# Ladder for energy calculation

J
Thread Starter

#### José Claudeni

I need to make a RSLogix500 ladder to calculate the energy consumed (KWh) by a pump. I have the following informations in integers values(N7: circuit current(0-500), voltage(0-5000) and power factor(-36 to 72). The PLC platform is Micrologix 1500. Thanks for now.

P

#### Paul Fletcher

power = V * I * cos(power factor) assuming you have true rms readings for current and voltage. You will then have to integrate the power to give you energy (kWh) This can be done using trapezoidal rule integration which is:

energy used for sample = power value at sample 1 + power value at sample 2 * time between samples / 2.

This is fairly easy if you have a timer that completes at a regular known interval (say 1 second) and you sample the kW value every second whilst remembering the previous value. Do the maths as above and you have energy in kWs. Accumulate the energy used in a register and divide by 3600 and that will give you kWh.

J

#### José Claudeni

Ops...A correction. The power factor(PF), a integer value, varies from 809(1000*cos-36) to 309(1000*sin72).

M

#### mjv

Single Phase Calculation of KW:
(I x E x PF) / 1000

Three Phase Calculation of KW:
(I x E x PF x 1.732) / 1000

S

#### Steve Myres, PE

Actually since half of each sample value gets added twice (once when it's sample 1, and once when it's sample 2), just use sample 1 times the length of the time interval, add to the totalization, and you'll get the same result
without having to store the value of sample 1.

N

#### Naggs

Dear Sir

These are formula's you have established for KW calculation:

Single Phase Calculation of KW:
(I x E x PF) / 1000

Three Phase Calculation of KW:
(I x E x PF x 1.732) / 1000

In the above equations, I is the current and E is the voltage . How do you determine current in the above equation . What is the formula for current

Thanx

Naggs