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Calculating Cost of Running Motors 2
- Thread starter RJB32482
- Start date
- Thread starter
- #3
I would use the following:
250HP * 746W/HP = 186.5kW
then 186.5kW * 24 hrs =4476kWh
assuming a rate of $0.055/KWh your cost per day is
4476kWh * 0.055 $/kWH = $246.18
per year is $246.18 * 365 = $89,855.7
Jason Buda, PE
Greenville, SC
jbuda54@yahoo.com
250HP * 746W/HP = 186.5kW
then 186.5kW * 24 hrs =4476kWh
assuming a rate of $0.055/KWh your cost per day is
4476kWh * 0.055 $/kWH = $246.18
per year is $246.18 * 365 = $89,855.7
Jason Buda, PE
Greenville, SC
jbuda54@yahoo.com
- Thread starter
- #7
OK so Here's the information I have for the motor from the spec Sheet:
Horsepower: 250 HP
Voltage: 2300V AC
Current Full Load: 59.5 Amps
RPMS: 1800
Service Factor: 1.0
Looking at our DCS, the motor is actually pulling about 26.5 amps. How will this change the calculation?
Thanks.
Horsepower: 250 HP
Voltage: 2300V AC
Current Full Load: 59.5 Amps
RPMS: 1800
Service Factor: 1.0
Looking at our DCS, the motor is actually pulling about 26.5 amps. How will this change the calculation?
Thanks.
I ommitted efficiency and should not have.
250HP * 746W/HP = 186.5kW
186.5kW / 0.9 =207.22kW
then 207.22kW * 24 hrs =4973.33kWh
assuming a rate of $0.055/KWh your cost per day is
4973.33kWh * 0.055 $/kWH = $273.53
per year is $246.18 * 365 = $99,839.67
Makes a 10k difference... Oops
Jason Buda, PE
Greenville, SC
jbuda54@yahoo.com
250HP * 746W/HP = 186.5kW
186.5kW / 0.9 =207.22kW
then 207.22kW * 24 hrs =4973.33kWh
assuming a rate of $0.055/KWh your cost per day is
4973.33kWh * 0.055 $/kWH = $273.53
per year is $246.18 * 365 = $99,839.67
Makes a 10k difference... Oops
Jason Buda, PE
Greenville, SC
jbuda54@yahoo.com
-
1
- #10
Cost of operation however should also factor in any power factor penalties from the Utility, unless the power factor is corrected somewhere or the util doesn't asses them.
The real way to do this if all you know is measured amps, nameplate and Utility rate info is (1.732*E*I*pf*eff/1000)*H*r*p, where;
E = measured line Volts, or use nominal if not measured. Just don't use motor nameplate volts because that's about the motor design, not what's existing.
I = measured Amps on highest phase
pf = power factor (assume .80pf if not corrected and not shown on the nameplate)
eff = efficiency (assume .95 for that size motor with approx. 1/2 load as indicated and if not shown on the nameplate)
H = operating hours if you want a cost per hour, leave this out or use 1.
r = Utility rate per kWh
p = power factor penalty if applicable and not corrected. Most Utils use a multiplier on top of your rate.
Eng-Tips: Help for your job, not for your homework Read faq731-376![[pirate]](/data/assets/smilies/pirate.gif "[pirate] [pirate]")
The real way to do this if all you know is measured amps, nameplate and Utility rate info is (1.732*E*I*pf*eff/1000)*H*r*p, where;
E = measured line Volts, or use nominal if not measured. Just don't use motor nameplate volts because that's about the motor design, not what's existing.
I = measured Amps on highest phase
pf = power factor (assume .80pf if not corrected and not shown on the nameplate)
eff = efficiency (assume .95 for that size motor with approx. 1/2 load as indicated and if not shown on the nameplate)
H = operating hours if you want a cost per hour, leave this out or use 1.
r = Utility rate per kWh
p = power factor penalty if applicable and not corrected. Most Utils use a multiplier on top of your rate.
Eng-Tips: Help for your job, not for your homework Read faq731-376
-
1
- Moderator
- #11
We better re-think the power factor.
At 90% Eficiency and 80% power factor, the reactive current would be 28.9 amps.
Since the current is less than this, the power factor must be much better than 80%.
At 95% efficiency the reactive current would be even greater.
The point is it is futile to try to estimate the power drawn
by a lightly loaded motor without knowing the power factor.
Rememember, the reactive current does not change with the load and as a result the power factor of a lightly loaded motor is very low.
250 HP x 746 = 186500, or 186.5 KW
2300V x 59.5A x 3^.5 = 237031 or 237 KVA
That leaves us quite a gap to be filled by power factor and efficiency.
