Help

[Jordanh1992]

Hey, I'm currently in training and I'm stuck on this question! I would really appreciate some help in how to calculate the answers thanks

A circuit to supply a load of 2.75kW at 230 volts 50Hz.

A BS 5467 PVC insulated and sheathed, steel wire armoured cable is to be installed clipped to a perforated metal cable tray, with a total cable length of 30m from the distribution board.

The distribution board is located at the supply intake position. For most of its length the cable is installed touching two other cables in an ambient temperature of 45 degrees.

If protection is by a BS 88 type fuse and the voltage drop must not exceed 5%, what is the minimum size of cable that can be used to comply with both current carrying capacity and voltage drop constraints?

 

[Octopus]

And your attempts at the answers is?

 

[telectrix]

help you will get. giving you the answer on a plate we not. start by caclculating the curreent. then take it step by step. post your thoughts and then we can see if you are going about it the right way.

 

[Mr.Si]

Do you have your regs books?

 

[Stroppy]

When I see this title "Domestic help" I keep thinking you are advertising for servants.

 

[valleybilly]

no wonder they wanna take calculators ouda classrooms !!!!!!!!! lol

 

[Stu]

Hello I'm stuck on this exact question too. Actually stuck and lacking confidence in finding the right information and applying it correctly.

I can start with finding the Reference Method as it only requires physical information.

Step 1 Ref Method E or F Perforated Tray

Next, I'll find Ib from Power and Voltage values.

Ib= 2750Watts/230Volts = 11.96A

In is a protection device or fuse of type BS88 found in page 56 of unite book and for this application is 16A.

In= 16A

It= In/(Ca x Cg)

I've used table 4B1 for Ca as ambient is 45'C and BS5467 is rated at 90'C so a Ca of 0.87

Ca= 0.87

Cg I tried table 4C4 but that directed me to table 4C1 item 3 (1tray). 3 circuits touching gives a value of 0.82

Cg= 0.82

Thus It = 16/(0.87x0.82) = 22.54A

Have I done OK so far as not sure.

Iz is confusing me as it looks like a value of 50mm2 (Table 4E3A)?

Aluminium would give 16mm2 using Table 4J4A

Iz= 85A

Am I going completely wrong here?

You helpful advice would be gratefully received

 

[trev]

Hey, I'm currently in training and I'm stuck on this question! I would really appreciate some help in how to calculate the answers thanks

A circuit to supply a load of 2.75kW at 230 volts 50Hz.

A BS 5467 PVC insulated and sheathed, steel wire armoured cable is to be installed clipped to a perforated metal cable tray, with a total cable length of 30m from the distribution board.

The distribution board is located at the supply intake position. For most of its length the cable is installed touching two other cables in an ambient temperature of 45 degrees.

If protection is by a BS 88 type fuse and the voltage drop must not exceed 5%, what is the minimum size of cable that can be used to comply with both current carrying capacity and voltage drop constraints?

Jordan you may think we're being horrible bar stewards but the rule here is for a homework question you have to have a go at it yourself and post your answer up before you get any help. Trust me mate you'll thank us for it later

 

[Stu]

Hello again. Had another look at the question that both I and the OP are stuck on (see first post).

Taken on board telectrix and trev replys in this thread and completed as I thought correct.

STEP 1

Ref. Method E or F (Perforated Tray)

STEP 2

[*] Ib = 2750W/230V = 11.96A

[*] In = BS88 type = 16A

STEP 3

[*] Ca = BS5467@90'C Ambient@45'C = 0.87

[*] Cg = (3 circuits) 4C4 -> see item 3 of Table 4C1 = 0.82

[*] It = In/(Ca x Cg) = 16/(0.87x0.82) = 22.43A

[*] Iz = 29A (Table 4E4A)

STEP 4

From Table 4E4A column 4 minimum CSA of BS5467 cable is 1.5mm(2) @ Iz (29A)

Conductor Size = 1.5mm(2)

STEP 5

Voltage Drop using Table 4E4B

[*] 1.5mm(2) VD = 31(mV/A/m)

[*] VD = (mV/A/m) x Ib x L x 0.001 = 31 x 11.96 x 30 x 0.001 = 10.76V

[*] VD = 10.76Volts < 5% 230V (11.5V)

The Voltage Drop is just acceptable as it's under 11.5 volts.
We Can use 1.5mm(2) cable.

######

How did I do. In my previous post I got confused with Iz. I'm hoping I got that bit right.

Cheers again Stu

 

[telectrix]

i may be wrong here, but should you not use the Ib of 11.96A to correct and calculate Iz, not the In od 16A.

so the calc. then becomes 11.96/(0.87 x 0.82) = 17.08A.

 

[Stu]

Hi telectrix. Thanks very much for quick reply.

Bit confused so had a re-look.

Iz is just the value greater or equal to It

It uses In in it's equation:

It=In/(Cc x Ci x Cg x Ca x Cf x Cd x Cs)

I think that part is right!

Following the examples in my book!