I have a design unit and 14 circuits to do. Can some one please help as I have completed 1 of my knowledge and before I crack on any help on this would be great. It's a rough copy so excuse the mess. Thanks. It's question 4. Thanks
- Thread starter Cornish1984
- Start date
Yea it's the design part on cable calculations. Was not sure if was on right path
I have m8 but I am a little dyslexic so it takes a while for it to sink in. I just was hoping I was on the right track. As I have missed 2 weeks with holiday
So far the information you have given as answers appears correct, there are a couple of minor points to note (in bold).
a. Design current is the expected load in watts divided by the supply voltage. You have stated 1900W/230V =8.26A
b. Based on the design current the circuit breaker must be rated higher than the design current, for a load of 8.26A the next highest standard breaker rating is 10A so the minimum is 10A but higher can be used, also the characteristics of the load may require a type B, C or D circuit breaker. If the load does not show high current peaks on starting (or stopping) then a B type breaker is a good choice.
As the circuit is part of a swimming pool and assuming it is not SELV or electrically separated then it would require 30mA RCD protection as well.
c. The cable appears to be twin and earth cable installed in conduit on the surface, not a usual choice, but this would be installation method B, however BS7671 does not give tabulated current ratings for twin and earth in conduit, but insulated and sheathed multicore cable tables are pretty close.
d. from your answer the only rating factor that applies is an expected ambient temperature of 40°C (Ca at 0.91)
e. Determining the minimum csa for current carrying capacity means applying the rating factors to the In (16A in your case) as a divisor and referencing the appropriate table and installation method. With a corrected design current of 17.58A the minimum csa live conductor that can carry this load is 2.5mm² (for multicore cable thermoplastic insulated and sheathed).
Determining the minimum csa for voltage drop would mean a rearrangement of the usual equation to say:
mV/A/m (min) = (Max volt drop permitted * 1000) / (Ib*L)
Max volt drop for lighting is 3% of supply voltage, for a 230V supply this is 6.9V, Ib is 8.26A, L=46m
mV/A/m(min) = 6900/8.26*46 = 6900/389.16 =18.16 mv/A/m
From the tables of volt drop the next size of cable that has a volt drop <=18.16 is 2.5mm² as 2.5mm² has a volt drop of 18 mV/A/m.
f. The actual volt drop can now be calculated as you have done and got 6.83 V just within the 6.9V limit for lighting.
g. For 230V final circuits <=32A: on TN system earthing 0.4s, on TT earthing 0.2s; for distribution circuits and >32A circuits: TN system earthing 5s, TT earthing 1s.
h. earth fault loop impedance is calculated from Zs = Ze = (R1+R2) as you have done.
Because you have used resistance values at 20°C this would be equivalent to a measured value and so the maximum Zs you would refer to would be from the OSG and for a B16 this would be 2.2Ω (0.8* the BS7671 value to correct for temperature of the conductors).
i. the maximum earth fault loop impedance stated in BS7671 is 2.73Ω, this is different from the value you have given (2.87) because it is taking into account the Cmin value of 0.95.
Overall pretty good, I think.
a. Design current is the expected load in watts divided by the supply voltage. You have stated 1900W/230V =8.26A
b. Based on the design current the circuit breaker must be rated higher than the design current, for a load of 8.26A the next highest standard breaker rating is 10A so the minimum is 10A but higher can be used, also the characteristics of the load may require a type B, C or D circuit breaker. If the load does not show high current peaks on starting (or stopping) then a B type breaker is a good choice.
As the circuit is part of a swimming pool and assuming it is not SELV or electrically separated then it would require 30mA RCD protection as well.
c. The cable appears to be twin and earth cable installed in conduit on the surface, not a usual choice, but this would be installation method B, however BS7671 does not give tabulated current ratings for twin and earth in conduit, but insulated and sheathed multicore cable tables are pretty close.
d. from your answer the only rating factor that applies is an expected ambient temperature of 40°C (Ca at 0.91)
e. Determining the minimum csa for current carrying capacity means applying the rating factors to the In (16A in your case) as a divisor and referencing the appropriate table and installation method. With a corrected design current of 17.58A the minimum csa live conductor that can carry this load is 2.5mm² (for multicore cable thermoplastic insulated and sheathed).