However, I would assume 100% power factor and use 2300 x 26.5 x 3^.5 /1000 = 105.6 KW If your power factor is less than 100% your cost will be less. Efficiency does not enter this calculation.
If the power factor of the motor is less than 100% it will reduce the cost of operating derived from this formula. Any penalties for low power factor will be assesed on a plant wide basis. Even if the power factor of this motor is 100 % you would still have to pay any percentage penalties assesed on a plant wide basis if the overall plant power factor is below the cutoff point. Does any one still run with out correction these days?
yours
At 90% Eficiency and 80% power factor, the reactive current would be 28.9 amps.
Since the current is less than this, the power factor must be much better than 80%.
At 95% efficiency the reactive current would be even greater.
The point is it is futile to try to estimate the power drawn
by a lightly loaded motor without knowing the power factor.
Rememember, the reactive current does not change with the load and as a result the power factor of a lightly loaded motor is very low.
250 HP x 746 = 186500, or 186.5 KW
2300V x 59.5A x 3^.5 = 237031 or 237 KVA
That leaves us quite a gap to be filled by power factor and efficiency.
However, I would assume 100% power factor and use 2300 x 26.5 x 3^.5 /1000 = 105.6 KW If your power factor is less than 100% your cost will be less. Efficiency does not enter this calculation.
If the power factor of the motor is less than 100% it will reduce the cost of operating derived from this formula. Any penalties for low power factor will be assesed on a plant wide basis. Even if the power factor of this motor is 100 % you would still have to pay any percentage penalties assesed on a plant wide basis if the overall plant power factor is below the cutoff point. Does any one still run with out correction these days?
yours
- Thread starter
- #12
The most accurate method of determining the power used by the motor is to measure it directly with a 3-phase wattmeter. If you don't have a wattmeter, borrow or rent one. That might be no more difficult than measuring actual power factor.
The second choice would be to measure the motor current and read the motor load on a graph of current vs. load. You may be able to get several current vs. load data points from the motor manufacturer. Efficiency and power factor data points would allow you to calculate the current data.
Another alternative is to get or estimate the motor's no load current and estimate load from measured A by interpolation between full-load A (FLA) and no-load A (NLA). PU Load = (A - NLA)/(FLA - NLA). NLA can be estimated an reactive amps calculated from FLA and FL pf.
The second choice would be to measure the motor current and read the motor load on a graph of current vs. load. You may be able to get several current vs. load data points from the motor manufacturer. Efficiency and power factor data points would allow you to calculate the current data.
Another alternative is to get or estimate the motor's no load current and estimate load from measured A by interpolation between full-load A (FLA) and no-load A (NLA). PU Load = (A - NLA)/(FLA - NLA). NLA can be estimated an reactive amps calculated from FLA and FL pf.
The posting by waross about rethinking the power factor reveals some possible problems with the measured 26.5 amp actual running current figure:
1. The current reading is not correct.
2. The motor was not actually loaded when the reading was taken.
3. The reading was taken on the line side of pf correction caps connected to the motor.
1. The current reading is not correct.
2. The motor was not actually loaded when the reading was taken.
3. The reading was taken on the line side of pf correction caps connected to the motor.
- Moderator
- #16
How a redneck will find the power factor with an amprobe, some wrapping paper, a pencil, string and a carpenters square.
Measure your capacitor amps. This is your reactive current. Measure your motor amps. This is the A component of KVA.
Measure your combined amps, motor current and capacitors. This is your partially corrected current.
Draw a vertical line in the center of the wrapping paper that is proportional to the capacitor current. We know this current is virtually 100% reactive so we know the phase angle will be at right angles to the Watts.
From the top of the reactive line, or vector use the string and the pencil to construct an arc proportional to the motor current.
From the bottom of the vertical line construct an arc to intersect the first arc.
Complete the triangle to the point of intersection.
Now extend the vertical line as required and use the carpenters square to draw a horizontal line from the point of intersection of the arcs to intersect the vertical line.
The horizontal line is proportional to the current component of the watts.
You may now either scale the in phase current directly from the drawing or use the law of cosines to calculate it.
Power factor, bottom line divided by the top line.
If you have a right angle triangle the power factor is unity.
If your horizontal line intersects the portion of the vertical line that represents the capacitor current, your power factor is over corrected.
If your horizontal line intersects the vertical line below the line segment representing the capacitor current, your power factor is under corrected. If the horizontal line intersects the vertical line ABOVE the line segment representing the capacitor current then the KW exceeds the KVA. You have either discovered perpetual motion or you (or your techs) have inadvertently transposed your motor current and line current.
