volume at inlet conditions:
v = ZRT / (144 P) ft3/lb
Obtain Z from compressibility charts, (or for specific gas or mixtures).
Actual cubic feet per minute, ACFM
Actual cubic feet per minute (ACFM) is a unit of volumetric capacity. The volumetric capacity of a gas at the inlet of compressor.
Specific gravity of natural gas compared with that of air is thus MW ng /MW a = 18.53/28.8 = 0.643
For Natural Gas (means C1 composition only for Approximation)
http://checalc.com/solved/naturalgasZ.html
Gas Gravity: 0.643
Pressure: 14.7 PSIA
Temperature: 60 °F
Nitrogen: 0 Mol %
Carbon Dioxide: 0 Mol %
Hydrogen Sulfide: 0 Mol %
--------------------------------------------------------------------------------
Sutton's correlations along with Wichert and Aziz corrections are used to calculate pseudo critical temperature and pressure for the natural gas mixture.
Pseudocritical Pressure: 671.08 PSIA
Pseudocritical Temperature: 363.33°R
Pressure, Pr: 0.0219
Temperature, Tr: 1.4303
Compressibility Factor, Z: 0.99748
Specific Heat Ratio, k
The value k is defined as the ratio of specific heats.
k = cp /cv
where,
Cp = specific heat at constant pressure
Cv == specific heat at constant volume
Also, k = Mcp / (Mcp - 1.99)
where,
Mcp = molal specific heat at constant pressure.
Mixture Specific Heat Ratio, km
xn is the mole percent of component of gas in mixture
Cpm = x1 Cp1 + x2 Cp2 + x3 Cp3 + ....
therefore,
km = CpM / (CpM - 1.99)
Note:
k = Cp / Cv = MW x Cp / (MW x Cp - 1.99)
for k = 1.0 to 1.1 gas easily compressed
for k > 1.35 gas difficult to compress
(large T incr. opposes compression process)
R = Gas constant - Btu/lb mol oR. Use the gas constant consistent with the units, 1.99 Btu/lb mol °R.
Molar gas mixture heat capacities, CpM
Mol % Cp at 100°F Btu/lb-mol-°R yCp
Helium 0.45 4.97 0.022
Nitrogen 14.65 6.96 1.020
Methane 72.89 8.65 6.305
Ethane 6.27 12.92 0.810
Propane 3.74 18.20 0.680
Butanes 1.38 24.32 0.336
Pentanes and heavier 0.62 40.81 0.253
CpM 9.426
Molal Heat Capacity, Cp
Cp = cp × M
Cp = Molal heat capacity at constant pressure
cp = Specific heat capacity at constant pressure
M = Molecular weight
Ratio of Molal Specific Heat of Gas Mixture at Average Temperature, Km
xn = mole fraction of gas n
Cpm = x1 Cp1 + x2 Cp2 + x3 Cp3 + ....
Km = Cpm / ( Cpm - 1.99 )
Head Equation with Compressibility For Polytropic Process, H
H = Z * (1545/M) * Ti * (Rc^Kf -1)/(Kf)
H = ft lb-force / ft-lb mass /lb
Z = Compressibility factor at inlet
M = Mole Weight
Rc = Ratio of Compression
Ti = Temperature at inlet in absolute, psia
Kf = (Km - 1)/Km
Km = ratio of specific heats for gas mixtre at T average
The CvM for the mixture is CpM − R = 9.43 – 1.99 = 7.44 Btu/lb-mol-°R
and specific heat ratio, k = 9.43/7.44 = 1.33.
The specific heat at constant pressure (Cp) may be used to calculate the specific heat ratio. The formula is:
Specific Heat Ratio = MW (Cp) / (MW (Cp) - 1.99)
Critical Pressure (Pc) and Critical Temperature (Tc) - These two values are used to calculate a value called compressibility (Z).
