synchronous motor problem

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

your voltage source...
you haven't detailed it...

Hi madeza....

thanx for your consern....

This is the only information, and based on the provided information, the problem should be solved...



Thanx...

It seems that it is a tough question to be solved....

any way.... thanx for passing by.....

Hi Eng.saeed,

Haven’t done this type of calc for years but I’d approach it like this:

Convert everything from HP to KW

Use the efficiency figures to calculate the actual power drawn in each case

Draw a triangle showing the original KW, KVA, KVARS PF (you have power and PF so you calculate the KVARs and KVA). i.e. before the new motor is attached


Draw another triangle showing the same but including the new motor power, don’t forget to factor in the efficiency (you are given the factory PF). You have power and pf so you can calculate VARS and VA.


You now have a description for when the motor is and isn’t attached, so you can draw a third triangle that describes the motor (using the difference in KW & KVARs ) from that you can see the motor PF and KVA.

Hope that makes sense, if not let me know and I’ll try again,,

Bob

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

Click to expand...

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...)

Hi xxargs....

Thanx for your help...It seems working Ok...I'm going to check it....



Thanx again

Added after 1 hours 5 minutes:

Thanx Engineer_Bob...

appreciating your help......

eng.saeed said:

Hi xxargs....

Thanx for your help...It seems working Ok...I'm going to check it....



Thanx again

Added after 1 hours 5 minutes:

Thanx Engineer_Bob...

appreciating your help......

Click to expand...

Thanks.

I written from head for understandable (hopfully) solution without skill of complex numbers on readers, this description is not 'academic' elegance or optimize solution.

If you have know on complex numbers and calculator/mathprogram with complex support, above calculation can done with less step, 'simple' and possible more elegance ;-).

Good calculator-simulator with full complex support on all functions is 'free42' (open source), is a simulator of real hp42S calculator and IMHO very good (possible best ever) choice to handle 'simple' electrical enginering problem as above.

Is really pitty is not possible to buy hp42S from hp now...


/Xxargs