Calculations again.... help please

[m4tty]

Hi,

A 230v 6KW single phase load is supplied by single core pvc cables in high impact pvc conduit fixed to the surface of a brick wall over a distance of 20 metres. The circuit is protected by a C Type CB. The temperature is 35 degrees C.

Determine

a) Design current

B) Installation method

c) CB rating

d) Correction Factor

e) Minimum cable rating

f) Minimum cable size

g) Volt Drop

a) 6000/230 = 26A

B) Reference Method B (From OSG P52)

c) 32A Type C

d) 0.94 (Page 124 OSG)

e) 32/0.94 = 34A

f) 6mm cable is good for 38A (Ref Method B)

g) 7.3 x 32 x 20

1000

= 4.672V which is less than 5% allowed for power circuits

Can someone let me know if ive done it right

Thanks in advance

Matt

 

[Banjo]

Hi,A 230v 6KW single phase load is supplied by single core pvc cables in high impact pvc conduit fixed to the surface of a brick wall over a distance of 20 metres. The circuit is protected by a C Type CB. The temperature is 35 degrees C.

Determine

a) Design current

B) Installation method

c) CB rating

d) Correction Factor

e) Minimum cable rating

f) Minimum cable size

g) Volt Drop

a) 6000/230 = 26A

B) Reference Method B (From OSG P52)

c) 32A Type C

d) 0.94 (Page 124 OSG)

e) 32/0.94 = 34A

f) 6mm cable is good for 38A (Ref Method B)

g) 7.3 x 32 x 20

1000

= 4.672V which is less than 5% allowed for power circuits

Can someone let me know if ive done it right

Thanks in advance

Matt

Hi mate,

only problem with your answer is finding the volt drop.

mV/A/m x Ib x Metres

1000

so... 7.3 x 26 x 20 = 3.91v

1000

Hope this helps

Stu

 

[m4tty]

Nice one cheers mate

 

[revjames]

Sorry to throw a spanner in and apologies if I'm wrong but.... I thought you applied the ambient and grouping factors to the In rating - in this case 32/0.94 which is 34A and then used this in the Vd calc so:

7.3 x 34 x 20/1000

5v

am I right? I have been doing loads of these calcs on my assignment and used my course notes which seemed to imply this method.

 

[Banjo]

Sorry to throw a spanner in and apologies if I'm wrong but.... I thought you applied the ambient and grouping factors to the In rating - in this case 32/0.94 which is 34A and then used this in the Vd calc so:7.3 x 34 x 20/1000

5v

am I right? I have been doing loads of these calcs on my assignment and used my course notes which seemed to imply this method.

Hi mate

Read Appendix 6 pg 124 O/S/G mate tells you there.

Stu

 

[revjames]

OK Stu you are correct, but what is the purpose in working out In/Ca Cg Ci Cc ? just dont get it....

 

[Banjo]

OK Stu you are correct, but what is the purpose in working out In/Ca Cg Ci Cc ? just dont get it....

It is to be applied as each installation has different effects such as temperature and grouping factor which need to be taken into consideration as this can affect the current capacity in the cable.

e.g A cable is installed with to many other cables in a tight area which could produce a lot of heat which will then increaseing the amount of resistance in the cable.

Hope this helps

Stu

 

[revjames]

I think I have worked it out.

OSG says that you work out It (tabulated value of cable size) by dividing In/Ca Cg Ci Cc and using that figure to select cable size. Then use that size cable as basis for working out Vd ie Vd A M/1000

I use the In/Ca Cg Ci Cc figure in the Vd calc. I think I arrive at the same conclusions as by the other method.