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n = 500 olives
2 dripper/olive
2 l/h (dripper)
(1) (499) (500)(n)
Flow>Qmin
Qmin (1) = 500 x 2 x 2 = 2000 l/h
We need more than this flow
To presurize the pipe
To work the dripper
?Howmuchwaterneed every tree
In our case we use 20 l/tree each day
Qmin (2) = 500 x 20 = 10.000 l/day
6 -7 WECANSUPPOSEINFIRSTAPROACHTHATWEONLYHAVE PSHATIRRIGATIONSEASON
Qmin (2) = 10.000 l/day
6,5 PSH/day ≅ 1.540 l/h
WEMUSTSELECTTHEMAXIMUNVALUEOFQmin
Qmin (2) ≅ 1.540 l/h
Qmin (1) = 500 x 2 x 2 = 2000 l/h
Q min= 2000 /l h
We must calculate the loss of pressure. It has three components
(1) Height losses
(2) Pipe losses
(3) Dripper losses
∆ = + ( ) + P H Ploss pipe ∆Dripper
(1) Height losses
H1
H2D
Nivel medio del aguaH1
H2D
Nivel medio del agua
H1 10H2 0
H1
H2D
Nivel medio del agua
(2) Pipe losses
Lenght int diameter speed m.c.a.(m) (mm) (m/s) tramo10 30 0,90 0,34
Exact value
Lenght % loss mca10 3 0,3
Estimated value
I = 1000 W/m2
The PV Voltage increase if cell Tª decrease
The PV Voltage decrease if cell Tª increase