Determining the minimum csa for voltage drop would mean a rearrangement of the usual equation to say:
mV/A/m (min) = (Max volt drop permitted * 1000) / (Ib*L)
Max volt drop for lighting is 3% of supply voltage, for a 230V supply this is 6.9V, Ib is 8.26A, L=46m
mV/A/m(min) = 6900/8.26*46 = 6900/389.16 =18.16 mv/A/m
From the tables of volt drop the next size of cable that has a volt drop <=18.16 is 2.5mm² as 2.5mm² has a volt drop of 18 mV/A/m.
f. The actual volt drop can now be calculated as you have done and got 6.83 V just within the 6.9V limit for lighting.
g. For 230V final circuits <=32A: on TN system earthing 0.4s, on TT earthing 0.2s; for distribution circuits and >32A circuits: TN system earthing 5s, TT earthing 1s.
h. earth fault loop impedance is calculated from Zs = Ze = (R1+R2) as you have done.
Because you have used resistance values at 20°C this would be equivalent to a measured value and so the maximum Zs you would refer to would be from the OSG and for a B16 this would be 2.2Ω (0.8* the BS7671 value to correct for temperature of the conductors).
i. the maximum earth fault loop impedance stated in BS7671 is 2.73Ω, this is different from the value you have given (2.87) because it is taking into account the Cmin value of 0.95.
Overall pretty good, I think.
Thanks Richard it's really appreciated thank you very much for your time A* 
Amazing how helpful people are on here and this is much appreciated. Thanks allot to every one. I will now grab a coffee and crack on for the rest of the day. 
Would you recommend using all type B breakers m8. I am cracking on today and there is no cost involved I can use any breakers I like and and chose rcbos for the extra protection. Thanks Telectrix
Also was it not 1.97 thanks
Also was it not 1.97 thanks
I know this is rough but I think I have cracked the 1st circuit wit type B rcbo
I am not sure why your Ze (not ZE) is 1.01Ω.
On your previous post the length of circuit was, I thought, 46m it is now 40m.
Remember to use the correct case for your symbols.
The first Zs value you give (in b) for the circuit breaker is the one from the previous edition of BS7671.
Work out your volt drop calculation again in sections e and f and select your cable size on this basis. This affects the later calculations.
Check what page in BS7671 gives you maximum disconnection times, which is what is asked for in g.
h starts the line below g. Remember you state at the top your design parameter is single core cables and so you have a choice of cpc size, it is not dictated by the twin and earth values.
you have labelled as H the answer for i.
An example answer for your reference since you have been working on this and are getting very close.
Swimming Pool Floor Plan – Pool Lights Ze = 1.01 Ω TNCS 0.35 Ω
Design Specifications
Ca = 40 °C (ambient temperature)
Wiring system – 90 °C insulated unsheathed cables in PVC conduit
a) Design Current (Ib) = Power rating of equipment / voltage
19 lights @ 1900 W total 1900/230 = 8.26 A
b) Circuit breaker selected 10 A Type B RCBO to BSEN61009-1 (with maximum Zs of 4.37 Ω)
In = 10 A
c) Installation Method – B – singles in PVC conduit
d) Rating factor that applies ambient temperature (Ca) @ 40 °C, Ca is 0.91
e) Minimum csa of cable for current carrying capacity
It >= In / factors It is the minimum rating for the cable to be selected
It >= 10/ 0.91 = 10.98 A
From table 4E1A 1mm² cable can carry 17 A for reference method B and so is the minimum csa for current carrying capacity.
Minimum csa for volt drop, volt drop limit for lighting is 3% of supply voltage = 6.9V
mV/A/m = (max Vd * 1000) / (Ib *L) = (6.9*1000) / (8.26*40) = 20.88 mV/A/m
From table 4E1A the nearest cable size with a mV/A/m <20.88 is 2.5mm² cable at 19 mV/A/m.
