kW per hour calculation
kW per hour calculation
(OP)
I'm monitoring power consumption on an electrically heated furnace. I'm logging the kW with a data logger. The kW readings/samples are each second. How do I estimate the kW usage per hour based on the readings of kW I get per second? Do I add up all the kW in one hour period?
Thanks,
Thanks,






RE: kW per hour calculation
Your logger is simply recording power samples kW. If you want to know the average rate of energy use over an hour, simply add the data and divide by the # of samples to get an answer in units of kW. If you want to know the energy used in that hour, multiply the average by one hour to get the energy used in kWh.
RE: kW per hour calculation
RE: kW per hour calculation
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RE: kW per hour calculation
KVARHr is a standard metering quantity.
The historical method of determining power factor and the resulting penalties has been a KWHr meter with the voltage coils phase shifted to register KVAR Hours.
At the end of the billing period the months KWHrs and the months KVARHrs would be used to calculate the average power factor for the month and the PF penalty.
Small phase shifting transformers were added to a standard three phase meter. The transformers shifted the voltage 90 degrees so that the meter would register KVARHrs.
Old school power factor correction consisted first, of determining how many KVARHrs of capacity were required to improve the average power factor to a few percent above the penalty cutoff. Usually around 92%.
Then we would look for large motors or banks of motors that ran continuously. Fans in a lumber kiln were a great location, because they typically ran 24/7 even f the mill was on a 40 hour work week. We would correct these motors to 100%, and some times add 50% or double the amount of capacity. We may measure the no-load current of the supply transformer and add the same capacity.
Then we would start looking for large motors to correct. The code prohibits over correcting if the capacitor current passes through the motor overload device. By connecting the capacitors ahead of the O/L heaters we could go 50% or 100% over correction. We would tally the expected monthly running hours of each motor time the KVARs of correction added.
You needed some familiarity with the plant and process to avoid over-voltages, but as the large over corrected motors came online there would be a number of smaller uncorrected motors coming online to absorb the surplus KVARs.
With experience and judgment we could eliminate the penalties without generating excess voltages.
Yours
Bill
Bill
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"Why not the best?"
Jimmy Carter
RE: kW per hour calculation
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Eng-tips forums: The best place on the web for engineering discussions.
RE: kW per hour calculation
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Eng-tips forums: The best place on the web for engineering discussions.
RE: kW per hour calculation
Bill
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"Why not the best?"
Jimmy Carter
RE: kW per hour calculation
Now, there's a tug boat struggling to bring a boat across the sea and, at the same time, with the captain struggling so that the crew doesn't drink the beer.
This situation gives rise to a few questions:
1. What, then, is the analogy between tug captain and the CEO of the utility company?
2. If the frequency is 50 Hz, would the beer be a lager or an ale type of beer?
3. Is the sea half full or half empty?
4. Considering the wind and the foam - when will the crew be half-drunk?
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: kW per hour calculation
In that regard this analogy stands the test of time. Grin
Bill
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"Why not the best?"
Jimmy Carter
RE: kW per hour calculation
Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: kW per hour calculation
The mug (approx 0.473176473 liter) is more representative at the distribution and consumption level, which seems more relevant for this thread imo.
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