Environmental Engineering Reference
In-Depth Information
Example 14.43
Problem: The motor horsepower required for a particular pumping job is 32 hp. If
your power cost is $0.06/kWh, what is the cost of operating the motor for 1 hr?
Solution:
Cost ($/hr)
=× ×
mhp
0.746 kW/hpCost($/kWh)
=
32
hp
×
0.746 kW/hp
×
$0.06/kWh
= $143
/hr
Example 14.44
Problem: The minimum motor horsepower requirement of a particular pumping
problem is 28 mhp. If the cost of power is $0.022/kWh, what is the cost to operate
the pump for 12 hr?
Solution:
Cost ($/hr)
=× ×
mhp
0.746 kW/hpCost($/kWh)
=
28
hp
×
0.746 kW/hp
×
$0.022/kWh
=
$0.46/hr
Totalc
ost
=
$0.46/hr
×
12 hr
= $552
Example 14.45
Problem: A pump is discharging 1200 gpm against a head of 60 ft. The wire-to-
water (overall) efficiency is 70%. If the cost of power is $0.022/kWh, what is the cost
of the power consumed during a week in which the pump runs 80 hr?
Solution:
Flow (gpm)Head (ft)
3960
×
120
0 gpm
×
60 ft
mhp
=
=
=
25.97 hp
×
Efficiency
3960
×
0.70
Cost ($/hr)
=× ×
mhp
0.746 kW/hpCost($/kWh)
=
25
.97 hp
×
0.746 kW/hp
×
$0.022/kWh
=
$0.43/hr
Tota
llcost
=
$0.43/hr
×
80 hr
= $.
34 40
Example 14.46
Problem: Given a brake horsepower of 18 hp, a motor efficiency of 86%, and a cost
of $0.014/kWh, determine the daily power cost for operating a pump.
 
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