Head = Z * (1545/Mw) * Ti * [(Rc^Kf) - 1]/(Kf)
Head = ft lb-force / ft-lb mass /lb
K = average ratio of specific heats
Kf = (K-1)/K
Z = Compressibility factor at inlet
Mw = mole weight
Rc = Ratio of Compression
Rc = p2 / p1
Ti = Temperature at inlet in absolute
The polytropic exponent can be determined when inlet and discharge pressure and temperature is known by the thermodynamic relationship given in equation below.
T2 / T1 = ( p2 / p1 )^Kf
T1 = Temperature at inlet in absolute
T2 = Temperature at outlet in absolute
p1 = Pressure at inlet in psia
p2 = Pressure at outlet in psia
z = PV/RT
z is the compressibility factor
average temperature = 0.5 (Tin + Tout)
"k" value for the gas mixture:
Fraction, y Mcp* (y)(Mcp)
Methane 0.6 9.15 5.48
Nitrogen 0.4 7.035 2.81
8.29
* at 150°F from average data tables
k = cp/cv = 8.29 / (8.29 - 1.99) = 1.315
Mass Flow, m̆
m̆ = 144 P Q / (R T Z)
m̆ = mass flow (lbs/min)
R = Universal Gas Flow Constant divided by M.W.
R = (1545 ft•lbf/(lb•mol)(°R)) / MW
T = Gas Temperature in °R (°F + 460)
Z = Compressibility Factor
P = Gas Pressure in psia
Q = Volumetric Flow in CFM (Cubic Feet per Minute)
Compressor Selection Calculation:
k factor of the gas at suction conditions:
k = MWcp / ( MWcp - 1.99 x Z )
Had = R x T x Z x Ba
Polytropic Head:
Hp = Had x φ
Gas Horsepower
The Gas horsepower is the amount of real horsepower going to the Compressor, not the horsepower used by the motor.
Due to hydraulic, mechanical and volumetric losses in a Compressor the actual horsepower available for work on or from the fluid is less than the total horsepower supplied.
Gas Horsepower for a Centrifugal Compressor
The Gas horsepower - GHP - for a Centrifugal Compressor can be expressed as:
GHP = ( γ Q h / 33000 ) / η
where,
GHP = Gas horsepower (horsepower, hp)
Q = volume flow rate at inlet condition (ft3/min, cfm)
γ = specific weight (lbf/ft3) (weight is force)
η = overall efficiency
GHP = W x Hp / (33000 x Effp)
Power requirement
The total power requirement of a compressor for a given duty is the sum of the gas power and the friction power. The gas power is directly proportional to head and mass flow and inversely proportional to efficiency. Mechanical losses in the bearings and, to a lesser extent, in the seals are the primary source of friction power.
For centrifugal compressors, the gas power can be calculated as
GHP = W Hp / (33000 ηp)
where
GHP = gas power, horsepower,
W = mass flow, lbm/min.,
and
Hp = polytropic head, ft-lbf/lbm.
Diagram 8:
CALCULATIONS FOR A CENTRIFUGAL COMPRESSOR
OPERATING CONDITIONS
Capacity: 36.9 MMSCFD
Capacity, m: 239 .832 Lb/min
Suction pressure, Ps: 537 psia
Suction temperature, Ts: 121 °F
Suction temperature, Ts: 580.67 °R
Discharge pressure, Ps: 800 psia
GAS DATA
Capacity: 25,623 SCFM
Molecular weight: 3.7308 MW
Gas constant: 414.12 R
Critical pressure, Pc: 235.54 psia
Critical temperature, Tc: 92.15 R
Ratio of specific heats, K: 1.375 @ 150 °F
Reduced pressure, PR: 2.28
Reduced temperature, TR: 6.30
Compressibility @ suction, ZS: 1.026
Compressibility @ discharge, ZD: 1.037
INLET VOLUME FLOW
Specific volume, vs: 3.1905 ft3/lb
Weight flow, m: 239 .832 lbs/min
Polytropic efficiency, EPOLY: 78.5%
Inlet capacity, QS: 765.2 ICFM
COMPRESSOR POLYTROPIC HEAD
Polytropic head, HPOLY: 106,069 ft-lbF/lbM
Number of stages: 8 (LP CASING) + 7 (HP CASING)
COMPRESSOR HORSEPOWER
Gas horsepower: 458.2 + 523.6 = 981.8 GHP
Bearing losses, BL: 26.3 + 26.3 = 52.6 HP
Seal losses: 14 + 14 = 28 HP
Compressor brake horsepower: 498.5 + 563.9 = 1062.5 BHP
DISCHARGE TEMPERATURE
Discharge temperature, Td: 666.9 °R
Discharge temperature , Td: 207.2 °F
IMPELLER DIAMETER & TIP SPEED
Impeller diameter, D2: 12.6 inches
Impeller tip speed, U: 676 FPS
ACOUSTIC VELOCITY
Acoustic velocity, Va: 3303 FPS
DIMENSIONLESS VALUES
Specific speed, Ns: 56.9 Dimensionless
Specific diameter, Nd: 2.7 Dimensionless
Flow coefficient, phi: 0.02177 Dimensionless
Head coefficient, μ: 0.4976 Dimensionless
Compressor Speed: 12,300 RPM
Specific volume
The flow at inlet conditions of temperature and pressure must be
calculated in order to correctly select a centrifugal compressor.
