Question on Power Factor. HELP

[fan_103]

Hi guys I am having problems with this question,can u help me?!?thnks

3. The loads taken from an a.c. supply consist of:
(a) a heating load of 15 kW;
(b) a motor load of 40 kVA at 0.6 power factor; and
(c) a load of 20 kW at 0.8 lagging power factor.
Calculate:
(i) the total load from the supply in kW and kVA and its power factor,
(ii) the kVA rating of the capacitor to bring the power factor to unity.
Draw the power triangle and show how the capacitor would be connected to the
supply and the loads.
Power factor=Cosine angle(between supply voltage and current)
ANS:59kW, 75.4 kVA, 0.782; 47 k Ar

 

[MATLABdude]

As per the forum rules, you need to show some work. I'll help you out, however: decompose each of these loads (into real and reactive power) using the power triangle, and then find the resultant of real and reactive powers to find your total apparent power.

EDIT: Power triangle:
http://en.wikipedia.org/wiki/Electric_power#Alternating_current

 

[Mene]

To calculate the total load from the supply, we first need to convert the motor load from kVA to kW. Since the power factor is 0.6, we can use the formula P=V*I*Cosθ to find the real power (kW). Plugging in the values, we get P=40*0.6=24 kW. Therefore, the total load in kW is 15+24+20=59 kW.

To calculate the total load in kVA, we simply add all the loads together, so we get 15+40+20=75 kVA.

The power factor can be calculated by dividing the real power (kW) by the apparent power (kVA). So in this case, the power factor is 59/75=0.782.

To bring the power factor to unity (1), we need to add a capacitor in parallel to the load. The kVA rating of the capacitor can be calculated using the formula Q=V^2*Xc, where Q is the reactive power (kVAR), V is the supply voltage, and Xc is the capacitive reactance. Since we want to bring the power factor to unity, Xc should be equal to the inductive reactance of the load, which is 40*0.6=24 ohms. Plugging in the values, we get Q=230^2*24=126160 kVAR. Converting kVAR to kVA, we get 126.16 kVA. Therefore, the kVA rating of the capacitor should be 126.16 kVA to bring the power factor to unity.

The power triangle for this scenario would look like this:

Supply voltage (V) ───┐
│
│
│
│
│
│
│
│
│
│
│
│
│
│
│
│
└──┬──┐
│ │
│ │
│ │
│ └──┐
│ │
│ │
│ │
│ │
│ │
│ │
│ │
└──────┘

The capacitor would be connected in parallel to the load, as shown in the power triangle.