fresh water pump?

Re: small wires No Charles, that\'s quite wrong.

I've had a look to see if I can find a suitable website but none that I have seen really explain this old chestnut. The best way I can try to point you in the right direction is to use the water analogy. Volts = pressure, Amps = flow rate and Resistance = some constant x the length of the pipe. Your argument, in water terms, is that if you make the water pipe longer (increasing the resistance to flow) that the flow will rise while the pressure remains constant! Clearly that is wrong. Your big error was in assuming that the power (Watts) would be constant. Actually, the variable in this electrical case is the resistance. If you increase the resistance keeping the Voltage constant then the current will fall and the power will fall. The voltage (like the head of water) is a constant and it is the voltage/pressure that drives the current/flow of water.

It is a very common mistake due to people thinking that you can treat any of the terms in Ohm's Law as variables. You can't. In most cases V is fixed and I results from V being applied across a circuit having a resistance R. Varying R does not vary V.

I'm sure that it can be explained much better than that and maybe someone has a link to a suitable resource.
 
Re: small wires No Charles, that\'s quite wrong.

Charles statement of volt drop=increased amperes, of which I believe correct

David are you trying to say that volt drop does not create higher amps. Can you show in a calculation what you believe to be correct.

Your energy loss that you mention can be lost in heat usually caused by a high resistance.

.
 
Re: small wires No Charles, that\'s quite wrong.

Let's consider a simple circuit --- an ideal battery of 10V
and a perfect resistance of 10 Ohms connected with perfect cable having zero resistance. The current flowing is given by I=V/R = 10/10 = 1A Now, suppose we give the cables some resistance and to keep the numbers easy, let's assume that the cables now have a resistance of 10 Ohms. This is equivalent to making the cables very much thinner. So now the total resistance is 20 Ohms (the original 10 Ohms of the resistor and the 10 Ohms of the cable). Now the current I=V/R = 10/20 = 0.5A So the current has fallen by half.

The power being delivered by the battery is given by W=IV. In the first example the power W = 1x10 = 10W The power in the second example, with the thinner wires, is given by power W = IV = 0.5 x 10 = 5W. The power delivered to the actual load, the resistor of 10 Ohms, is only 2.5W, or 1/4 of the original power.
 
Re: small wires No Charles, that\'s quite wrong.

What about the resistance/wattage of the water pump ?

What size are the cables suppling the pump ?

What is the volt drop value of the cables ?

Are you saying the voltage at the supply is the same as at the motor ?..

This is a litte more than your basic Ohms Law lesson, sorry !

.
 
Re: small wires No Charles, that\'s quite wrong.

[ QUOTE ]
This is a litte more than your basic Ohms Law lesson, sorry !

[/ QUOTE ]Yes, there is. You have to substitute a real 12V battery for the ideal 10V one, you have to add the resistance of the cables and get the characteristics of the motor, and plot a load line. I haven't got the time to do that for you. Without doing any of that I can tell you that if you connect the pump with smaller cables it will run slower, pump less and it will draw a lower current. Which was the point, surely? If you don't believe me, try it.
 
Re: small wires

Now, now Charles be nice!!
I dont squander water, but my guests (landlubbers) tend to.
I have more than adequate wiring, plus of course mine are 24 volts, so lower amperage anyway! But of course same brushes etc.
Also, arcing is not normally a product of more amps, we run 750 amps through our DC motors aboard here, and there is virtually no visible arcing.
Also, Charles, you must have forgotten, I KNOW you liveaboard for long periods.
 
Re: small wires No Charles, that\'s quite wrong.

It has been well explained but this may (or may not) help:

Remember that the current flowing through a series circuit remains constant. So with the same battery voltage for both thick and thin cabled systems. Then:

Counting cable resistance we now have a simple series circuit comprising: +ve cable resistance, load resistance and -ve cable resistance.

This total circuit resistance will determine how much current will flow through the circuit. Thinner cabling will increase the total circuit resistance hence the current flow WILL REDUCE.

Current = V/R - so if R increases current must reduce
Power = V2/R - so if R increases power must reduce

If unsure, its best to think of electron flow in terms of water flow: current is the water flow rate, resistance is the diametre of the hose and the water pressure is the voltage!

Now you must assume a water pump is keeping the pressure (voltage) constant, regardless of hose size (Resistance).

If we now use a small hose (cable) little water (current) will flow out of the end. Replace it with a bigger hose (cable) now there is less resistance and a lot of water (current) will come out of the end!

Same principle of drinking with one straw (small cable higher resistance) and then with two. With the same suction pressure applied, one straw will give you a small flow, two straws will give you a much greater flow. Less resistance = higher flow rate!

Clear as mud!

Finally, never eat a cake when working with electrics. If you drop it, then tread on it, the current can run up your leg (old ones are the best) /forums/images/graemlins/cool.gif
 
We use our Jabsco water pump multiple times every day of the year. Like all marine kit it's best to size up. e.g. if you have two water outlets buy a pump for four or six, four outlets eight or ten.

Pump life is based on cycles so fit a domestic sized pressure tank (or at least the biggest you can find, not the dinky nonsense chandlers sell) and the pump cycles drop dramatically. The pump will last years.
 
We have a FEIT A88-24volt pump that might be slightly higher than most on power consumption but I would probably say that it's belt and braces of all pumps. This copes with 2 heads/galley/washing machine and deck hose without any problems with pressure. Also there is a Jabsco 18220 as a secondary that quite capable run on it's own. They have both done eleven years up to now with no problems. I would think the FEIT pump to be fairly expensive though.

.
 
I spoke to Jabsco this morning about their smallest pump, the PAR Max 1. I'm interested in using it for my refrigerator coolant circuit as it has a very low current draw. As in the post further up, I run it on 6 volts via an Isotherm voltage reducer. I wanted to know whether they considered it suitable in this duty for a nominal 50% of the time.

They suggested that the pump should give me several thousand hours in this application.
 
If it helps you we have 2 Flojet's Pt/no. 4105 512 on our frdges which are supposed to be specially designed for fridge or a/c cooling.
We have a change-over seacock and switch and only run one which can serve both fridges.
The pump switches in as and when one or both fridges cut in. This ensures we always have a spare to fall back on if the worst happens.
I do periodically change them over so they both get a regular run.

.
 
You must log in or register to reply here.

Other threads that may be of interest

MagicalArmchair
Servicing the fresh water pump of a d2-55A
Replies
0
Views
74
MagicalArmchair
MagicalArmchair
cyberkryten
We finally bought a boat - Merry Fisher 895s2
2
Replies
26
Views
785
Bouba
Bouba

Share this page

Top