The minimum csa cable that can accommodate both current carrying capacity and volt drop is 2.5mm² cable.
f) Actual volt drop
Vd = (mv/A/m * Ib * L) /1000 = (19 * 8.26 * 40)/1000 = 6.28 V
g) From p.55 table 41.1 BS7671 10A Type B RCBO BSEN61009-1 disconnection time of 0.4s for a final circuit <=32A on TNCS.
h) Using 2.5mm² line conductor and 1.5mm² cpc.
(R1/m+R2/m) * L / 1000 = (7.41 + 12.10) * 40 /1000 = 0.78 Ω at 20 °C
Compensating for a conductor temperature of 90 °C multiply by 1.28 = 0.998 Ω
Zs = Ze + (R1+R2) = 1.01 + 0.998 =2.01 Ω @ conductor temperature of 90 °C
i) The maximum Zs given in BS7671 on p.58 for a 10 A RCBO to BSEN61009-1 with 0.4 s disconnection time is 4.37 Ω.
On your previous post the length of circuit was, I thought, 46m it is now 40m.
Remember to use the correct case for your symbols.
The first Zs value you give (in b) for the circuit breaker is the one from the previous edition of BS7671.
Work out your volt drop calculation again in sections e and f and select your cable size on this basis. This affects the later calculations.
Check what page in BS7671 gives you maximum disconnection times, which is what is asked for in g.
h starts the line below g. Remember you state at the top your design parameter is single core cables and so you have a choice of cpc size, it is not dictated by the twin and earth values.
you have labelled as H the answer for i.
An example answer for your reference since you have been working on this and are getting very close.
Swimming Pool Floor Plan – Pool Lights Ze = 1.01 Ω TNCS 0.35 Ω
Design Specifications
Ca = 40 °C (ambient temperature)
Wiring system – 90 °C insulated unsheathed cables in PVC conduit
a) Design Current (Ib) = Power rating of equipment / voltage
19 lights @ 1900 W total 1900/230 = 8.26 A
b) Circuit breaker selected 10 A Type B RCBO to BSEN61009-1 (with maximum Zs of 4.37 Ω)
In = 10 A
c) Installation Method – B – singles in PVC conduit
d) Rating factor that applies ambient temperature (Ca) @ 40 °C, Ca is 0.91
e) Minimum csa of cable for current carrying capacity
It >= In / factors It is the minimum rating for the cable to be selected
It >= 10/ 0.91 = 10.98 A
From table 4E1A 1mm² cable can carry 17 A for reference method B and so is the minimum csa for current carrying capacity.
Minimum csa for volt drop, volt drop limit for lighting is 3% of supply voltage = 6.9V
mV/A/m = (max Vd * 1000) / (Ib *L) = (6.9*1000) / (8.26*40) = 20.88 mV/A/m
From table 4E1A the nearest cable size with a mV/A/m <20.88 is 2.5mm² cable at 19 mV/A/m.
The minimum csa cable that can accommodate both current carrying capacity and volt drop is 2.5mm² cable.
f) Actual volt drop
Vd = (mv/A/m * Ib * L) /1000 = (19 * 8.26 * 40)/1000 = 6.28 V
g) From p.55 table 41.1 BS7671 10A Type B RCBO BSEN61009-1 disconnection time of 0.4s for a final circuit <=32A on TNCS.
h) Using 2.5mm² line conductor and 1.5mm² cpc.
(R1/m+R2/m) * L / 1000 = (7.41 + 12.10) * 40 /1000 = 0.78 Ω at 20 °C
Compensating for a conductor temperature of 90 °C multiply by 1.28 = 0.998 Ω
Zs = Ze + (R1+R2) = 1.01 + 0.998 =2.01 Ω @ conductor temperature of 90 °C
i) The maximum Zs given in BS7671 on p.58 for a 10 A RCBO to BSEN61009-1 with 0.4 s disconnection time is 4.37 Ω.
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