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I have been trying to solve homework questions on three phase systems but I am got stuck in two questions while solving solving them.

Three similar impedances are connected in star across 2000V, 50Hz three phase supply. Power absorbed is 300KW. Current taken is 100A lagging. Find the values of circuit parameters in each phase.

While Solving this, I calculated : V_ph =2000/√3=1154.7V and the impedence to be: Z=Vph/100=11.54 ohms.

The power factor is:2000/3XVphXIph which comes out to be 0.28.

My question is how can i caluculate the power for each impedence.As far as I know ,I only know equation for total power

A three phase 400V motor has an output of 20HP with efficiency of 90% and at a power factor of 0.8. Determine the reading on each of the two wattmeters connected to measure the input. Also find the line current.

So input Power is 90% of 20 that is 18 Watts

So I presume W1+W2=18 watts

My book gives the formula: tan(θ)=√3(w2-w1/w2+w1) or tan(θ)=-√3(w2-w1/w2+w1) where w1 and w2 are wattmeter readings in the wattmeter. My question is how do i identify which equation to use.

Can anyone please help me with these doubts and also tell me what I am doing wrong.

EDIT:

input power is 100HP=74.57kW.

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    \$\begingroup\$ Z = Vph / 100 = 1.154 ohms? Tip: as spaces into your equations to let them breath. \$\endgroup\$ – Transistor May 16 at 7:12
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Three similar impedances are connected in star across 2000 V, 50 Hz three phase supply. Power absorbed is 300 kW. Current taken is 100 A lagging. Find the values of circuit parameters in each phase.

I calculated : V_ph = 2000 / √3 = 1154.7V and the impedence to be: Z = Vph / 100 = 1.154 ohms.

Double-check that last bit.

The power factor is: 2000 /3 × Vph × Iph which comes out to be 0.28.

That's not the correct way to calculate PF. You need to calculate the per-phase power (kW) and per-phase apparent-power (kVA) and go from there.

My question is how can I calculate the power for each impedence. As far as I know ,I only know equation for total power.

As shown above.


A three phase 400 V motor has an output of 20 HP with efficiency of 90% and at a power factor of 0.8. Determine the reading on each of the two wattmeters connected to measure the input. Also find the line current.

So input Power is 90% of 20 that is 18 watts.

Alarm bell 1: You reckon you can supply it with 18 W and get 20 W out!

Alarm bell 2: You think that 20 HP = 20 W. (It doesn't so you need to look that up.)

So I presume W1 + W2 = 18 watts.

My book gives the formula: tan(θ) = √3(w2 - w1 / w2 + w1) or tan(θ) = -√3(w2 - w1 / w2 + w1) where w1 and w2 are wattmeter readings in the wattmeter. My question is how do i identify which equation to use.

It depends on the relative polarity of the two meters. Since it's a motor you know the load is inductive and can base your result on that.


From the comments:

Why can't we use: P = 3 x V_ph x I_ph x p.f.?

You can but you wrote "The power factor is:2000/3XVphXIph" or \$ PF = \frac V {3V_{ph}I_{ph}} \$. It should be \$ PF = \frac P {3V_{ph}I_{ph}} \$ where \$ V_{ph} is the phase to star-point voltage (not phase to phase).

Yes, you can calculate from the total power.

Also I didn't get how u concluded that the load is inductive.

It's a motor.

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  • \$\begingroup\$ Why can't we use:P=3 x V_ph x I_ph x p.f? \$\endgroup\$ – Scáthach May 16 at 13:02
  • \$\begingroup\$ You need to calculate the per-phase power (kW) and per-phase apparent-power (kVA) and go from there. I am not familiar with these terms.Can you tell what they are.Are u telling to calculate the apparent power of each phase and total power and take their ratio.If so ,yes I can do that but can't i calulate it directly from total power \$\endgroup\$ – Scáthach May 16 at 13:03
  • \$\begingroup\$ Also I didn't get how u concluded that the load is inductive \$\endgroup\$ – Scáthach May 16 at 13:11
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor May 16 at 13:35
  • \$\begingroup\$ so both wattmetres are postive but still either of the equation can hold on \$\endgroup\$ – Scáthach May 16 at 14:10
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For the first question. The magnitude of the impedance is \$ 11.5470 \,\,\Omega \$, not \$ 1.15470 \,\,\Omega \$. The single-phase/per-phase apparent power is

\$ |S_{1\phi}| = |\tilde V_\phi| |\tilde I_\phi| = (1\,154.7005 \text{ V})(100 \text{ A}) = 115\,470.0538 \text{ VA} \tag*{} \$

The power factor of each load impedance is

\$ \text{PF} = \dfrac{P_{1\phi}}{|S_{1\phi}|} = \dfrac{P_{3\phi}}{3|S_{1\phi}|} = \dfrac{300 \text{ kW}}{3(115\,470.0538 \text{ VA})} = 0.8660 \text{ lagging} \tag*{}\$

or

\$ \text{PF} = \dfrac{P_{3\phi}}{|S_{3\phi}|} = \dfrac{P_{3\phi}}{\sqrt{3} |\tilde V_{LL}| |\tilde I_L|} = \dfrac{300 \text{ kW}}{\sqrt{3} (2\,000 \text{ V}) (100 \text{ A})} = 0.8660 \text{ lagging} \tag*{}\$

which is different from your result. Check the equation you used for the PF. The PF can be expressed as the ratio of active power to apparent power.

As you could see, the (active) power of each impedance is a third of the total active power, i.e. \$\text{100 kW}\$.

For the second question. The given power is the nominal output power of the motor; of course, it is only "active" power, i.e. watts. You can find the nominal input active power as

\$ \eta = \dfrac{P_o}{P_i} \implies P_i = \dfrac{P_o}{\eta} = \dfrac{\text{20 HP}}{0.9} = \text{22.2222 HP} = 16\,577.7777 \text{ W} \tag*{} \$

If you want, you can use the PF to get the corresponding nominal input apparent power. This is the total/three-phase active power consumed by the motor, and so is the sum of the readings of the wattmeters (assuming the two-wattmeter method) at full load. Thus, the current at full load is

\$ P_i = \sqrt{3} |\tilde V_{LL}| |\tilde I_L| \cos{\theta} \implies |\tilde I_L| = \dfrac{P_i}{\sqrt{3} |\tilde V_{LL}| \cos{\theta}} = \dfrac{16\,577.7777 \text{ W}}{\sqrt{3} (400 \text{ V}) (0.8)} = 29.9100 \text{ A} \tag*{} \$

As for the individual reading of each wattmeter, the \$\pm\$ sign in the equation you posted depends on which of the three phases was taken as the common (reference) phase for the negative terminal of the voltage coils.

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    \$\begingroup\$ We normally don't solve homework problems directly but guide the OP through the process, let them update the question with their work and we can update our answers to guide them through the next steps. This way we weed out the "gimme the answers" guys and also avoid giving answers to people doing "open-book" exams during Covid-19 restrictions. \$\endgroup\$ – Transistor May 16 at 7:32
  • \$\begingroup\$ He asked how to compute the PF and I told him how. \$\endgroup\$ – Alejandro Nava May 16 at 7:35
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    \$\begingroup\$ Exactly my point. You didn't just tell him how, you did the calculation too. \$\endgroup\$ – Transistor May 16 at 7:37

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