The flow at inlet conditions is calculated as follows:
vs = zs (1545 /M) [Ts / (144 P1)
CALCULATIONS FOR A CENTRIFUGAL COMPRESSOR
OPERATING CONDITIONS
Capacity: 36.9 MMSCFD
Capacity, m: 239.832 Lb / min
Suction pressure, Ps: 537 psia
Suction temperature, Ts: 121 °F
Suction temperature, Ts: 580.67 °R
Discharge pressure, Ps: 800 psia
GAS DATA
Capacity: 25,623 SCFM
Molecular weight: 3.7308 MW
Gas constant: 414.12 R
Critical pressure, Pc: 235.54 psia
Critical temperature, Tc: 92.15 °R
Ratio of specific heats, K: 1.375 @ 150 °F
Reduced pressure, PR: 2.28
Reduced temperature, TR: 6.30
Compressibility @ suction, ZS: 1.026
Compressibility @ discharge, ZD: 1.037
INLET VOLUME FLOW
Specific volume, vs: 3.1905 ft3/lb
Weight flow, m: 239 .832 lbs/min
Polytropic efficiency, EPOLY: 78.5%
Inlet capacity, QS: 765.2 ICFM
COMPRESSOR POLYTROPIC HEAD
Polytropic head, HPOLY: 106,069 ft-lbF/lbM
Number of stages: 8 (LP CASING) + 7 (HP CASING)
COMPRESSOR HORSEPOWER
Gas horsepower: 458.2 + 523.6 = 981.8 GHP
Bearing losses, BL: 26.3 + 26.3 = 52.6 HP
Seal losses: 14 + 14 = 28 HP
Compressor brake horsepower: 498.5 + 563.9 = 1062.5 BHP
DISCHARGE TEMPERATURE
Discharge temperature, Td: 666.9 °R
Discharge temperature , Td: 207.2 °F
IMPELLER DIAMETER & TIP SPEED
Impeller diameter, D2: 12.6 inches
Impeller tip speed, U: 676 FPS
ACOUSTIC VELOCITY
Acoustic velocity, Va: 3303 FPS
DIMENSIONLESS VALUES
Specific speed, Ns: 56.9 Dimensionless
Specific diameter, Nd: 2.7 Dimensionless
Flow coefficient, phi: 0.02177 Dimensionless
Head coefficient, μ: 0.4976 Dimensionless
Compressor Speed: 12,300 RPM
Specific volume, vs
The flow at inlet conditions of temperature and pressure must be calculated in order to correctly select a centrifugal compressor.
