Calculating Efficiency: Find Useful Energy Output

[Hussman]

I have built a plumbing test rig for my project. I have a couple of electrical devices on this rig and need to find out the efficiency of them/the rig. I have looked online to find equations and have found all the relevant material except for the useful output energy. in all the examples they seem to just give out this information and not tell you how to obtain it.

HOW DO YOU WORK OUT THE useful energy output?


I have a 3 port valve with a power consumption of 6.5W and switching current of 2.2A

There is almost a cylinder thermostat with a switching current of 3A

I am also using a 20amp junction box (not sure if this effects anything) and a 3amp plug.


using P= V x I i worked out the pwer for the wholse system to be 1.196 watts

% energy efficiency = (Useful energy produced × 100) / Total energy used


THANKS!

 

[jambaugh]

It is not clear to me what efficiency you want to determine. Efficiency applies to transfer of power and energy not simply to its utilization. For example the valve simply uses energy it doesn't convert it to some other useful form. So in this sense it is zero percent efficient. But if does its job. A valve can in principle be arbitrarily "efficient" in that the theoretical limit of the power needed to open or close a valve is infinitesimal (unless you want to get into some deep quantum theory). But as a practical matter you can't find frictionless surfaces and zero mass perfectly rigid components. So for practical valves there is a tradeoff between the cost of making it use less energy and the reliability within the working parameters to the cost of the energy required to operate it.

As far as the efficiency of your whole rig, you'll just have to add to your input power the average power consumption of the valve, (averaged over the time of a full operational cycle during which it may only sometimes be drawing power). The junction box should have little effect. The circuit resistance of it should be inconsequential if its within rated values (and if it isn't efficiency isn't your worry, safety is.)

Do be careful with power for AC devices. RMS current time RMS voltage doesn't always give you power. If the voltage peaks at a different time in the AC cycle than the current then you are dealing with less power. (This mistake is BTW the basis for many a "free energy" device.)

You say a plumbing test rig, what form does the output power take? If you're pumping fluid the power will be volumetric flow rate times pressure. That is the fluid analogue to the current times voltage.

 

[Hussman]

jambaugh thanks for your reply

I cannot go into too much detail about the device/system but as the valve is motorised I though it would be possible to work out loss of energy through heat and voltage output ect and work out how much power is used/wasted

As for the pressure it involves no pump as it would be plumbed into the pipes leading off a combi boiler to the taps in the kitchen. So the water pressure will be that of the boilers output

To be honest I am looking for equations and workings to put into my report. Though the efficiency would be an easy one!

It has a 22mm copper pipe input to the valve and 2 15mm outputs so i guess i can do some fluid pressures for that

 

[jambaugh]

If this is the case you shouldn't use the term "efficiency". Just say "power utilization" in your report.

 

[rezeew]

To calculate the useful energy output, you will need to first determine the purpose of each electrical device on your plumbing test rig. This will help you identify the specific energy output that is considered useful for your project. Once you have identified the useful energy output, you can use the formula you mentioned: % energy efficiency = (Useful energy produced × 100) / Total energy used.

For example, if the purpose of the 3 port valve is to regulate the flow of water, then the useful energy output would be the energy required to open and close the valve. This can be measured by the power consumption of 6.5W and switching current of 2.2A. Similarly, for the cylinder thermostat, the useful energy output would be the energy required to regulate the temperature of the water.

The 20amp junction box and 3amp plug may not directly affect the useful energy output, but it is important to take into consideration the total energy used in your calculation. This will give you a more accurate measurement of the efficiency of your test rig.

In summary, to work out the useful energy output, you will need to identify the purpose of each electrical device and the specific energy output that is considered useful for your project. Then, you can use the formula provided to calculate the energy efficiency of your test rig. I hope this helps!