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Single Phase Water Pump
2

Single Phase Water Pump

Single Phase Water Pump

(OP)
how to wire 3 nos. 1.6kW water pump if the source is single phase (240V)?

atached herewith are two wiring diagrams made by myself. i doubt n not sure whether the connections were right.
Pls comment and guide me upon this.

thanks.
 

RE: Single Phase Water Pump

I can't comment on the appropriate use of an RCCB, we have no requirement like that here. You will need t consult your local installation codes for that.

But as to your two diagrams, the 2nd one is all wrong. You have the motor protection MCBs in series with each other. It will never work. The first drawing is the correct way to do it from a circuit path standpoint.


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RE: Single Phase Water Pump

(OP)
so if i parallel the connection for each motor, it will work isn't it?

what about the current? as we all know that motor starting current is high, almost 7 times full load current if using DOL starter.

would there any other way of wiring the 3 pumps?
 

RE: Single Phase Water Pump

It is not indicate to use an overcurrent protection for all motor together but to provide individual protection on each one. Let's say, only 2 [or even 1] motor will be in operation the 32 A overcurrent protection will not protect the motor at all.
As jraef said, the second diagram does not work if the MCBs are provided with overcurrent protection or they will burn if no protection is provided.
The 3*10 sqr.mm copper will not be more warm  than permitted[70 oC]  even for all 3 pumps together, but you have to check the voltage drop-mainly for DOL starting current.
I should recommend not more than 10% drop on the common line and not more than 20% on motor terminal for one motor starting while the other 2 are in operation.
Short circuit current shall be not more than 3000 A.
If the total length [included common] will not be more than180 m one motor will start well.
The common length has to be not more than 30 m.
If one of these conditions will be over the cable cross section has to be increase.
For cable data see manufacturer catalog.For instance this one:
http://www.cse-cables.com/power-cable/swa-cable.asp
 

RE: Single Phase Water Pump

(OP)
Actually the distance from main to the pumps is about 500m.
So here I attached another option of wiring the 3 pumps.

FYI the 3 pumps will not operate at once but in sequence.
Say pump 1 is on, then after 10sec the next pump turns on, same goes to the third pump.

Base on my calculation, the cable to be used is 3x1C - 16mm sq PVC/SWA/PVC cable with approx. 4% Voltage drop

RE: Single Phase Water Pump

If motor Irated=9A [PF=0.87,EFF=85%] then through the common line [from source up to J.B.] will flow 3*9=27 A.
When motor will start will take 7*9=56 A. When the third motor will start then through the common line [from source up to J.B.] will flow 2*9+56=74 A.
If the common cable is 2*16 sqr.mm then the voltage drop per A and per m will be 0.00275 V .So for 450 m and 74 A the common drop will be :0.00275*74*450=91.6 V[91.6/240*100=38.2%].
Even for steady state [usually all 3 pumps in function at rated current] the common drop will be :0.00275*27*450=33.4 V[33.4/240*100=13.9%].
You need to reduce the impedance to 25% [0.00275/4=0.69/1000 V/m/A].
According to the manufacturer catalog we have to choose 2* 70 sqr.mm copper.
Actually the voltage drop is less a few percent [as the conductor will be less warm -only 70 instead of 90-so less resistance, but the reactance will increase a bit as PVC insulation is thicker. So 0.69 mV/A/m will be fair].
If the cable is 2*70 sqr.mm copper then the voltage drop per A and per m will be 0.00069 V .So for 450 m and 74 A the common drop will be :0.00069*74*450=23 V[23/240*100=9.6%] This drop does not disturb the 2 motors already in operation.
If the individual cable -from J.B. up to the motor terminal- will be 2*10 sqr.mm the voltage drop will be: 0.00275*7*9*50=8.66 V.
If we add the common 23 V the total is 31.7 V[=31.7*100/240=13.2%] at the 3rd motor terminals when this starts.
And for steady state [usually all 3 pumps in function at rated current] the common drop will be :0.00069*27*450=8.38 V[8.38/240*100=3.46%].
If the individual cable -from J.B. up to the motor terminal will be 2*10 sqr.mm the voltage drop[steady state] will be: 0.00275*9*50=1.24 V. If we add the common 8.38 V the total is 9.6 V[=9.6*100/240=4%].
 