The flow at inlet conditions is calculated as follows:
vs = zs (1545 /M) [Ts / (144 P1)
vs = 1.026 [ 1545 / 3.73] [ 580.67°R / (144 x 537 psia)]
Vs = 3.191 ft3/lb
Inlet flow, Qs
Qs = (m) (v)
Qs = (239.8 lbs/min x (3.191 ft3/lb)
Qs = 765.2 ICFM
Ratio of Compression, RC
Rc = P2 / P1
Rc = 800 psia / 537 psia
Rc = 1.489
Polytropic exponent = [n - 1]/n
Polytropic exponent = [ (1.375 - 1) / (1.375 x 0.785) ]
Polytropic exponent = 0.347
Compressor head, H
Hpoly = [(Zs + Zd) / 2] [ 1545 / MW ] (Ts) { Rc^[(n - 1)/n] } / [(n - 1)/n]
Hpoly = [ (1.026 + 1.037) / 2 ] [ 1545 / 3.73 ] (580.67°R) [ ( 1.489 (0.347) - 1 ) / 0.347 ]
Hpoly = 106,069 ft-lbsf / lbm
Number of stages
stages = [ 106,069 ft / 9000 ft/stage ] = 11.8 rounded to 12
Since the compressor manufacturer selected a compressor with 15 impellers, the remaining calculations will be made with 15 impellers. Fifteen impellers were selected in order to reduce the polytropic head developed per impeller, the corresponding rotating speed, and thus keep the tip speed of each impeller low enough to meet the ratio of actual stress to yield stress. The result was lower stresses in the impeller. The impeller stress limit was 60 percent of material yield @ maximum continuous speed. Consequently, the impellers selected produce less than 9000 ft per impeller guideline. The actual value was 106,069/15 = 7071 ft/impeller.
A two casing compressor train design arrangement will accommodate the 15 impellers selected. Single casings are limited to nine stages for nonintercooled straight through designs. The two casings are designated "LP casing" and "HP casing" for low pressure casing and high pressure casing respectively. The LP casing has seven impellers and the HP casing has eight impellers.
Compressor Horsepower
The polytropic head is used to calculate gas horsepower and then bearing and seal losses to obtain the total brake horsepower of the compressor.
GHP= (m) x (Hpoly) / [ 33,000 x (Epoly) ]
Compressor BHP = compressor GHP + bearing losses + seal losses
Bearing losses = BL (N / 1000)^2
Seal losses = SL (N / 1000)^2
Bearing and seal loss calculations relate compressor frame size (which is a function of capacity in actual cfm) and speed to power. Mechanical losses are not a function or percentage of gas horsepower and must be calculated. Many bearing manufacturers publish data, curves and the like relating mechanical losses to speed, clearance, and oil film thickness for various types of journal and thrust bearings. If such design data is available, the engineer would need to estimate the shaft size to utilize the information.
Ghp (LP casing) = (239.8 lbs/min) x (56570 ft) / [ 33,000 x (0.785) ]
Ghp (LP casing) = 458.2 hp
BL (LP casing) = 0.174 (12,300/1000)^2
BL (LP casing) = = 26.3 hp
SL (LP casing) = 0.093 (12,300/10000)^2
SL (LP casing) = 14 hp
Bhp (LP casing) = 498.5 hp
Ghp (HP casing) = 523.6 hp
BL (HP casing) = 26.3 hp
SL (HP casing) = 14 hp
Bhp (HP casing) = 563.9 hp
Total Bhp (LP casing + HP casing) = 1062.50 hp
Polytropic discharge temperature, Td (poly)
Td (poly) = [Ts, °R] [ Rc^(n-1)/n ]
Td (poly) = (121 +459.67°R) x 1.489^0.347
Td (poly) = 666.9 °R = 207.2°F
The impeller diameter is selected as a value consistent with the casing size, ICFM required, and rotating speed. For this application, a 12 to 13 in impeller diameter would be appropriate while limiting the rotating speed to 10 to 11,000 rpm. The compressor manufacturer selected a 12.6 in diameter for all 15 stages and a speed of 12,300 rpm. Therefore, the remaining calculations are based on these values.
Impeller Tip speed, U
U = 2π [ D/2 ] [ N / (60 x 12) ]
U = 2π [ 12.6 / 2 ] [ 12,300 RPM / (60 x 12) ]
U = 676 fps
Acoustic Velocity, Va
Acoustic velocity @ inlet Va, ft/sec = [ k g R Ts Zs ]^0.5
Va = [ (1.375) (1545/3.73) (580.67°R) (32.16fps^2) (1.026) ]^0.5
Va = 3303 fps
The Mach number, U / Va, will be approximately 0.2 using an acoustic velocity of 3300 fps. (The speed of sound in air is approximately 1200 fps.)
