Quick calculations check

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Fishn sparky

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Washington State
Good afternoon gents,

Can anyone check my calculations on these (2) questions:

1. Residential Heat Pump compressor has current draw of 34A when operating. Compressor has a PF of 70%. Back up heat strip is rated at 10 KW. What is the total current draw when heat strip comes on while the compressor is operating?

My answer: 69.8A (Still new to the postings so don't have the cool symbols yet so I will not show my work)

2. Working in an industrial plant. I have been instructed to double the capacitance connected to a machine. The markings on the capacitor are not visible. The capacitor is connected to a 560V line and has a current draw of 6A. What size capacitor is needed to be connected in parallel with the existing capacitor to double the capacitance?

My answer: 28.42 micro farads (Still new to the postings so don't have the cool symbols yet so I will not show my work)


Thanks to all who view! Happy Thanksgiving!
 
1. Try again. Hint: What does power factor have to do with it?

2. Not enough info to calculate the value of existing.

Welcome to the forum if I missed you first go round... and Happy Thanksgiving to you, too.
 
ya got cap right, think you took a sine when ya shudda taken cos for current - off just a little as .7 near 45 deg
 
junkhound said:
ya got cap right, think you took a sine when ya shudda taken cos for current - off just a little as .7 near 45 deg
Click to expand...

sin(45) = cos(45) = .707. So with a PF of .7 the difference is very small indeed!

More to the point, the current for 10kW heat strip cannot be determined without knowing the line voltage. It is probably 240, but could be 208.
And is the fan on the compressor unit included in the calculation?
 
Both look good
assuming for 1 you:
Source = V / 0 deg
you determined Z comp (after calc'ing i)
then R for heating
calc'ed Zeq = Zcomp || Rhtg

i = V/Zeq = 69.8 / 20.4 deg

if not i = 34 + 10000/240 = 75.7
 
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69.824 at 20.35 degrees so your #1 is correct. Did not check #2 since you seemed to be getting conflicting answers {on #1 only}. Those that said you got #1 wrong were, in fact, wrong :p
 
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Smart $ said:
1. Try again. Hint: What does power factor have to do with it?.
Click to expand...
The currents are at different phase angles.

Think about if you had 10 amps inductor current and 10 amps resistive current? See what I mean? pf matters.
 
mivey said:
69.824 at 20.35 degrees so your #1 is correct. Did not check #2 since you seemed to be getting conflicting answers. Those that said you got #1 wrong were, in fact, wrong :p
Click to expand...
Awww... come on, now. Since when do we use vector math for residential load calculations. :lol:
 
Smart $ said:
Awww... come on, now. Since when do we use vector math for residential load calculations. :lol:
Click to expand...

NEC appears to allow it. :angel:
And it gives you a lower number.

Real world, you can set an upper limit on current by using scalar math, and maybe oversize the circuit. May not cost any more than the time you spend doing vectors. :happyyes:
 
mivey said:
69.824 at 20.35 degrees so your #1 is correct. Did not check #2 since you seemed to be getting conflicting answers {on #1 only}. Those that said you got #1 wrong were, in fact, wrong :p
Click to expand...

I rounded 69.8/20.4 deg lol (I did get 69.82xx / 20.35xx)

second is correct
assuming single phase, f = 60 Hz
Xc = V/i
C = 1/(2 Pi Xc f)
to double C put same value in parallel
 
Ingenieur said:
I rounded 69.8/20.4 deg lol (I did get 69.82xx / 20.35xx)

second is correct
assuming single phase, f = 60 Hz
Xc = V/i
C = 1/(2 Pi Xc f)
to double C put same value in parallel
Click to expand...
I missed yours when I went to get my calculator. 28.4205255521 uF for #2:D
 
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