Compressor System Modeling PDF

Mathcad – Assignment
System Calculation
Dr Malcolm W. Renton
Compressor Drive System
• Block Diagrams:
Output
Input Compressed Gas
Electricity Drive System Temperature, Pressure
3-phase AC and Flowrate
Input
Armature Control
3-phase controlled
rectifier
DC Motor Output
Field Control
1-phase uncontrolled Gearbox
rectifier
Reciprocating
Compressor
AC-DC Rectifier
• Converts Alternating Current to Direct Current
• Two basic types:
– Half-Wave
– Full-Wave
• Two modes of operation:
– Uncontrolled
– Controlled
• Used throughout industry, commerce and
domestic installations
Half Wave and Full Wave
• Half-Wave:
• Full-Wave:
Uncontrolled Rectifier
• Diodes conduct when cathode (K) voltage is
greater than the anode (A).
• Governing Equation:
• Single-Phase supply EDC = 0.9E AC
Three-Phase Uncontrolled rectifier
Output
Vo
Voltage
VK A
1.73Vm 1.5Vm
o  
/3 2/3
• Governing Equation:
• Three-Phase Supply: VKA = 1.35EAC
• or EDC = 1.35EAC
Controlled Rectifier
• Transistors or Thyristors instead of Diodes
• Timed switching, controls output D.C. Level
• Single-phase supply EDC= 0.87EmaxAC cos(a)
– where a is the firing angle of the thyristor
3-phase Controlled Rectifier
• Three or Six Thyristors instead of Diodes
• Governing Equation
• EDC = 1.35 EAC cos(a)
Firing Angle Control
• Increasing
firing angle: 0°
15°
45°
Rectifier Arrangements
• Simple Uncontrolled Bridge >
• Power Rectifier >

DC Motors
DC Machine (Motor)
• Comprises two windings (coils) through which
current flows:
– Armature (has resistance Ra and current Ia)
– Field (has resistance Rf and current If)
• Different characteristics of motor due to the
connection arrangement of the two windings:
– Shunt
– Series
– Separate
– Compound (cumulative or differential)
Separately Excited
• Schematic Diagram:
• Equivalent circuit:
Governing Equations
• Basic Eqn : Ea = V - Ia.Ra and Vf = If.Rf
• Proportionality :
– T a f.Ia and Ea a f.N
• Also flux f a If thus

– T a If.Ia and Eo a If.N
• Vary voltage (field or armature) to vary speed
• Reverse voltage – reverse speed

DC Motor Constants
• Constant k associated with the machine
internal arrangement:
– Number of coils (z)
– Number of pole pairs (p) – (no. of poles/2)
– Coil arrangement (c)
• 2p for lap wound
• 2 for wave wound
• k = (z.2.p)/(60c)
• Hence: T =k.f.Ia and Ea = k.f.N
• Also flux f = kmagIf where kmag is the flux/amp.
DC Motor Power
• A motor converts electrical energy to mechanical
energy
• Equating the previous proportional relationships:
• Electrical Ea.Ia = T.w Mechanical
• Note: This is the developed power of the motor

• The output power is less,
– due to mechanical efficiency
– i.e. Friction and windage losses
Gearboxes
• Devices to change speed, direction and or alignment.
Gearbox Equation
• The gear ratio (e.g. 3:1) is the ratio of
input/output speed
– also the inverse ratio of gear teeth on the gearwheels
– and the inverse ratio of input/output torque
• In an ideal gearbox Power in = Power out
• When efficiency is less than 100%, power out

and output torque are reduced
– but speed remains the same!
Crank Shaft
• Physical connection between a rotating shaft and a
reciprocating piston
Camshaft Geometry
• Relationships associated with displacement, velocity

and acceleration can be developed from this diagram
Reciprocating Compressor
• Device that uses pistons to compress fluid
• Operation depends on camshaft speed, physical

dimensions of the pistons, fluid intake
conditions and Gas Laws
Properties
• Pressure (p) in bar or Pa
• Temperature (T) in degrees Kelvin (K)
• Gas Constant R in (Nm/kgK)
• Polytropic index (n) – characteristic of fluid
• Mass flow rate (m) in kg/min
• Volume flow rate (V) in m³/min
• Stroke (l) - length of piston in metres
• Bore (d) – diameter of piston in metres
Governing Equations
 2
Piston Volume = d l  n 1
4
T2  p2  n
=  
PV = m
 RT T1  p1 
n
indicated work = m R(T2  T1 ) kJ / min
n 1
Reciprocating Compressor
• V arrangement
Complete System
Assignment
• For rectifier firing angles from 0° to 60°

determine:
– Motor speed and output torque
– Compressor mass flow rate
– Compressor output temperature and pressure
• Use plot(s) to show the variation of these with
firing angle of the rectifier.
Equipment Details
• AC Supply V3 phase
– 415V, 3-phase, 50 Hz Note V1 phase =
3
• Motor
– 6-pole, lap-wound separately-excited
– Ra= 0.2 W, Rf =200 W
– Mechanical efficiency = 95%
– kmag = 0.2 Wb/A
– Ia = 80 A at full volts (0° firing angle)
• assume Ia varies in proportion with E
– Number of conductors = 720
Equipment Details
• Gearbox
– Gear ratio 3:1
– Efficiency = 92%
• Compressor
– Efficiency = 90%
– 2 cylinder each with a bore of 200 mm
– Stroke/bore ratio = 1.9:1
– Intake at atmospheric conditions 1.013 bar, 15°C
– R= 287 (Nm/kgK) and n = 1.35

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Mathcad – Assignment

• Simple Uncontrolled Bridge >

• Power Rectifier >

• Also flux f a If thus

• Vary voltage (field or armature) to vary speed

• Reverse voltage – reverse speed

• Note: This is the developed power of the motor

• In an ideal gearbox Power in = Power out

• When efficiency is less than 100%, power out

• Relationships associated with displacement, velocity

• Operation depends on camshaft speed, physical

• For rectifier firing angles from 0° to 60°

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