Specific speed, NS
NS = [N] [ Q ]^0.5 / H^0.75
NS = [12,300 RPM] [765.2/60 cfS]^0.5 / [ 7071 ft ^ 0.75 ]
NS = 56.963
Capacity, Q, is in cubic feet per second, cfs.
Specific diameter, DS
DS = D H^0.25 / Q^0.5
DS = [12.6 / 12 ft] [ 7071 ft ]^0.25 / [ 765.2/60 CfS]^0.5
DS = 2.696
The Ds vs Ns diagram relates various turbomachinery impeller geometry configuration to stage efficiency. A recycle gas compressor will have a low flow coefficient with backward leaning impeller blades.
Flow coefficient, Φ
Φ = [700.3] [Qs ] / [N] [D^3]
Φ = [700.3] [765/2 ICPM ] / [12,300 RPM] [12.6]^3
Φ = 0.02177
Head coefficient, μ
μ = (Hstg) (g) / U^2
μ = [7071 ft] [ 32.16 ] / [676 fPS]^2
μ = 0.4976
NOMENCLATURE
MW Molecular weight
Cp Constant pressure specific heat
Cv Constant volume specific heat
Tc Critical temperature
Pc Critical pressure
k Ratio of specific heats
Tr Reduced temperature
Pr Reduced pressure
m weight flow, lb/min
n number of moles per minute
Q capacity, cfm (usually inlet cfm, icfm)
v specific volume, ft3/lb
Z Compressibility, dimensionless
T Temperature, °R
P Pressure, psia
P1 Inlet Pressure, psia
P2 Outlet Pressure, psia
icfm Inlet cubic feet per minute
scfm Cubic feet per minute, at 60 °F, 14.7 psia
mmscfd scfm x 10^6
H Specific head, ft-lb/lbm
R Ratio (of compression)
R Gas constant, dimensionless
E Efficiency, percent
GHP Gas horsepower, hp
BHP Brake horsepower, hp
BL Bearing loss coefficient, dimensionless
SL Seal loss coefficient, dimensionless
D Diameter (impeller diameter), inches
N Rotating speed, rpm and natural frequency in log decrement analysis
g acceleration constant, 32.16 fps^2
Nst Number of stages
V velocity, fps
U Tip speed (impeller tip speed), fps
MN Mach number, dimensionless
NS Specific speed, dimensionless
Ds Specific diameter, dimensionless
Greek letters
δ log decrement
Ω damping exponent
Φ Flow coefficient, dimensionless
μ Compressor head coefficient, dimensionless
Subscripts
s property at suction conditions
d property at discharge conditions
c compression
c a critical property
r reduced property (pressure, temperature, and compressibility)
poly polytropic value
a acoustic
1 property at inlet condition(s)
2 property at outlet condition(s)
L Loss (used for bearing and seal loss computation)
Q = inlet volume flow
H = head
N = speed (rev/min)
rp = absolute pressure ratio (P2/P1)
ΔT = change In temperature
HP or kW = power
Flow Calculations
Most common compressor flow conditions are often expressed in:
1. Weight flow-lb/min, lb/h (kg/min. kg/h)
2. SCFM-60°F. 14.7 psia and dry
3. number of mols/h
None of these flows can be used directly in calculating compressor performance. All must be converted to
ACFM-actual cubic feet per minute.
This is also commonly referred to as ICFM-inlet cubic feet per minute.
