excel graph in time domain for V(t), I(t), P(t), VAR(t), and apparent power

Status
Not open for further replies.
B

badgers

Member
http://forums.mikeholt.com/showthread.php?t=124592
seemed to be on the right track but I couldn't find the base equations
I have a spreadsheet with a pretend load on a 120V single phase line 0.8PF
RMS value Peak
Volt 120 169.7056275
Current 1.04167 1.473139127
power 100 141.4213562
VA 125 176.7766953
VAR 75 106.0660172
angle 36.87

Here are my equations now
60*360= 21600
TD Scope.jpg

V(t)= 169.7*SIN(RADIANS(21600*B2))
I(t) = 1.04*SIN(RADIANS(21600*B2-36.87))
P(t) = V(t)*I(t)
VAR(t)=????
 
B

badgers

Member
I am assuming the VAR curve is 90 lagging current times time domain voltage
VAR=V(t)*SIN(RADIANS(21600*B2-90))*1.04 VAR Time domain.jpg
the problem I think of is that the VAR should cross zero with the current not the voltage but maybe I am wrong.
PS this is a lagging load.
 
B

badgers

Member
The attached pictures are getting shrunk
White Is voltage
Blue is current
Yellow is real power
Red is supposed to be VAR
how do I post a bigger picture.
 
G

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
badgers said:
I am assuming the VAR curve is 90 lagging current times time domain voltage
VAR=V(t)*SIN(RADIANS(21600*B2-90))*1.04....
the problem I think of is that the VAR should cross zero with the current not the voltage but maybe I am wrong.
PS this is a lagging load.
Click to expand...
Your problem is that the VAR is the reactive portion of the apparent power, or looking at it a different way, is the difference between the apparent power (IRMS*VRMS) and the real power which is the integral of I(t)*V(t).

The reactive current is just the component of the true current that is 90 degrees out of phase with the voltage. Your angle term is correct in the equation, but the coefficient multiplying the current needs to be a constant which is 1.04*cos (36.87) instead of multiplying by 1.04.

The real power, yellow curve, is properly shown as zero when either I(t) or V(t) is zero, and is negative (power flowing back to the source) between the two close zero points.

VAR should also cross zero where either V(t) or IREACTIVE(t) is zero.
 
G

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
In totally realistic terms, it means that real power is flowing back out to POCO during that part of the cycle. That extra power is then flowing back from POCO above and beyond the resistive component during the other part of the cycle for a net of zero averaged over one full cycle.
What is left is the "real" power that you get billed for.

Tapatalk!
 
B

badgers

Member
Smart, I notice on your graph in the spreadsheet power never goes negative. Can you help me understand what drives the power curve up to prevent it going negative?
 
Besoeker

Besoeker

Senior Member
Location
UK
badgers said:
http://forums.mikeholt.com/showthread.php?t=124592
seemed to be on the right track but I couldn't find the base equations
I have a spreadsheet with a pretend load on a 120V single phase line 0.8PF
RMS value Peak
Volt 120 169.7056275
Current 1.04167 1.473139127
power 100 141.4213562
VA 125 176.7766953
VAR 75 106.0660172
angle 36.87

Here are my equations now
60*360= 21600
View attachment 10202

V(t)= 169.7*SIN(RADIANS(21600*B2))
I(t) = 1.04*SIN(RADIANS(21600*B2-36.87))
P(t) = V(t)*I(t)
VAR(t)=????
Click to expand...
I don't think you can do VAR(t) or VA(t)
The instantaneous value of volts times Amps is Watts.

Here's one I did earlier....

Dec13pf0-7_zpsf5b88639.jpg

The power was actually calculated from the current times the voltage.
All f(t)
 
Besoeker

Besoeker

Senior Member
Location
UK
badgers said:
Smart, I notice on your graph in the spreadsheet power never goes negative. Can you help me understand what drives the power curve up to prevent it going negative?
Click to expand...
Unity power factor does that. A resistive load.
When the voltage is positive, the current is also positive and V*I is positive.
Similarly, when the voltage is negative, the current is also negative and V*I is still positive..

5*5 = 25
-5 * -5 = 25
 
Smart $

Smart $

Esteemed Member
Location
Ohio
badgers said:
Smart, I notice on your graph in the spreadsheet power never goes negative. Can you help me understand what drives the power curve up to prevent it going negative?
Click to expand...
Real power is never negative (at least where there is only one source). If you look at apparent power and reactive power, both go negative; former is a result of latter.
 
Besoeker

Besoeker

Senior Member
Location
UK
Smart $ said:
If the scope is limited to a single instant. Just as RMS and average values would be superfluous for a single instant.
Click to expand...
VAR(t), mentioned by the OP in the OP, is a function wrt time. Just as v(t) is. Except that VAR(t) isn't a meaningful concept.
 
Status
Not open for further replies.
Top