# Fuses and Circuit breakers

Discussion in 'Homework Help' started by Biggsy100, Jan 16, 2015.

1. ### Biggsy100 Thread Starter Member

Apr 7, 2014
88
1
Hi, please could someone have a read through my mini report and critiq as need be.
It does need editing and prettying up, but for now i'd like to focus on the highlighted detail. Many thanks

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Apr 5, 2008
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3. ### Biggsy100 Thread Starter Member

Apr 7, 2014
88
1
So my report is no good?

Apr 5, 2008
19,908
4,138
Hello,

I did not say it is bad.
On the first page the drawing is not shown correctly.
Also the text in the symbols are splt or going through the symbols.
Text and the drawing are running over each other making the text unreadable.

Bertus

5. ### Biggsy100 Thread Starter Member

Apr 7, 2014
88
1
Ok, if you can give me feedback on just the written content (In yellow) i'd be grateful

6. ### WBahn Moderator

Mar 31, 2012
24,555
7,691
Circuit breakers and fuses, unless installed on the equipment, are generally intended to protect the wiring and not the equipment. Think about it -- the circuit breakers are installed in the facility independent of what will eventually be connected as a load, which might change many times over the life of the circuit. Consider your home. Most of your circuits are probably on 15A breakers. Yet most of the things that you plug into them will be destroyed long before they reach 15A of current draw.

Your calculation work largely supports this view.

As for your calculations, some of them don't make a lot of sense. You seem to be starting off saying that you want a load capacity of 12MW (12,000kW) from a single phase supply of 400kV. Sounds excessive, doesn't it. But let's go with it. You then get that you need to support 30A of current in order to provide the desired load capacity. Okay. You then say that you want to size the wire so that the needed current capacity is only 80% of the wire's ampacity. Okay. But then you take the maximum load that you NEED to be able to deliver to your load, namely the 30A, and multiply THAT by 80% to the a maximum loading of 24A. What gives? If your maximum loading is 24A, then you can't deliver the 30A that you are designing for!

In the next line, you switch to 12kV instead of the 400kV you used in the initial computation. Why the change?

Apr 5, 2008
19,908
4,138
Hello,

I have made a screenshot with the drawing as I see it on my computer:

Bertus

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8. ### Biggsy100 Thread Starter Member

Apr 7, 2014
88
1
ok I see, below is the initial question I am working on. As I have already mentioned there are some mistakes. It is purely in it's skeleton phase until I get some 'meat on it' so to speak. I just need to know if it is heading in the right direction.

design a suitable protection system for a 12 kW load. The installation is supplied from a 400 V source having a measured impedance of 0.040 W that can be assumed worst-case. Your design solution should comprehensively include:

· Relevant design calculations,

· Details of the chosen protection system

· The design outcomes,

· The implications for the circuit cables to be used,