These conversions are:
ACFM = w • v
ACFM = SCFM • Pa • T1 • Z1 / [ P1 • Ts • Zs ]
ACFM = no. of mols/min • MW • v
w = weight flow, lb/min (kg/min)
v = inlet specific volume, ft3/lb (m3/kg)
Ps = standard pressure, 14.7 psi (1.013 bar) absolute
P1 = inlet pressure, psi (bar) absolute
Ts = standard temperature, 520°R
T1 = inlet temperature, °R
Z1 = inlet compressibility
Zs = standard compressibUity, always 1.0
MW = molecular mass
DELAVAL ENGINEERING GUIDE TO COMPRESSOR SELECTION
Given: k = 1.275, MW; 18.12, Z = 0.98
P1 : 124.5 psia, P2 = 500 psia,
m̆ = 5470 lbm/min
T1 = 90°F = 550°R
Steps:
Inlet Volume Flow, ACFM
v1 = 0.98 [1544 / 18.12] 550 / 124.5 (144)
= 2.56 ft3/ lbm
ACFM1 = mv1
ACFM1 = 5470 (2.56)
= 14000 ft3/min
Adiabatic Head, Had
Had = 0·98 [1544/18.12] (550) {500/124.5]^[1.275-1/1.275] - 1}/ [1.275 - 1]/1.275
= 74,460 ft
Discharge Temperature, T2
ΔT = T1 [ r^( ( k - 1 ) / k) - 1 ] / ηad
assume ηad = 0.75
r = P2/P1
ΔT = 550 {[500/124.5]^[1.275-1/1.275] - 1}/ 0.75
= 256°
T2 = T1 + ΔT
= 256° + 90°
= 346°
T2 = 346°F (no intercooling required)
Frame Size
From Figure 5, inlet flow is close to maximum of
31 frame and wen within the range of 37 frame
Impeller Wheel Diameter, D
Wheel diameters
31 - 19.25"
37 - 22.875"
Head Per stage
From Figure 6, maximum head per stage = 11,000 ft
Minimum number of stages = 74,460/11,000 = 6.77
or 7 stages.
Tip Speed, U
U = [ 74,450 (32.2) / (7)(0.46) ]^0.5
= 863 ft/sec
Flow coefficient, Φ
Φ for 31 frame = 3.056 (14000) / [ 863 (19.25)^2 ]
Accord1ng to Figure 7, a 31 frame is marginal
Φ1 for 37 frame = 3.056 (14000) / [ 863 (22.875)^2
= 0.095
Φ2 is calculated from Q2
Q2 = mV2
v2 = Z2RT2 / [ 144 P2 ]
(from example on Page 27, z~found on Figure 2 from TR and PR)
V2 = 0.99 [1544 / 18.12] (806) / [ 144 (500) ]
= 0.944 ft3/lbm
Q2 = 5470 (0.944) = 5166 ft3/min
Φ2 = 0.035
Stage Efficiency, η
From Figure 7:
ηΦ1 = 0.775; ηΦ2 = 0.775; η avg. = 0.775
Wheel Efficiency,
Determine impeller efficiency correction from Figure 8:
1.0075 ( 0.775) = 0.781
Speed
RPM = 229 (863) / 22.875 = 8640 RPM
Horsepower
GHP = 74460 (5470) / 33,000 (0.781)
= 15803 hp
Mechanical losses = 81 hp.
BHP = 1.02 (15.803) + 81 = 18,200 hp
Casing Split
A discharge pressure of 500 psia corresponds to a 550 psi MWP casing. Therefore. casing is horizontal ty split. ModeJ selected is a seven-stage, 37-frame horizontally split: 7C37
Selection and Performance Calculation of a Centrifugal Compressor Train
Gas Compressor, Two Casings
Given:
Capacity
m̆t = 23.66 kg/s
Suction pressure
p1 = 0.92 bar abs
Suction temperature
T1 = 333K
Relative humidity
φ1 = 0%
Discharge pressure
p2 = 16.1 bar abs
Dry molecular mass
Mt = 17.03 kg/kmol
Isentropic exponent cp/cv
k = 1.29
Compressibility factor
Z = 1
Calculation
1. Determination of the absolute humidity x (from T1, p1, φ1, Mt) with Diagram 1
x = 0
2. Determination of the wet molecular mass Mf (from x, Mt) with Diagram 2
Mt = Mf = 17.03 kg/kmol
3. Calculation of the wet mass flow m̆f = m̆t (1 + x)
m̆f = m̆t = 23.66 kg/s
4. Determination of the max. permissible peripheral speed umax (from Z, k, T1, Mf) with Diagram 3
Electric motor umax = 320 m/s
Turbine umax = 290 m/s
For further calculation, motor drive has been selected.
