NCEES #526
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Flyer_PE
Jr. PE / Sr. Company Pilot
Losses at 0% output are all core losses. Core losses remain relatively constant regardless of load.
The losses at 100% load are the sum of the core losses and the copper losses. The copper losses are determined by subtracting the core losses from the total.
The losses at 100% load are the sum of the core losses and the copper losses. The copper losses are determined by subtracting the core losses from the total.
Sorry, I didn't state my question clear. I meant I understand the equation I^2R=lo-load losses but I don't understand how they come up with (2,950-420)/(I/100)^2 as the I^2R value.
Flyer_PE
Jr. PE / Sr. Company Pilot
At 0% output current, the losses are 420 W.
At 100% output current, the losses are 2950 W.
The copper losses at 100% are then 2950-420 = 2530 W
What you are then solving for is what percentage of full load current results in copper losses of 420 W.
I2 = 420/2530 = 0.17
I = 0.41
At 100% output current, the losses are 2950 W.
The copper losses at 100% are then 2950-420 = 2530 W
What you are then solving for is what percentage of full load current results in copper losses of 420 W.
I2 = 420/2530 = 0.17
I = 0.41
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Flyer_PE
Jr. PE / Sr. Company Pilot
They didn't get it backwards.
The 0.17 value is essentially a per-unit value for the quantity I2R
Power out of the transformer is proportional to I, not I2. At rated voltage, 41% current out of the transformer results in 41% rated power out of the transformer.
The question is wrong for the answer they gave. The question asks: (1) the %, (2) of nameplate kVA (POWER), (3) at which MAX efficiency occurs. No mention of current anywhere in the question.
Let's look at this:
Ohms = 0.5
Current = 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10
% of Current = 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 % of nameplate
Current^2 = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Power: 0.5, 2, 4.5, 8, 12.5, 18, 24.5, 32, 40.5, 50
% of kVA: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
% of kVA was the question. The Power is proportional to the square of the current, not to the current. Moreover, in bold above, when power is at 16% of nameplate rating, the current is at 40% of nameplate rating. And the square of the current is at 16% of maximum. The power is not proportional to the square of the power loss, but to the square of the current. Which brings to another point. They screwed up the next problem as well, problem 527.
Let's look at this:
Ohms = 0.5
Current = 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10
% of Current = 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 % of nameplate
Current^2 = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Power: 0.5, 2, 4.5, 8, 12.5, 18, 24.5, 32, 40.5, 50
% of kVA: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
% of kVA was the question. The Power is proportional to the square of the current, not to the current. Moreover, in bold above, when power is at 16% of nameplate rating, the current is at 40% of nameplate rating. And the square of the current is at 16% of maximum. The power is not proportional to the square of the power loss, but to the square of the current. Which brings to another point. They screwed up the next problem as well, problem 527.
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And this brings me to another point. There a lot of trick questions. By that I mean, either people who publish this do not know how to express themselves clearly, or they are intentionally misleading. For example, problem 531 asks: (1) the reactive power losses (Notice plural), (2) in the transmission line (Notice singular for the line).
Then, the problem is solved for 3 lines. This is patently tricky. Testing should not be like that. I took a lot of tests, even a bar exam, and this is patent BS.
Then, the problem is solved for 3 lines. This is patently tricky. Testing should not be like that. I took a lot of tests, even a bar exam, and this is patent BS.
Flyer_PE
Jr. PE / Sr. Company Pilot
Power dissipated through a resistance is proportional to I2. However the I2 relationship is arrived at by algebraic manipulation of P=VI and ohms law.
For every generator I'm aware of S=VI* (single phase) or S=sqrt(3)*VI* (three phase)
In the given problem the first thing solved for is basically a per-unit value for I2 since you're dealing with power dissipated through a resistance.
Power = VI. V is constant at 1.0 pu (rated voltage). %Power = sqrt(I2)
I understand, and you are right. However, I disagree on this problem. This problem presumes that the resistance is constant; not voltage. Go to the solution. The solution is based on the formula that:
R = 420 W(Pnl)/I^2 = (2950 W - 420 W - P fl)/100^2
So while their ratio (Power/I^2) explicitly explains that P varies in proportion to the square of the current, they nevertheless disregarded their own formula. It's silly. It's plain wrong.
R = 420 W(Pnl)/I^2 = (2950 W - 420 W - P fl)/100^2
So while their ratio (Power/I^2) explicitly explains that P varies in proportion to the square of the current, they nevertheless disregarded their own formula. It's silly. It's plain wrong.
