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ISENTROPIC EFFICIENCY CALCULATIONS for POSITIVE-DISPLACEMENT ROTARY-LOBE BLOWERS PowerPoint Presentation
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ISENTROPIC EFFICIENCY CALCULATIONS for POSITIVE-DISPLACEMENT ROTARY-LOBE BLOWERS. Roger E. Blanton, P.E. District Sales Manager Kaeser Compressors, Inc. EFFICIENCY. All air and gas compressors are less than 100% efficient. P1 V1 T1. Position 1. P2 V2 T2. Position 2. EFFICIENCY.

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slide1

ISENTROPIC EFFICIENCY CALCULATIONS

for

POSITIVE-DISPLACEMENT

ROTARY-LOBE BLOWERS

Roger E. Blanton, P.E.

District Sales Manager

Kaeser Compressors, Inc.

efficiency
EFFICIENCY
  • All air and gas compressors are less than 100% efficient
efficiency1

P1 V1 T1

Position 1

P2

V2

T2

Position 2

EFFICIENCY

P2 > P1

V2 < V1

T2 > T1

efficiency2
EFFICIENCY
  • Entropy is a measure of the energy loss
compression model
COMPRESSION MODEL
  • Constant entropy (isentropic) process is ideal
  • No heat loss and perfectly reversible
isentropic compression
ISENTROPIC COMPRESSION
  • Follows this relationship,

P2 /P1 = (v1/ v2)k » Pvk= a constant

where: v = specific volume [ft3/lb]

P1, P2 = initial and final pressure [Pabs]

k = Cp/Cv = specific heat ratio = 1.4 for air

compressor work
COMPRESSOR WORK
  • For continuous isentropic compression of a gas obeying the previous relationship, work per unit mass (ft·lb/lb) is

Wis = vdP = P1 * v1 * [k / (k-1)] * [(P2 / P1)((k-1) / k) – 1]

BHF = Big Hairy Formula !!!

power
POWER
  • hpis = Wis * m / 60 / 550

where m is mass flow rate in lb/min

efficiency3
EFFICIENCY
  • Isentropic efficiency of a compressor is the ratio of the ideal isentropic work to the actual work required

is = Wis / Wact = hpis / hpact

example calculation
EXAMPLE CALCULATION
  • Brand K model 42 @ 2000 rpm
  • P1 = 14.7 psia
  • P2 = 10 psig (24.7 psia)
  • T1 = 68° F
  • ρ1 (density) = 0.075 lb/ft3
  • K = 1.4 (air)
  • BHP = 13.2 (from performance table)
  • CFM = 222 (from performance table)
  • m = 222 * 0.075 = 16.65 lb/min
example calculation cont d
EXAMPLE CALCULATION (cont’d)
  • v1 = 1/ ρ1 = 1 / 0.075 = 13.33 ft3/lb
  • k / (k-1) = 1.4 / (1.4 – 1) = 1.4 / 0.4

= 3.5

  • P2 / P1 = 24.7 / 14.7 = 1.68
  • (k-1) / k) = (1.4 -1) / 1.4 = 0.4 / 1.4

= 0.286

  • (P2 / P1)((k-1) / k) = 1.680.286 = 1.16
example calculation cont d1
EXAMPLE CALCULATION (cont’d)

Back to the Big Hairy Formula

Wis = P1 * v1 * [k / (k-1)] * [(P2 / P1)((k-1) / k) – 1]

Wis = 14.7 * 13.33 * 3.5 * [1.16 -1] * 144

= 15,801 ft·lb/lb

example calculation cont d2
EXAMPLE CALCULATION (cont’d)
  • hpis = Wis * m / 60 / 550
  • hpis = 15801 * 16.65 / 60 / 550

= 8 hp

example calculation cont d3
EXAMPLE CALCULATION (cont’d)
  • is = hpis / hpact
  • is = 8 / 13.2 = 61% efficient
conclusion
CONCLUSION
  • Repeat this procedure for different blowers at the same pressure
conclusion1
CONCLUSION
  • Perform this calculation on blowers bid for a project to evaluate efficiency
  • Look at blower efficiency, not just price when making a decision
conclusion2
CONCLUSION
  • Perform a Life Cycle Cost (LCC) analysis
  • Typically a 20 year project horizon is considered
  • Include installation, estimated maintenance, and operating costs
  • Discount the expenses to net present value (NPV) for comparison
conclusion3
CONCLUSION
  • Operating cost greatly outweighs first cost
  • Blower efficiency is important for customer satisfaction