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Observation
An air leak test was performed at the facility. During the 25 min (1,500 sec) test, pressure dropped from 110 psig to 20 psig. The air system consists of one 300 hp centrifugal compressor with a 500-gallon (66.84-ft3) accumulator, and 1860 ft of 4-inch diameter pipe and 4500 ft of 8-inch diameter pipe.
Recommendation
Repair leaks or replace the compressed air system. A non-leaking system will reduce energy usage and cost associated with operating the air compressor. Implementation costs are calculated as leak repair of piping system.
Calculations
Once the air compressor was turned off, the pressure dropped from 110 psig to 20 psig over the 25-minute air decay test. Using these measurements, the mass flow rate of the air due to the leaks was found.
Calculations to find the volume of the air system:
Va = 66.84 ft3
Vp = π/4 ´ d2 ´ l
Vs = Va + Vp
Vs = 66.84 + 1,734.36
Vs = 1801.2 ft3
Where
Va = volume of accumulator, [ft3]
Vp = volume of piping, [ft3]
d = diameter of piping, [ft]
l = length of piping, [ft]
Vs = total volume of air system, [ft3]
Calculations to get from 110 psig to 20 psig:
Mass of air in accumulator and piping at atmospheric pressure:
M1 = Vs ´ D1
M1 = 1801.2 ´ 0.075
M1 = 135.09 lbm
Where
D1 = density of air at atmospheric pressure, [lbm/ft3]
M1 = weight of air in system at atmospheric pressure, [lbm]
Mass of air in tank at 110 psig:
D2 = (P2 + Po) / (P1 + Po) ´ D1
D2 = (110 + 14.7) / (0 + 14.7) ´ 0.075
D2 = 124.7 / 14.7 ´ 0.075
D2 = 0.636 lbm/ft3
M2 = D2 ´ Vs
M2 = 0.636 ´ 1801.2
M2 = 1145.56 lbm
Where
Po = atmospheric pressure correction, [psia]
P1 = gauge atmospheric pressure, [psig]
P2 = gauge working pressure, [psig]
D2 = density of air at 110 psig, [lbm/ft3]
M2 = weight of air in system at 110 psig, [lbm]
Mass of air in tank at 20 psig (after 25 minutes):
D3 = (P3 + Po) / (P1 + Po) ´ D1
D3 = (20 + 14.7) / (0 + 14.7) ´ 0.075
D3 = 34.7 / 14.7 ´ 0.075
D3 = 0.177 lbm/ft3
M3 = D3 ´ Vs
M3 = 0.177 ´ 1801.2
M3 = 318.81 lbm
Where
P3 = gauge pressure @ 25 minutes drop, [psig]
D3 = density of air at 20 psig, [lbm/ft3]
M3 = weight of air in system at 20 psig, [lbm]
Mass flow rate for the decay test of the total compressed air system:
MF = ((M2-M3)/ T)
MF = ((1145.56 – 318.81) / 1500)
MF = 0.551 lbm/sec
Where
T = time from 110 psig to 20 psig, [seconds]
Energy Savings
ES = {MF ´ CP ´ Ta ´ H ´ K1 ´ K2 ´ [(PL /PI)g - 1/g – 1]} / (EA ´ EM)
ES = {0.551 ´ 0.24 ´ 520 ´ 8760 ´ (1/3,412) ´ 3,600 ´
[(124.7 / 14.7)(/1.4 – 1]} / (0.82 ´ 0.944)
ES = 691,373 kWh/yr
Where
CP = specific heat of air, 0.24 [Btu/lbm-°R]
Ta = temperature of line air, 520 [°R]
H = annual operating hours of compressors, 8,760 [hr/yr]
K1 = conversion factor, 1 / 3,412 [kW/Btu]
K2 = conversion factor, 3,600 [sec/hr]
PL = line pressure, 124.7 [psia] (110 psig)
PI = atmospheric pressure, 14.7 [psia]
EA = air compressor isentropic efficiency, 0.82
EM = motor efficiency, 0.944
Energy Cost Savings
ECS = ES ´ CE
ECS = 691,373 ´ 0.04263
ECS = $29,473.23/yr
Where
CE = cost of electricity, [$/kWh]
Fees Cost Savings
EFCS = ECS ´ TRE
EFCS = 29,473.23 ´ 0.038
EFCS = $1,119.98/yr
Where
TRE = tax rate on electrical consumption, [3.80%]
Cost Savings
The cost savings associated with eliminating the air leaks are computed below:
CS = ECS + EFCS
CS = 29,473.23 + 1,119.98
CS = $30,593.21/yr
Implementation Cost
The following implementation costs are assumed for 309 air leaks and labor of 1 hour per leak. Assume that the company will utilize two people for one week with total plant overhead being $30.00/hour. In addition, a miscellaneous cost for materials ($10 per leak) is considered to repair each leak.
NL = 
NL = 
NL = 309
Where
MF = mass flow rate, [lbm/sec]
ץ = specific heat ratio, no units
R = gas constant for air, [Btu/lbm ´ mol ´ °R]
A = cross-sectional area of leak, [in2]
PL = line pressure, [psia]
Ta = temperature of line air, [°R]
G = conversion constant
C = discharge coefficient for a square-edge orifice, constant
Non-Capital Cost
Capital Cost
Thus,
IC = NCC + CC
IC = 9,270.00 + 3,090.00
IC = $12,360.00
Where,
IC = Implementation cost, [$]
NCC = Non-Capital cost, [$]
CC = Capital cost, [$]
Payback
Payback = IC / CS
Payback = 12,360.00 / 30,593.21
Payback = 0.4 years (4.8 months)
