of 72 to 77 percent. This research showed about 25 percent of electrical energy used for pumping was wasted due to poor pumping plant efficiencies alone. some of the more common
causes of unsatisfactory performance and their remedies are as follows: 1. Impellers that are out of adjustment. 2. Pump bowls designed for a higher pumping rate 3. Damaged impellers 4.
Incorrect power unit selection 5. Failure to perform required maintenance Chavez et al., 2010.
In recently years pressurized irrigation systems expand in Iran. For provide pressure these systems consume more energy than gravity systems therefore investigation of efficiency and
waste energy at agricultural pumping plant in present situation and promotion ways of energy efficiency is necessary.
2. MATERIAL AND METHODS
Energy efficiency and consumption in pressurized irrigation pumping plants were measured from 2007 to 2009 in 44 fields in Hamedan, Kerman, and Khorasan Razavi provinces, which
were preformed 22-drip and 22-sprinkler irrigation in these fields. Numbers of irrigation systems were as follow: 17 sprinklers irrigation in Hamedan, 2 sprinklers and 12 trickles
systems in Kerman and 3 sprinklers and 10 trickles systems in Khorasan Razavi. All of the irrigation pumping plants in Kerman, and Khorasan Razavi provinces were electro-motor but in
Hamedan 6 of irrigation pumping plants operated with diesel engine. All of electro-motors had 4 poles, 50 Hz frequency and 1500-rpm synchronous speed.
Instruments were used to measure including: 1. Multimeter model HIOKI 3280-10 Japan made 2. Chronometer 3. Tachometer model Pantec DTM-30 England made. 4. Scaled container 5.
Pito pipe with pressure gauge. 6. WSC flume. Wells discharge base on situations measured with WSC flume, jet method, cross-section
speed and multiply sprinkler discharge by number of sprinklers. Electro-motor or diesel engine rotated speed was measured with tachometer to compare with nominal rotated speed of motor
or engine. Current and voltage were measured in three entrance wires from board with multimeter.
Pumping plants performance compared with Nebraska Pumping Plant Performance Criteria NPPPC. The performance criteria developed by Schlencener and Sulek 1959 for evaluating
the operating efficiency of irrigation pumping plants has become widely used. The criteria is relationship of output horsepower-hours and water horsepower-hours per unit of fuel
consumption Schneider and New, 1986. Power output kWh from diesel engines and electric motors and amount of useful power output from pumping plant kWh per unit of energy-
consumed base on NPPPC are shown in table 2 Anon,1997; Harrison Skinner, 2009; Anon, 2009. The Nebraska Pumping Plant Performance Criteria for power unit motor or engine and
overall efficiencies motor+drive+pump are shown in Table 3New Schneider, 1988; Fippes Neal, 1995. The Nebraska overall efficiencies are based on considering 75 efficiency for
a pump, 95 efficiency for drive, 88 efficiency for electrical motor and 33 efficiency for diesel engine New Schneider, 1988.
Table 2. The Nebraska pumping plant performance criteriaNPPPC
Energy source
NPPPC for power unit
NPPPC for pumping station
Energy units
Diesel 3.3
2.5 kWh.L
-1
Gasoline 2.3
1.7 kWh.L
-1
Propane 1.8
1.4 kWh.L
-1
Natural gas 1.7
1.3 kWh.m
-3
Electricity 0.88
0.67 kWh. kWh
-1
English Unit convert to SI unit
Table 3. The Nebraska pumping plant efficiency criteria
Power unit Power unit
efficiency Overall
efficiency Electrical
88 66
Diesel 33
24 Natural gas
24 17
Overall efficiency calculated from the equation:
P
out
= Output power from pump kW P
in
= Input power to motor kW Researcher gave different quantities for one litter of diesel energy content among them
37.6 MJL New Schneider, 1988, 37.9 MJL Fippes Neal, 1995 and 36.1 MJL Weddington Canessa, 2006. In this research used standard diesel energy content that is
36.2 MJL Weddington Canessa, 2006. Therefore, a litter of diesel per hour produce 10.07 kW power, then diesel engine power consumption per hour estimate from volume of diesel
consumption per hour multiply 10.07 kW. Electric motors power consumption calculated by voltage and current measurement at the beginning and the end of test duration from the
equation:
ϕ cos
3
I
V P
i
P
i
= Three-phase power W V= RMS voltage, mean line-to-line of 3 phases
I = RMS current, mean of 3 phases cos
φ = Power factor as a decimal according to the motor data sheet Pump output power calculated from the equation:
102 QH
P
P = Output power from pump kW Q = Discharge LS
-1
H = Head m Two methods can use for calculation fuel amount consumption. In the first method, energy
consumption that need base on NPPPC minus actual energy consumption and result give extra fuel consumption per hour. In the second method, at first determine pumping plant
performance from equation 4, then calculate pumping plant performance rating Eq. 5 and finally estimate fuel wasted per hour from equation 6 Dron, 2004; Smajstrla et al.,2005.
Duration Test
the for
Consumed Fuel
Duration Test
out P
e Performanc
Criteria e
Performanc Plant
Pumping Nebraska
e Performanc
Rating e
Performanc
Fuel Wasted Hour = Energy Current Fuel Consumption Rating × 1-Performance Rating The shaft output power calculated by several ways, such as nameplate method, in put power
measurements method, slip method, current method, statistical method, segregated loss method and shaft torque method Anon, 2007; Hsu et al., 1998. Base on this study data, shaft
100
in out
P P
OPE
output power calculated with nameplate, slip and current methods. Since torque meter was not available, these methods used to calculate shaft output power. Estimated error is about 1, 6.5,
8 and 10 percent in shaft torque, current, slip and nameplate methods respectively Hsu et al., 1998.
Motor load by current method calculate with the following equationAnon, 2007:
100 r
V V
r I
I Load
Load = Output power as a of rated power I = RMS current, mean of 3 phases
I
r
= Nameplate rated current V = RMS voltage, mean line-to-line of 3 phases
V
r
= Nameplate rated voltage Motor efficiency determine from the equation:
i P
Load ir
P η
η = Efficiency as operated in P
ir
= Input power at full-rated load in kW P
i
= Measured three-phase power in kW Load= Output power as a of rated power
3. RESULTS
