Quick calculations check

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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!
 

Smart $

Esteemed Member
Location
Ohio
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.
 

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
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
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
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

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?
 

Ingenieur

Senior Member
Location
Earth
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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mivey

Senior Member
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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mivey

Senior Member
1. Try again. Hint: What does power factor have to do with it?.
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.
 

Smart $

Esteemed Member
Location
Ohio
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
Awww... come on, now. Since when do we use vector math for residential load calculations. :lol:
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Awww... come on, now. Since when do we use vector math for residential load calculations. :lol:

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:
 

Ingenieur

Senior Member
Location
Earth
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

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
 

mivey

Senior Member
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
I missed yours when I went to get my calculator. 28.4205255521 uF for #2:D
 
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