All About Circuits Forum three phase transformer
 Register Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read

 Homework Help Stuck on a textbook question or coursework? Cramming for a test and need help understanding something? Post your questions and attempts here and let others help.

#1
05-11-2012, 11:20 PM
 PG1995 Senior Member Join Date: Apr 2011 Posts: 753
three phase transformer

Hi

Regards
PG
Attached Images
 hughesExamp34.9_3.jpg (288.2 KB, 37 views)
#2
05-12-2012, 06:36 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,962

Q1: The normal convention used for 3-phase systems is that, unless specifically indicated by the wording, the stated supply voltage is taken to be the line-to-line value.

Q2: It doesn't matter if the supply is star or wye connected. If the system voltage is quoted as a certain value, the source [generator / transformer] line-to-line value would be that same. The individual phase winding voltages at the source would differ depending on whether the source is configured as delta or wye. But the source configuration [be it wye or delta] has no bearing on the voltages observed in the transformer windings being considered here.

Q3:The problem states that the secondary is configured as star [wye]. The primary winding configuration is delta. So there is 3300V across any line-to-line primary [delta connected] winding. The transformed winding voltage on the secondary must be induced according to the primary-to-secondary winding ratio. So the secondary windings all have 283V. But the secondary windings are in star [wye], so the secondary line-to-neutral [star point] voltage is 283V. Hence the secondary line-to-line voltage is √3*283V or 490V.
 The Following User Says Thank You to t_n_k For This Useful Post: PG1995 (05-12-2012)
#3
05-12-2012, 08:45 PM
 PG1995 Senior Member Join Date: Apr 2011 Posts: 753

Thanks a lot, t_n_k. I understand it now.

Someone who has reached this thread might find the following link useful:

Regards
PG
#4
05-14-2012, 11:46 PM
 PG1995 Senior Member Join Date: Apr 2011 Posts: 753

Hi

Thanks for the help.

Regards
PG
Attached Images
 HughesExamp34.6_&_34.7_combined.jpg (247.6 KB, 6 views)
#5
05-16-2012, 02:14 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,962

On the matter of efficiency I'm not sure where the author's use of the n^2 comes from - although I might be better informed if I had the complete text in front of me.

I would approach the problem from a more "fundamental" perspective.

Consider the primary input voltage to be Vp with current Ip and power factor cos(θ).

The fractional efficiency would be given by the general relationship

$\eta=\frac{P_{out}}{P_{in}}=\frac{P_{in}-P_{loss}}{P_{in}}=1-\frac{P_{loss}}{P_{in}}$

The losses would comprise the [assumed constant] no-load magnetization losses Po plus the [variable] winding losses.

In other words

$P_{loss}=P_o+I_p^2R_t$

Where if N (not the 'n' mentioned above) is the primary-to-secondary turns ratio

$R_t=R_p+N^2R_s$

In your example Rp=0.42Ω, Rs=0.0019Ω and N=11000/400=27.5
Hence

$R_t=0.42+(27.5)^2*0.0019=0.42+1.4369=1.8569 \ \Omega$

We can then write the fractional efficiency as

$\eta=1-\frac{(P_o+I_p^2R_t)}{P_{in}}$

But

$P_{in}=V_pI_pcos(\theta)$

Hence

$\eta=1-\frac{(P_o+I_p^2R_t)}{V_pI_pcos(\theta)}=1-\frac{P_o}{V_pI_pcos(\theta)}-\frac{I_pR_t}{V_pcos(\theta)}$

to find the maximum efficiency we differentiate η with respect to the variable input current Ip at some arbitrary power factor.

Or

$\frac{\partial \eta}{\partial I_p}=\frac{P_o}{V_pI_p^2cos(\theta)}-\frac{R_t}{V_pcos(\theta)}$

We find the maximum (or minimum) by equating the derivative to zero.

Hence

$\frac{P_o}{V_pI_p^2cos(\theta)}-\frac{R_t}{V_pcos(\theta)}=0$

Which reduces to the condition for maximum efficiency

$\frac{P_o}{V_pI_p^2cos(\theta)}=\frac{R_t}{V_pcos( \theta)}$

or after simplifying

$P_o=I_p^2R_t$

Which re-iterates the statement in the text that maximum efficiency occurs when the winding I^2R losses equal the no-load losses.

From this one then can deduce the actual primary current to meet this condition.

We can also note (along with the text) that the maximum efficiency condition is independent of power factor. However the actual efficiency value at that condition will depend on the power factor. As an exercise you might try to determine what that maximum efficiency value might be.

So in the case of your example problem 34.7 with Po=2.9kW and Rt=1.8569Ω we have the value of

$I_p=\sqrt{$$\frac{2900}{1.8569}$$}=39.519 \ A$

At Vp=11kV this gives the primary VA input as 11000*39.519=434.71kVA.

Assuming a constant power factor of 0.8 through the transformer [*] then the input power would be 347.77kW. With the losses of 5.8kW this gives the load power as 341.97kW which differs slightly from the text value. In my case the maximum efficiency [at 0.8 pf] would then be 98.33%.

My final comment is that this is all a bit arbitrary. The text results imply a change in overall efficiency from full-load efficiency to maximum efficiency as a difference of 0.3%, which barely merits a mention at all. Efficiency is certainly a matter of great importance at rated operating conditions but the small difference from maximum η to the actual η value at rated conditions is miniscule.

Also be advised this analysis is also underpinned by certain assumptions [*] and approximations and does not reflect the exact "truth" over the possible load operating range having regard to such matters as load power factor and fault conditions.

