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Jordanh1992

A 230 volt 14 kW domestic storage heater, installed in a house is to be supplied by a 15m run of PVC insulated and sheared cable, clipped to a surface. The circuit is to be protected by a BS 3036 fuse and the value of external impedance is 0.85 ohms. Calculate the minimum size cable that can be used for this circuit to comply with shock protection.
Ib: 60.87
In: 100
It: 60.87
Iz: 65 No correction factors...
csa= 10mm2
VD = 4.4 x 60.87 x 15/1000 = 4.02 (VD is below 11.5)
Max Zs = 0.53 and that is already higher than the Ze? What should I do...
 
so...has any thought been given to the earthing arrangements here....as if its a TN system...you wouldn`t even start as the Ze has failed....if its a TT then an RCD to 30mA would be required for fault....at which point you can use the formula 50/0.03=1666.66
so, i give you 1667 ohms....so, as far as disconnection times are concerned....
 
you dont need to know the earthing arrangements to complete the question.
and i wouldnt be using 100A fuse for a 60a load , is 63 or 80a not available ?
 
you dont need to know the earthing arrangements to complete the question.
and i wouldnt be using 100A fuse for a 60a load , is 63 or 80a not available ?
biff....i`v given him an example of how to carry out an adiabatic on his other thread he`s got going....now, let him suss it out for himsen....
 
A 230 volt 14 kW domestic storage heater, installed in a house is to be supplied by a 15m run of PVC insulated and sheared cable, clipped to a surface. The circuit is to be protected by a BS 3036 fuse and the value of external impedance is 0.85 ohms. Calculate the minimum size cable that can be used for this circuit to comply with shock protection.
Ib: 60.87
In: 100
It: 60.87
Iz: 65 No correction factors...
csa= 10mm2
VD = 4.4 x 60.87 x 15/1000 = 4.02 (VD is below 11.5)
Max Zs = 0.53 and that is already higher than the Ze? What should I do...


ok Im doing 2395 in june someone will correct me if im wrong.

Here is what i come up with.

[FONT=&quot]If= Uo/Zs 230/0.85=270.85

[/FONT]
[FONT=&quot][/FONT] S= √(I²×t) [FONT=&quot][FONT=&quot]270.85[/FONT] [/FONT][FONT=&quot] x [/FONT][FONT=&quot][FONT=&quot]270.85 [/FONT]x 5 = 366798.61[/FONT]


[FONT=&quot]square root of that is: [/FONT]605.63

divide that by k factor of 115= 5.26mm

So the next cable size is 6mm





[FONT=&quot] [/FONT]
 
Jeeze I'm not asking for the answer or I would have said that...
iv been trying to work this out for over a week, I'm asking for advise... Where to look, what to take into account, what I might be doing wrong! I'm not an expert, but if you wanna be awkward then that's fine. And nope the tables show that the only protective device is an 100A
 
Calm down Jordan :)

You're not stupid and you are persistent so you'll get there in the end.

The question asks about shock protection so it's asking about disconnection time not cable temperature. Forget Adiabatic (in this case) and work out disconnection time for the 3036 fuse.

Are you SURE it's not a 1.4kW storage heater? (they exist, 14kW storage heaters don't , as far as I know, exist).

Hope that's helpful
Laurie
 
Calm down Jordan :)

You're not stupid and you are persistent so you'll get there in the end.

The question asks about shock protection so it's asking about disconnection time not cable temperature. Forget Adiabatic (in this case) and work out disconnection time for the 3036 fuse.

Are you SURE it's not a 1.4kW storage heater? (they exist, 14kW storage heaters don't , as far as I know, exist).

Hope that's helpful
Laurie

Have to disagree with you it asks this

Calculate the minimum size cable that can be used for this circuit to comply with shock protection?
 

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