Answers to design activities and design exercise

Extracting design data

1.  State how we calculate design current: divide power rating by circuit voltage (I = P/V)

2.  State how we determine In (size of protective device): Protective device needs to be equal to or greater than the design current (In ≥ Ib).

3.  Where can you find protective device ratings available: Appendix B, tables B1-B6, of the OSG, dependant on type and the rule mentioned above.

4.  Write out the formula for tabulating the current-carrying capacity of a cable

$\text{It}\ge \frac{\text{In}}{\text{Ca}\text{Cg}\text{Ci}\text{Cf}}(\text{Appendix}\text{F}\text{OSG})$

5.  When using a BS 3036 fuse (Cf Correction) what correction factor should we use during the cable calculation stage = 0.725 (Appendix F OSG)

6.  What correction factor should we use for 3 PVC/PVC twin core cables that are grouped together and installed using Reference Method C? (Look in Appendix F) = 0.79

7.  What correction factor should we use for any type of cable that is surrounded by thermal insulation for 0.5m = 0.50 (Appendix F OSG)

8.  What correction factor should we use for a PVC/PVC twin core cable that is exposed to an ambient temperature of 35° = 0.94 (Appendix F OSG)

9.  $\text{It}\ge \frac{20}{0.8\times 0.65}\text{It}=38.46\text{A}$

10.  With reference to Appendix F of the OSG, what ambient temperature does not need correcting = 30° (correction factor = 1, dividing by 1 does not affect any number)

11.  What is the correct description of Reference Method C = Clipped direct (Table 7.1 (ii) OSG)

12.  Given the reference method above and using table F6 of the OSG, determine what size cable is required to carry 47A = 6mm²

13.  State the tolerances for voltage drop regarding Lighting & Power? (Look in Appendix F of the OSG or volt drop in the index) = 3% for lighting (6.9V) & 5% for power (11.5 V)

14.  What is the formula used for calculating voltage drop? (Look in Appendix F of the OSG or volt drop from the index)

$\text{Voltage}\text{drop=}\frac{(\text{mv}/\text{A}/\text{m})\times \text{Ib}\times \text{L}}{1000}$

15.  Using Table F6 of the OSG, the Reference Method at step 11, size of cable at step 12, a design current of 27A, a circuit that is 20m long, calculate if this Power circuit complies with volt drop requirements?

$\text{Voltage}\text{drop}=\frac{(7.3)\times 27\times 20}{1000}$

VD = 3.942V well within 3%, 6.9V allowed for a lighting circuit.

16.  What do we mean by shock protection:

Shock protection is ensuring that the earth loop impedance is low enough, to generate a large enough fault current to operate a protective device within a specified time.

17.  Look in the OSG Table B6 and determine the maximum Zs value for a 40A Type B circuit Breaker = 0.88Ω

18.  Do we have to correct the value by multiplying by 0.8: No values in the OSG can be read directly the same applies to GN3. BS7671 values however do have to be corrected.

19.  Given the value above in step 17, if the measured loop impedance value Zs of a circuit is 0.91Ω, what does this mean regarding shock protection? It means shock protection has NOT been met since 0.91Ω > 0.88Ω in other words the measured value is greater than the tabulated value?

20.  Given the Zs value above calculate the PFC?

$\begin{array}{ccc}\text{I\hspace{0.17em}}=\frac{\text{V}}{\text{R}}& \text{I\hspace{0.17em}}=\frac{230}{0.91}& \text{I\hspace{0.17em}}=252.75\text{A}\end{array}$

21.  What do we mean by Thermal constraint?

When an installation is installed with PVC flat profile cable which incorporates a cpc which is smaller than the live conductors. Thermal Constraint is a calculation to ensure that the size of the cpc is big enough to withstand a vey large fault current over a short period of time.

22.  A 6mm² PVC flat profile cable has a 2.5mm² CPC and has a K constant of 115. If the PFC of the circuit is 1.7KA and the circuit needs to be disconnected in 0.4s, calculate if thermal constraint requirement is met.

$\begin{array}{l}\text{S\hspace{0.17em}}=\frac{\surd {\text{I}}^{\text{2}}\text{t}}{\text{K}}\hfill \\ \text{S\hspace{0.17em}}=\surd \frac{{1700}^2\times 0.4}{115}\hfill \\ \text{S\hspace{0.17em}}=9.34{\text{mm}}^2\hfill \end{array}$

A 6mm² PVC flat profile cable has a cpc of 2.5mm² (table 7.1 of the OSG). Therefore thermal constraint is NOT met because cpc needs to be 9mm² but circuit is fitted with 2.5mm² conductor.

Design exercise

A 10KW industrial oven is to be fed from a domestic property to a consumer unit which is located in a detached garage 15m away. The earthing system is TN-C-S and the wiring type to be used is PVC/PVC flat profile cable. This will be clipped direct (Installation method C) to the surface and will be directed away from any thermal insulation and routed through roof trusses. The circuit is to be protected by a Type B circuit breaker. The earth loop impedance Zs has been measured and recorded as 0.81Ω. Using Ohms law this means that the pfc is 283.9A. The cable is grouped with one other circuit in an area with an ambient temperature of 35 °C.

Work out the following:

1)  Design Current Ib

$\begin{array}{c}\text{I\hspace{0.17em}}=\frac{10000}{230}\\ \text{I\hspace{0.17em}}=43.47\text{A}\end{array}$

2)  Size of protective device In

45A

3)  Having established any correction factors, calculate the Tabulated current carrying capacity It

Correction factors Cg = 0.85, Ca = 0.94

Calculate the tabulated current carrying capacity (It)

$\begin{array}{cc}\text{It}\ge \frac{45}{0.85\times 0.94}& \text{It}=56.3\text{A}\end{array}$

4)  Size of cable required

10mm²

5)  Does volt drop comply?

$\text{Voltage}\text{drop}=\frac{(4.4)\times 15\times 43.47\text{A}}{1000}$

VD = 2.87V complies with volt drop requirement (<11.5V)

6)  Is shock protection requirements met?

OSG Table B6 maximum loop impedance value for Type B 45A breaker is 0.78. Shock protection requirements is NOT met given that 0.81 > 0.78 Ω (tabulated value).

7)  Thermal Constraint ok?

Referring to table 7.1 (i) of the OSG (under the cable size column) a 10mm² cable has a 4mm² cpc. The circuit is a final circuit therefore will need to be disconnected in 0.4s.

 $\begin{array}{l}\text{S\hspace{0.17em}}=\frac{\surd {283.95}^2\times 0.4}{115}\hfill \\ \text{S\hspace{0.17em}}=1.56{\text{mm}}^2\hfill \end{array}$

 Thermal constraint requirement is met because 1.56 < 4mm²