RE: Single Phase Water Pump

If more accurate formula will be used for voltage drop the conclusion is you don't need 3*70 for the common line BUT 3*25 SQR.MM COPPER WILL BE GOOD ENOUGH.
I think the formula VoltageDrop=I[A]*Z[ohm] is very conservative, indeed.
The accurate relation is sophisticated. But we may use another approxim. formula:
VoltageDrop=I*R*cosfi+I*X*sinfi we get for common line :
REMARK: the resistance from the catalog is calculated for 90oC[XLPE insulation] then for 70oC we have to multiply by 0.79.
I=54 A; cosfi[pf]=0.435 [for 2 motor in operation 1 starting]; sinfi=0.9;
R=1.5/1000*450=0.675 ohm ; X=0.16/1000*450=.072 ohm
VoltageDrop=54*0.675*0.435+54*.072*0.9 we get for common line :19.35 V [8%]

 

RE: Single Phase Water Pump

Sorry, the current is still 74 A ! I=74 A; cosfi[pf]=0.435 [for 2 motor in operation 1 starting]; sinfi=0.9.
So we have to increase ONCE AGAIN the cable cross section to 3*35 SQR.MM .The resistance will be  1.35 ohm/km [at 90 oC] and the reactance 0.155 .At 70 oC and 0.45 km will be R=1.35*.79*.45=0.48 ohm and X=0.155*.45=0.072.
VoltageDrop=74*0.48*0.435+74*.072*0.9 we get for common line :20.24 V [8.436%].
[Calculated by mean of accurate formula will be 9.27% up to J.B. and 10.27% up to motor terminal in function].
 So if 2 motor are in operation the third motor starting will not disturb.
Finally, from source to J.B. has to be 3*35 sqr.mm copper PVC insulation [450 m]
And from here 3*10 sqr.mm [50 m, up to motor terminals] will be o.k.
 

RE: Single Phase Water Pump

As I have now time to check what I posted I found a weird fragment:
2*9+7*9=74
As I remember from the first class 2*9+7*9=81.But as I checked further it is correct!
What happened? I forgot to mention that the starting current is proportional with the supply voltage .So if when motor starts the voltage drop at list with 10%.Then a factor of 0.9 has to be use in order to reduce the starting current. The calculation will be then so: 2*9+7*.9*9=74 A. I am sorry for the inconvenient!
 

RE: Single Phase Water Pump

(OP)
TQ 7anoter4 for the solution...

BTW, I hve checked the pump data and found the rated current is max of 11.1A.

Thru similar calculation as yours (at 3rd motor starting,I=99.9A), I got common drop of 31.5V (12.94%) using 70mm sq copper. Drop at motor terminal (10mm sq copper) would be 42.18V (17.58%) plus common drop. As for individual drop plus common drop equal to 11.87V (4.95%).

With your advise of using approx. formula (35mm sq copper XLPE); I=(2*11.1A)+(2*11.1A*.9)=92.13V (10.43%)
However, if I take I=99.9A (from the earlier 70mm sq copper XLPE) the Vdrop fall to 8.96% (20.11A).

So then how would you say bout that?
Should I go for 35mm sq XLPE or 70mm sq XLPE?
Or for safe side, I choose 50mm sq XLPE based on calculation on 35mm sq. Should that be okay?
What bout PVC insulated instead of XLPE?

~really2 confuse right now...huhu~
  

RE: Single Phase Water Pump

Quote:

Drop at motor terminal (10mm sq copper) would be 42.18V (17.58%)
Remember, when you start getting starting voltage drops of 17.58% the conductors are acting as a primary resistance starter. With reduced voltage at the motor terminals the current will be reduced also and the actual voltage drop will not be as severe as simple calculations may indicate.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Single Phase Water Pump

If the common cable will be 3*50 sqr.mm copper [only 2 conductors loaded] [SWA] PVC insulation will be good enough as the conductor temperature will be less than 70 oC for 33.3 A steady load.
The voltage drop up to motor terminal for steady load of 11.1 A/motor will be 5.8% [a little more than usual maximum permitted [4-5%].Even if we shall increase the motor cable from 10 sqr.mm copper to 50 sqr.mm this voltage drop will be still 5.23%.
 

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