50Hz HV Transmission line question

• David J
In summary, the conversation discusses a question regarding a 50 Hz high-voltage transmission line and the calculation of power absorbed from the supply by the transmission line on open circuit. The complex ABCD equations are used in the calculation, with the parameter values from TABLE C given. The question also addresses the issue of reactive power and power factor.
David J
Gold Member

Homework Statement

[/B]
This question was raised on the 11th January 2015 but that thread appears to be no longer active now. There are 2 parts to the question (a) and (b). Initially I am only trying to work on (a)

Figure shows a 50 Hz, high-voltage, transmission line. The relationships between the sending and receiving end voltages and currents are given by the complex ABCD equations:

where 'S' stands for sending-end and 'R' stands for receiving-end

(a) Given the parameter values in TABLE C and an open-circuit received voltage measured as 88.9 kV, calculate the values of
and
and hence the power
absorbed from the supply by the transmission line on open circuit.

The Table C values are :-

##A_1=0.8698##
##A_2=0.03542##
##B_1=47.94\Omega##
##B_2=180.8\Omega##
##C_1=0 S##
##C_2=0.001349 S##
##D_1=0.8698##
##D_2=0.03542##

Homework Equations

[/B]
##V_s=V_R\left(A_1+jA_2\right)+I_R\left(B_1+jB_2\right)##

##I_s=V_R\left(C_1+jC_2\right)+I_R\left(D_1+jD_2\right)##

The Attempt at a Solution

[/B]
##V_s=V_R\left(A_1+jA_2\right)+I_R\left(B_1+jB_2\right)##

The receiving end is open circuit so there cannot be any current flow so ##I_R=0##

##V_s=88900\left(0.8698+j0.03542\right)+0\left(47.94\Omega+j180.8\Omega\right)##

##V_s=\left(77325.22+j3148.84\right)+0##

So ##V_S=77325.22+j3148.84## or ##77389.31\angle2.33^0##

##I_s=V_R\left(C_1+jC_2\right)+I_R\left(D_1+jD_2\right)##

##I_s=88900\left(0+j0.001349\right)+0\left(0.8698+j0.03542\right)##

So ##I_s=j119.93##

To find the absorbed power

##P_{SO}=V_SI_S##

So ##\left(77325.22+j3148.84\right)\left(j119.93\right)##

So ##P_{SO}=-377640.38+j9273613.63##

In the post from 2015 this same calculation was made and I have shown it below

Now the power should be calculated by

P=Real{VsIs*}

P=Real{(77325.22+j3148.84)(j119.93)}

P=377565w

In the earlier post answer they only included the real value (P=377565w) which I suppose is ok so my calculation, in that case, is pretty close apart from the fact that my value is negative.
The question is asking for a value of power in watts so do I just assume that even though the calculation appears negative it is actually positive ??

My question is why is my value different ?? Have I made a mistake somewhere with this calculation ??

Appreciated as always

Remember that when calculating power using complex values for voltage and current that you should use the complex conjugate of the current

ok so the complex value of current is ##0+j119.93## so the complex conjugate of ##0+j119.93## would be ##-j119.93## ?

In that case

##\left(77325.22+j3148.84\right)\left(j119.93\right)##

becomes

##\left(77325.22+j3148.84\right)\left(-j119.93\right)##

which gives me ##377640.38 + (-j9273613.63)##

am I heading in the right direction with the above ? It makes more sense but I am still not getting the same values as others got and also, if this is correct then what do I do with the imaginary part, ##-j9273613.63## ?

David J said:
ok so the complex value of current is ##0+j119.93## so the complex conjugate of ##0+j119.93## would be ##-j119.93## ?
Yes.
In that case

##\left(77325.22+j3148.84\right)\left(j119.93\right)##

becomes

##\left(77325.22+j3148.84\right)\left(-j119.93\right)##

which gives me ##377640.38 + (-j9273613.63)##

am I heading in the right direction with the above ? It makes more sense but I am still not getting the same values as others got and also, if this is correct then what do I do with the imaginary part, ##-j9273613.63## ?
Your result looks okay to me. Given the original three significant figure of the received voltage you should be good with three figures for the power. I'd attribute any differences to rounding/truncation of intermediate values. If you can supply a link to the original thread I can take a look to see what happened there.

The imaginary part of the power represents reactive power. While the power supply has to provide this power to its load, the load cannot convert it to useful work (only real power does work). Power companies don't like high reactive power components because it represents wasted "effort" to supply it for no benefit to the customer (or themselves). The associated concept is "power factor" which represents how much a given load skews the phase of the current waveform from the voltage waveform. You don't have to do anything with this component of the power in the context of the current problem.

Good Morning and thanks for the info. I worked out a final answer of 378000 watts or 378KW

The initial post I was looking at can be reached via the link below. It was created on the 11th January 2015. I think it may actually still be active but there are no recent updates to any of the questions

Thanks again for your help with this

What is a 50Hz HV transmission line?

A 50Hz HV (high voltage) transmission line is a type of power line used to transmit electricity at high voltages (typically above 100kV) and a frequency of 50Hz. It is used to transport large amounts of electricity over long distances from power plants to distribution centers.

Why is 50Hz used for HV transmission lines?

50Hz is used for HV transmission lines because it is the standard frequency used in most countries around the world. This allows for the efficient transfer of electricity across international borders and between different power grids.

What are the main components of a 50Hz HV transmission line?

The main components of a 50Hz HV transmission line include conductors (wires), insulators, towers or poles, and transformers. Conductors carry the electricity, insulators prevent the electricity from escaping, and towers or poles support the conductors. Transformers are used to step up or step down the voltage as needed.

How does a 50Hz HV transmission line work?

A 50Hz HV transmission line works by using conductors to carry high voltage electricity from a power plant to a distribution center. The electricity flows through the conductors and is stepped up or down in voltage by transformers as needed. The electricity is then distributed to homes and businesses through lower voltage distribution lines.

What are the advantages of using 50Hz HV transmission lines?

Some advantages of using 50Hz HV transmission lines include: the ability to transport large amounts of electricity over long distances, high efficiency in transferring electricity, and the ability to connect multiple power grids together. Additionally, 50Hz is a relatively low frequency which results in lower costs for equipment and maintenance compared to higher frequency transmission lines.

Replies
13
Views
6K
Replies
15
Views
2K
Replies
1
Views
2K
Replies
10
Views
2K