5. Determination of the total polytropic head h*pT (from k, p2, p1, Z, Mf, T1) with Diagram 4
h*pT = 722.8 kJ/kg
6. Determination of the max. polytropic head obtainable per casing hpG max (from umax) with Diagram 5
hpG max = 300 kJ/kg
7. Calculation of number of casings i i = hpT/hpG max, with hpT = h*pT fT, whereby fT has to be estimated with Diagram 6
i= 2 with fT = 0.73
8. Determination of the pressure ratio per casing p2/p1G with Diagram 7
p2/p1G = 4.27
9. Determination of the polytropic head per casing h*pG (from k, p2/p1G, Z, Mf, T1) with Diagram 4
h*pG = 293 kJ/kg
From now on if two or more casings are necessary, the calculation has to be made for each casing separately (one after the other).
First casing
10. Determination of the influence of intercooling on the required shaft power (from p2/p1G, K, = T, T1 and estimated number of intercoolings per casing j) with Diagram 6
f= 0.91 with ΔT =0 and j = 1
11. Calculation of the fictive polytropic head hpG = h*pG f
hpG = 293 0.91
= 266.6 = 267 kJ/kg
12. Determination of the number of stages z per casing and the definite peripheral speed u (from hpG, z → u) with Diagram 8 (round off z to whole number and correct peripheral speed correspondingly)
z = 6
u = 304 m/s
13. Determination of the actual suction volume V1 (from .mf, p1, T1, Mf, Z) with Diagram 9
V̌1 = 41.8m3/s
14. Selection of the compressor size (nominal diameter D) as a function of V̌1 with Diagram 10
D = 112 cm
15. Type designation (from steps 10, 12, 14)
RZ 112-6
16. Calculation of the speed n = 60 u / [π D] = (D in meters)
n = 60 304 / [π 1.12]
= 5184 r/min
17. Determination of the power input P (from hpG, mf) with Diagram 11
P = 8100kW
18. Determination of the discharge temperature T2 (from p2/p1 between intercooling, k, T1) with Diagram 12 whereby T1 is the suction temperature after preceding intercooling and pressure ratio p2/p1 between intercooling has to be determined with Diagram 7
T2 = 413K
with T1 = 333K
and p2/p1 = 2.1
Two casings are necessary, the calculation is for Second casing
10. Determination of the influence of intercooling on the required shaft power (from p2/p1G, K, = T, T1 and estimated number of intercoolings per casing j) with Diagram 6
f= 0.91 with ΔT = 0 and j = 1
11. Calculation of the fictive polytropic head hpG = h*pG f
HpG = 293 0.91
= 266.6 = 267
12. Determination of the number of stages z per casing and the definite peripheral speed u (from hpG, z → u) with Diagram 8 (round off z to whole number and correct peripheral speed correspondingly)
z= 6
u = 304 m/s
13. Determination of the actual suction volume V1 (from .mf, p1, T1, Mf, Z) with Diagram 9
V̌1 = 10.2 m3/s
14. Selection of the compressor size (nominal diameter D) as a function of V̌1 with Diagram 10
D = 56cm
15. Type designation (from steps 10, 12, 14)
RZ 56-6
16. Calculation of the speed n = 60 u / [π D] = (D in meters)
n = 60 304 / [π 0.56]
= 10368 r/min
17. Determination of the power input P (from hpG, mf) with Diagram 11
P = 8100kW
18. Determination of the discharge temperature T2 (from p2/p1 between intercooling, k, T1) with Diagram 12 whereby T1 is the suction temperature after preceding intercooling and pressure ratio p2/p1 between intercooling has to be determined with Diagram 7
T2 = 413K
with T1 = 333K
and p2/p1 = 2.1
.
.
Good work
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