eng.saeed said:
Hi everybody....
I have a problem that I could not solve, and really need a help...
It says:
In a factory, the following are the loads:
Induction motors: 1000 hp
0.7 average power factor
0.85 average efficiency
Lighting and heating load: 100 KW
A 3Φ synchronous motor is installed to provide 300 hp to a new process. The synchronous motor operates at 92% efficiency. Determine the KVA rating of the synchronous motor if the overall factory power factor is to be raised to 0.95.
Determine the power factor of the synchronous motor..
Thanx a lot
hmm, seems a 'simpe' school problem ???
Ok...
1000 hp = 1000 * 0.736 Watt = 736 kW mechanical power
(1 european horse power = 736 Watt)
736 kW / 0.85 = 865.9 kW = electrical power to make 1000 hp
power factor cos(fi) = 0.7
fi = acos(0.7) = 45.54 degree
P = 865.9 kW
S = P / cos(fi) = 865.9 kW / 0.7 = 1237 kVA
Q = S * sin(fi) = 1237 kVA * 0.71 = 883.4 kVAr
Light and heat draw 1000 kW with powerfactor of 1
we cannot add 1237 kVA and 1000 kW directly for 2237 kVA (wrong answer) , but we can add real part (P) and imaginary part (Q) separatly, give:
P = 865.9 kW + 1000 kW = 1865.9 kW
Q = 883.4 kVAr + 0 = 883.4 kVAr
and from this calculate new KVA value S
S =sqrt( P^2 + Q^2) = sqrt(1865.9^2 + 833.4^2) = 2043.5 kVA
and power factor:
cos(fi) = P/'S' = 1865.9 / 2043.5 = 0.91 (or 25.3 degree phase angle)
---
new syncron machine load give:
300 hp * 0.736 = 220 kW
92% effency = 220 kW / 0.92 = 239.1 kW
overall power factor is given 0.95 and old power factor calculated before is a 0.91.
we know from old factory load
S = 2043.5 kVA
P = 1865.9 kW
Q = 833.4 KVAr
we add old P above with new motor load:
Ptot = 1865.9 kW + 239.1 kW = 2105 kW
we also know new overall power factor as cos(fi) = 0.95
and 'S' with new power factor and know Ptot give
Stot = Ptot / cos(fi) = 2105 kW / 0.95 = 2215.8 kVA
calculate phase angle:
fi = acos(0.95) = 18.2 degree
sin(fi) = sin(18.2) = 0.312
and
Qtot = Stot * sin(fi) = Stot * 0.312 = 2215.8 kVA * 0.312 = 691.8 kVAr
old load have Q = 883.4 kVAr and total Qtot have now 691.8 kVAr
new machine give Qm = Qtot - Q = 691.8 - 883.4 kVAr = -191.5 kVAr (capacitive reactive load - it's possible to make with syncron machine!)
new syncron machine:
P = 239.1 kW
Q = -191.5 kVAr
S = sqrt(239.1^2 + -191.5^2) = 306.3 kVA
power factor cos(fi) = P/S = 239.1/306.3 = 0.78
phase angle -38.7 degree ie. capacitive load
factory before new motors:
P = 1.865.9 kW
Q = 882.4 kVAr
S = 2043.5 kVA
cos(fi) = 0.91
phase angle = 25.3 degree (inductive load)
factory after new motor
P = 2105 kW
Q = 691.8 KVAr
S = 2215.8 kVA
cos(fi) = 0.95
phase angle = 18.19 degree (induktive load)
---
Check power factor with add 239.1 kW pure resistive load to old factory load:
P = 1865.9 + 239.1 kW = 2105 kW
Q = 883.4 kVAr
S = sqrt( 2105^2 + 883.4^2 ) = 2283 kVA
cos(fi) = P/S = 2105 / 2283 = 0.92
for rice overall power factor to 0.95 with new load - added load must give reactive capacitive load.
(hopfully not written or calculate wrong...)