Last edited by t_n_k; 05-16-2012 at 06:46 AM.
 The Following User Says Thank You to t_n_k For This Useful Post: PG1995 (05-16-2012)
#6
05-17-2012, 04:01 PM
 PG1995 Senior Member Join Date: Apr 2011 Posts: 753

Thank you very much for the help, t_n_k.

I have read your post several times since yesterday and now I have good understanding of the topic of maximum efficiency. But I'm kind of still struggling with the queries about the efficiency on half load and n^2 in my last post. This is the book I'm using (please replace asterisks with s-c-r-i-b-d dot com): http://www.***********/sipplefire/d/...y-10th-Edition. You can find the example problem 34.7 on page #723. Thanks.

Regards
PG
#7
05-18-2012, 12:58 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,962

It's interesting to note the derivation of the maximum efficiency condition at equation 34.19 is essentially identical to the process I followed in my previous post. So if you understand my post then the book should make equal sense.

Anyway leaving that aside for the moment I'll attempt to answer your questions concerning examples 34.6 & 34.7.

I'll start with example 34.7. With respect to the use of the fraction n I quote from the text.

"Let n=fraction of full-load apparent power in (kVA) at which the efficiency is a maximum."

To avoid any potential confusion it should be noted that the proposed value n is not the efficiency.

There is a small error made by the authors in moving from example 34.6 to 34.7. They quote the I^2R loss at full load as 3.86kW in example 34.7 whereas they derive it as 3.84kW in example 34.6.

The key to understanding what the authors are getting at in example 34.7 lies in the significance of equation 34.19 which for this example means that the maximum efficiency occurs when the winding losses equal the core losses of 2.9kW.

$I^2R_{Zc}=P_c$

This is the same as my equation

$P_o=I_p^2R_t$

Both equations tell me that the maximum efficiency condition occurs when the total primary & secondary winding I^2R losses equal the core magnetization [or no-load] losses.

The winding losses are proportional to the square of the currents in both primary & secondary windings. The equations above assume that we have lumped all the winding losses together by referring them to either the primary or the secondary - a process with which you should become familiar.

So at rated load the winding losses are actually 3.84kW. But maximum efficiency does not occur at rated load - rather it occurs at a somewhat lower value. If the primary current at rated load is Ir and the primary current at maximum efficiency is then they have a ratio

$k=\frac{I_{\eta}}{I_r}$.

Key Point: The ratio of the winding losses at the two conditions would therefore be proportional to the ratio of the square of currents.

$k^2=\frac{I_{\eta}^2}{I_r^2}$

Assuming the load voltages are essentially the same at these two conditions then we may write

$KVA_{$$\ max \ load$$}=I_rV_{load}$

KVA at maximum efficiency

$KVA_{$$\ max \ \eta$$}=I_{\eta}V_{load}$

Where $V_{load}$ would be the load voltage referred to the primary side. We could, with equal validity, refer everything to the secondary and this would produce the same result.

From the author's definition of n stated above we may write

$n=\frac{KVA_{$$\ max \ \eta$$}}{KVA_{$$\ max \ load$$}}=\frac{I_{\eta}V_{load}}{I_rV_{load}}=\frac{I_ {\eta}}{I_r}$

But we already know from above that .....

The ratio of the winding losses at the two conditions would therefore be proportional to the ratio of the square of currents

$k^2=\frac{I_{\eta}^2}{I_r^2}$

So in the same manner we can write

$n^2=\frac{I_{\eta}^2}{I_r^2}=k^2$

Which links back to the authors' original definition of the fraction n.

The ratio of the winding losses at maximum efficiency to the losses at full load is therefore n^2 [not n].

Last edited by t_n_k; 05-18-2012 at 06:10 AM.
#8
05-18-2012, 02:11 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,962

The half rated load condition occurs when the secondary load kVA value is decreased from 100% [500kVA] to 50% [250kVA].

The total winding resistance referred to the primary is 1.8569Ω.

The primary current at 100% rated load would be 500kVA/11kV=45.45A

The primary current at 50% rated load would be 250kVA/11kV=22.73A

The total [primary referred] winding loss at rated load current would be

The total [primary referred] winding loss at 50% rated load current would be

#9
05-18-2012, 01:52 PM
 PG1995 Senior Member Join Date: Apr 2011 Posts: 753

t_n_k, thank you very much. It's really very kind of you. I need to read the material several times before making any follow-on queries if there are any. Once again, thanks.

Best wishes
PG

 Tags phase, transformer

 Related Site Pages Section Title Textbook Tesla polyphase induction motors : Ac Motors Worksheet AC metrology Worksheet Advanced electromagnetism and electromagnetic induction Worksheet Impedance matching with transformers Worksheet Step-up, step-down, and isolation transformers Textbook Step-up and step-down transformers : Transformers Textbook Mutual inductance and basic operation : Transformers Textbook Practical considerations : Transformers Textbook Winding configurations : Transformers Textbook Special transformers and applications : Transformers

 Similar Threads Thread Thread Starter Forum Replies Last Post The Electrician General Electronics Chat 12 04-18-2012 06:08 AM Novice Powers The Projects Forum 25 03-02-2012 02:19 PM Tman General Electronics Chat 4 04-27-2011 09:11 AM stevegarren General Electronics Chat 3 12-11-2008 03:09 AM mksa Homework Help 2 05-09-2006 05:41 AM

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums     General Electronics Chat     The Projects Forum     Homework Help     Electronics Resources Software, Microcomputing, and Communications Forums     Programmer's Corner     Embedded Systems and Microcontrollers     Computing and Networks     Radio and Communications Circuits and Projects     The Completed Projects Collection Abstract Forums     Math     Physics     General Science All About Circuits Commmunity Forums     Off-Topic     The Flea Market     Feedback and Suggestions

All times are GMT. The time now is 12:50 PM.