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Observation
During the tour of the facility, a number of air leaks were noticed. Air leaks can cause unwanted drops in system pressure, shorten the life of all supply system equipment and cause the compressor to run for longer periods of time, thus, causing unwanted energy use.
Recommendation
Repair leaks in 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. In the present compressed air system, 36 % of the compressor energy usage is going to supplying leaks.
Calculations
To estimate the potential savings, our team obtained an air leak decay test from the plant, performed during a scheduled plant shutdown. During the 1 min (60 sec) test, pressure dropped from 108 psig to 107 psig. The air system in this plant consists of five 200 hp reciprocating compressor, two 150 hp rotary screw compressors, two 300 hp rotary screw compressors, and one 350 hp rotary screw compressor with two accumulators. The plant is presently only operating under one 300 hp rotary screw compressor. The accumulators are as follows: one 170.99 ft3, and one 502.66 ft3 accumulator. The volume of the piping consists of 10,560 ft of 10-inch diameter pipe.
Calculations to find the volume of the air system:
Va = 673.65 ft3
Vp = π/4 ´ d2 ´ l
Vp = π/4 ´ 0.8332 ´ 10,560
Vp = 5,755.00 ft3
Vs = Va + Vp
Vs = 673.65 + 5,755.00
Vs = 6,428.65 ft3
Where
Va = volume of accumulators, [673.65 ft3]
Vp = volume of piping, [5,755.0 ft3]
d = diameter of piping, [0.833 ft]
l = length of piping, [10,560 ft]
Calculations to get from 108 psi to 107 psi:
Mass of air in accumulator and piping at atmospheric pressure:
M1 = Vs ´ D1
M1 = 6,428.65 ´ 0.075
M1 = 482.15 lbm
Where
D1 = density of air at atmospheric pressure, [0.075 lbm/ft3]
Mass of air in tank at 108 psig:
D2 = (P2 + Po) / (P1 + Po) ´ D1
D2 = (108 + 14.7) / (0 + 14.7) ´ 0.075
D2 = 122.7 / 14.7 ´ 0.075
D2 = 0.626 lbm/ft3
M2 = D2 ´ Vs
M2 = 0.626 ´ 6,428.65
M2 = 4,024.34 lbm
Where
Po = atmospheric pressure correction, [14.7 psia]
P1 = gauge atmospheric pressure, [0 psig]
P2 = gauge working pressure, [108 psig]
D2 = density of air at 108 psig, [0.626 lbm/ft3]
M2 = mass of air in system at 108 psig, [4,024.34 lbm]
Mass of air in tank at 107 psig (after 1 minute):
D3 = (P3 + Po) / (P1 + Po) ´ D1
D3 = (107 + 14.7) / (0 + 14.7) ´ 0.075
D3 = 121.7 / 14.7 ´ 0.075
D3 = 0.621 lbm/ft3
M3 = D3 ´ Vs
M3 = 0.621 ´ 6,428.65
M3 = 3,992.19 lbm
Where
P3 = gauge pressure @ 1 minute drop, [107 psig]
D3 = density of air at 107 psig, [0.621 lbm/ft3]
M3 = mass of air in system at 107 psig, [3,992.19 lbm]
Mass flow rate for the decay test of the total compressed air system:
MF = ((M2-M3)/ T)
MF = ((4,024.34 – 3,992.19) / 60)
MF = 0.536 lbm/sec
Where
T = time from 108 psig to 107 psig, [60 sec]
Energy Savings (ES)
ES = {MF ´ CP ´ Ta ´ H ´ K1 ´ K2 ´ [(PL /PI)g - 1/g – 1]} / (EA ´ EM)
ES = {0.536 ´ 0.24 ´ 520 ´ 2,080 ´ (1/3,412) ´ 3,600 ´
[(122.7 / 14– 1)/1.4) – 1]} / (0.82 ´ 0.942)
ES = 158,418 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 ´ 5 ´ 52 = 2,080 hr/yr]
K1 = conversion factor, [1 kW / 3,412 Btu]
K2 = conversion factor, [3,600 sec/hr]
PL = line pressure, [122.7 psia (108 psig)]
PI = atmospheric pressure, [14.7 psia]
ץ = specific heat ratio, [1.4] no units
EA = air compressor isentropic efficiency, [0.82]
EM = motor efficiency, [0.942]
Energy Cost Savings (ECS)
ECS = ES ´ CE
ECS = 158,418 ´ 0.07117
ECS = $11,274.61 /yr
Where
CE = cost for electricity, [$0.07117 /kWh]
Fees Cost Savings (FCS)
FCS = ECS ´ TRE
FCS = 11,274.61 ´ 0.049
FCS = $552.46 /yr
Where
TRE = tax rate on electrical consumption, [4.9%]
Cost Savings (CS)
The cost savings associated with eliminating the air leaks are computed below:
CS = ECS + EFCS
CS = 11,274.61 + 552.46
CS = $11,827.07 /yr
Implementation Cost (IC)
The following implementation costs are based on the assumption of 1/32” diameter air leaks. From the air leak decay test it is calculated that there are 305 – 1/32” diameter air leaks (note: number used only for manpower calculations – not the exact number of leaks). We estimate it will take 1 labor hour to repair each 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 = 305
Where
NL = total number of leaks in the system [305]
MF = mass flow rate, [0.536 lbm/sec]
R = gas constant for air, [1,717 ft2/(s2 - ºR)]
A = cross-sectional area of leak, [0. in2]
PL = line pressure, [122.7 psia]
Ta = temperature of line air, [520 °R]
G = conversion constant, [32.174 ft/s2]
C = discharge coefficient for a square-edge orifice constant [0.80]
Payback
Payback = IC / CS
Payback = 12,200.00 / 11,827.07
Payback = 1.03 yrs (12 months)