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Flyer_PE
Jr. PE / Sr. Company Pilot
The problem may be silly from a real-world perspective, but it isn't wrong.
Their solution is based on the the following:
1. Peak efficiency occurs when the copper losses are equal to the core losses.
2. Core losses remain relatively constant regardless of transformer load. (420 W in this problem)
Step 1: Find the copper losses at 100% power:
At 0% power, all losses are core losses. (420 W)
At 100% power, the losses are the total of the core and copper losses (2950 W).
Copper losses at full power are then the total minus the core losses (2950-420 = 2530).
Step 2: What value of I2R results in a copper loss of 420 W.
Since we have no information for this transformer other than the losses at 0% and 100% power, we have to deal strictly with ratios.
at X% power: I2R = 420/2530 = 0.17
The 0.17 value is representative of I2, not I. I = sqrt(0.17) = 0.41 pu.
Let's do this step by step and solve this problem correctly.
1. Draw a transformer with shunt resistance (aka no load), in parallel with armature resistance, which is in series with the coil.
2. Label them Rnl (shunt) and Ra (for coil series resistance).
3. The current coming into the node splits into (1) I shunt, or Inl, and (2) Ia or I coil.
4. Power losses are due to the current flowing through (1) Rnl, shunt, and (2) Ra.
5. The power losses are (1) Inl^2*Rnl at the shunt which is constant, and (2) Ia^2*Ra.
6. The maximum power transfer occurs when Pnl = Pa. That condition is Ia (part load)^2 * Ra = 420 = Inl^2 * Rnl.
7. The second condition is at full load. At full load, Ia (full load)^2 * Ra = 2950 - 420. The 420 being assign to the shunt. So, Ia (full load) ^ 2 * Ra = 2530.
8. So from steps 6 & 7, we have. Ra = 420 / I part load ^2 = 2530 / I full load ^ 2.
9. Ratio of I ^2 part/ I ^2 full = 420/2530 = 17% (the ratio of the square of currents is 17%)
10. The ratio of currents then is the square root of 17% = 40 %.
11. Now lets go back at the power loss in the Ra.
12. When current through Ra is at 40 % of the full load current, the power loss increases by 17%.
13. Now, let's look at the current at the node. If we disregard the Inl (the shunt current), the current I coming into the system is equal to the current Ia, which increased by 40 %. So, I in = 0.4 Ia full load.
14. The nameplate power, the pwoer into the transformer is P or S = IV. Since the the current I in = .4 fl, the S = 0.4 IV.
15. The Nameplate power has increased by 40%, proportional to current, at the maximum efficiency.
16. I'm an asshole, and FLYER you are CORRECT. I apologize.
1. Draw a transformer with shunt resistance (aka no load), in parallel with armature resistance, which is in series with the coil.
2. Label them Rnl (shunt) and Ra (for coil series resistance).
3. The current coming into the node splits into (1) I shunt, or Inl, and (2) Ia or I coil.
4. Power losses are due to the current flowing through (1) Rnl, shunt, and (2) Ra.
5. The power losses are (1) Inl^2*Rnl at the shunt which is constant, and (2) Ia^2*Ra.
6. The maximum power transfer occurs when Pnl = Pa. That condition is Ia (part load)^2 * Ra = 420 = Inl^2 * Rnl.
7. The second condition is at full load. At full load, Ia (full load)^2 * Ra = 2950 - 420. The 420 being assign to the shunt. So, Ia (full load) ^ 2 * Ra = 2530.
8. So from steps 6 & 7, we have. Ra = 420 / I part load ^2 = 2530 / I full load ^ 2.
9. Ratio of I ^2 part/ I ^2 full = 420/2530 = 17% (the ratio of the square of currents is 17%)
10. The ratio of currents then is the square root of 17% = 40 %.
11. Now lets go back at the power loss in the Ra.
12. When current through Ra is at 40 % of the full load current, the power loss increases by 17%.
13. Now, let's look at the current at the node. If we disregard the Inl (the shunt current), the current I coming into the system is equal to the current Ia, which increased by 40 %. So, I in = 0.4 Ia full load.
14. The nameplate power, the pwoer into the transformer is P or S = IV. Since the the current I in = .4 fl, the S = 0.4 IV.
15. The Nameplate power has increased by 40%, proportional to current, at the maximum efficiency.
16. I'm an asshole, and FLYER you are CORRECT. I apologize.
Flyer_PE
Jr. PE / Sr. Company Pilot
:laugh: It's all good.
FWIW, you have a long way to go before you get labeled that in my book.