From this and a voltage check you will be able to both closely estimate the Watts that the motor is drawing, and the raw power factor and the corrected power factor of the motor.
Factors that effect the accuracy of this method.
The voltage should remain constant during the current checking.
The load should remain constant during the test.
Unbalanced voltages will have an effect on the test.
If the voltages are unbalanced, do a seperate diagram for each phase and then add up the vertical values, add up the horizontal values and use Pythagoris' theorem to calculate the hypotenews, hipotenoose, Aww the third side of the triangle.
When this old boy started doing power factor surveys a power factor meter was something we couldn't afford to even whisper about. I did more than one power factor survey, years ago with an amprobe, a 5 KVAR three phase capacitor and a roll of wrapping paper. The accuracy of this method is quite good. If you want more precision get a bigger piece of paper and expand your scale. Or brush off the law of cosines.
With this data you can extract the numbers you need to calculate the actual watts, which is what your cost is based on. This method inherently measures both load Watts and Loss watts so efficiency will not have to be factored in.
yours
Measure your capacitor amps. This is your reactive current. Measure your motor amps. This is the A component of KVA.
Measure your combined amps, motor current and capacitors. This is your partially corrected current.
Draw a vertical line in the center of the wrapping paper that is proportional to the capacitor current. We know this current is virtually 100% reactive so we know the phase angle will be at right angles to the Watts.
From the top of the reactive line, or vector use the string and the pencil to construct an arc proportional to the motor current.
From the bottom of the vertical line construct an arc to intersect the first arc.
Complete the triangle to the point of intersection.
Now extend the vertical line as required and use the carpenters square to draw a horizontal line from the point of intersection of the arcs to intersect the vertical line.
The horizontal line is proportional to the current component of the watts.
You may now either scale the in phase current directly from the drawing or use the law of cosines to calculate it.
Power factor, bottom line divided by the top line.
If you have a right angle triangle the power factor is unity.
If your horizontal line intersects the portion of the vertical line that represents the capacitor current, your power factor is over corrected.
If your horizontal line intersects the vertical line below the line segment representing the capacitor current, your power factor is under corrected. If the horizontal line intersects the vertical line ABOVE the line segment representing the capacitor current then the KW exceeds the KVA. You have either discovered perpetual motion or you (or your techs) have inadvertently transposed your motor current and line current.
From this and a voltage check you will be able to both closely estimate the Watts that the motor is drawing, and the raw power factor and the corrected power factor of the motor.
Factors that effect the accuracy of this method.
The voltage should remain constant during the current checking.
The load should remain constant during the test.
Unbalanced voltages will have an effect on the test.
If the voltages are unbalanced, do a seperate diagram for each phase and then add up the vertical values, add up the horizontal values and use Pythagoris' theorem to calculate the hypotenews, hipotenoose, Aww the third side of the triangle.
When this old boy started doing power factor surveys a power factor meter was something we couldn't afford to even whisper about. I did more than one power factor survey, years ago with an amprobe, a 5 KVAR three phase capacitor and a roll of wrapping paper. The accuracy of this method is quite good. If you want more precision get a bigger piece of paper and expand your scale. Or brush off the law of cosines.
With this data you can extract the numbers you need to calculate the actual watts, which is what your cost is based on. This method inherently measures both load Watts and Loss watts so efficiency will not have to be factored in.
yours
- Moderator
- #17
Sorry friends;
"From the bottom of the vertical line construct an arc to intersect the first arc."
SHOULD HAVE READ
From the bottom of the vertical line construct an arc PROPORTIONAL TO THE COMBINED AMPS to intersect the first arc.
This is not a joke. This method with a good quality clamp-on meter can give better accuracy than a poor quality power factor meter.
If I had to do it again, I would do a few plots on paper to verify my results and make sure I had my formulas the right side up, and then set up a spreadsheet to crunch the rest of the numbers for me. I don't think I even had a square root calculator the first time I used this method, let alone a computer. Memory may be failing me though. It was close to the time of my first calculator.
yours
"From the bottom of the vertical line construct an arc to intersect the first arc."
SHOULD HAVE READ
From the bottom of the vertical line construct an arc PROPORTIONAL TO THE COMBINED AMPS to intersect the first arc.
This is not a joke. This method with a good quality clamp-on meter can give better accuracy than a poor quality power factor meter.
If I had to do it again, I would do a few plots on paper to verify my results and make sure I had my formulas the right side up, and then set up a spreadsheet to crunch the rest of the numbers for me. I don't think I even had a square root calculator the first time I used this method, let alone a computer. Memory may be failing me though. It was close to the time of my first calculator.
yours
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