PG Diploma Solar Semester Two Paper 2 Exam Section B

Numerical Problems (01–30)

01. A PV module has short circuit current Isc=9.03I_{sc} = 9.03Isc​=9.03 A. Calculate the maximum current for cable sizing for a single string array.
Model Answer:
Formula: Imax=Isc×1.25I_{max} = I_{sc} \times 1.25Imax​=Isc​×1.25
Imax=9.03×1.25=11.2875I_{max} = 9.03 \times 1.25 = 11.2875Imax​=9.03×1.25=11.2875 A
Maximum current = 11.29 A approximately.
Marks: 3

02. A PV string has Isc=8.5I_{sc} = 8.5Isc​=8.5 A. Find the design maximum current.
Model Answer:
Imax=8.5×1.25=10.625I_{max} = 8.5 \times 1.25 = 10.625Imax​=8.5×1.25=10.625 A
Maximum current = 10.63 A approximately.
Marks: 3

03. If the rated current carrying capacity of a cable is 20 A and the total derating factor is 0.75, find the derated current carrying capacity.
Model Answer:
Derated CCC = Derating factor × Rated CCC
= 0.75 × 20
= 15 A
Marks: 3

04. A cable has rated current carrying capacity of 30 A and the derating factor is 0.8. Will it be suitable for a circuit with Imax=22I_{max} = 22Imax​=22 A?
Model Answer:
Derated CCC = 0.8 × 30 = 24 A
Since 24 A > 22 A, the cable is suitable.
Marks: 3

05. A cable has rated CCC of 25 A and derating factor 0.6. Check if it is suitable for Imax=16I_{max} = 16Imax​=16 A.
Model Answer:
Derated CCC = 25 × 0.6 = 15 A
Since 15 A < 16 A, the cable is not suitable.
Marks: 3

06. A cable has Impp = 8 A, resistance $R=0.4\Omega$, one-way length $L=20$ m, and total input voltage = 200 V. Calculate percentage voltage drop using the formula given in the notes.
Model Answer:
Vdrop=(Impp×R×L)×100Total input voltageV_{drop} = \frac{(Impp \times R \times L)\times100}{Total\ input\ voltage}Vdrop​=Total input voltage(Impp×R×L)×100​
=(8×0.4×20)×100200= \frac{(8 \times 0.4 \times 20)\times100}{200}=200(8×0.4×20)×100​
=64×100200= \frac{64 \times100}{200}=20064×100​
$=32\%$
Voltage drop = 32%. This is too high.
Marks: 3

07. For a solar circuit, Impp = 5 A, R = 0.2 Ω, L = 10 m, and input voltage = 100 V. Find the voltage drop percentage.
Model Answer:
Vdrop=(5×0.2×10)×100100V_{drop} = \frac{(5 \times 0.2 \times 10)\times100}{100}Vdrop​=100(5×0.2×10)×100​
=10×100100=10%= \frac{10 \times100}{100} = 10\%=10010×100​=10%
Voltage drop = 10%.
Marks: 3

08. A cable has reference resistance Rref=0.5 ΩR_{ref} = 0.5 \ \OmegaRref​=0.5 Ω, temperature coefficient $\alpha =0.004$, operating temperature T=45∘CT = 45^\circ CT=45∘C, and reference temperature Tref=25∘CT_{ref} = 25^\circ CTref​=25∘C. Find the operating resistance.
Model Answer:
R=Rref[1+α(T−Tref)]R = R_{ref}[1+\alpha(T-T_{ref})]R=Rref​[1+α(T−Tref​)]
$=0.5[1+0.004(45-25)]$
$=0.5[1+0.004\times 20]$
$=0.5[1+0.08]$
$=0.5\times 1.08$
$=0.54\Omega$
Operating resistance = 0.54 Ω.
Marks: 3

09. A cable’s reference resistance is 0.8 Ω at 20°C. If $\alpha =0.004$ and operating temperature is 50°C, calculate operating resistance.
Model Answer:
$R=0.8[1+0.004(50-20)]$
$=0.8[1+0.004\times 30]$
$=0.8[1+0.12]$
$=0.8\times 1.12=0.896\Omega$
Operating resistance = 0.896 Ω.
Marks: 3

10. A 1 MW solar plant generates 4,000 kWh per day. Find the monthly output for 30 days.
Model Answer:
Monthly output = 4,000 × 30 = 1,20,000 kWh.
Marks: 3

11. A 1 MW solar plant generates 4,000 kWh per day. Find the yearly output for 360 days.
Model Answer:
Yearly output = 4,000 × 360 = 14,40,000 kWh.
Marks: 3

12. If the average cost of a 1 MW plant is Rs 4.87 crore, what is the approximate cost per kW?
Model Answer:
1 MW = 1000 kW
Cost per kW = 4.87 crore / 1000
= 0.00487 crore/kW
= Rs 48,700 per kW approximately.
Marks: 3

13. A rooftop consumer invests Rs 6,00,000 in a solar system and recovers the investment in 6 years. Find the average annual recovery.
Model Answer:
Annual recovery = 6,00,000 / 6 = Rs 1,00,000 per year.
This aligns with residential payback logic discussed in the notes.
Marks: 3

14. A commercial user installs a solar system costing Rs 15,00,000 and expects payback in 5 years. Find the average annual savings required.
Model Answer:
Annual savings = 15,00,000 / 5 = Rs 3,00,000 per year.
Marks: 3

15. A solar project budget includes: Panels = Rs 3 crore, Inverter = Rs 1 crore, Combiner and junction boxes = Rs 20 lakh, Protective gear = Rs 10 lakh, SCADA = Rs 7 lakh, Erection = Rs 50 lakh. Find the total excluding land.
Model Answer:
Total = 3,00,00,000 + 1,00,00,000 + 20,00,000 + 10,00,000 + 7,00,000 + 50,00,000
= Rs 4,87,00,000
= Rs 4.87 crore approximately.
Marks: 3

16. If 1 MW requires 4 to 5 acres of land, how much land is needed for a 2 MW plant?
Model Answer:
For 2 MW, land required = 2 × (4 to 5 acres)
= 8 to 10 acres.
Marks: 3

17. A plant has 3 strings, each with Isc=9I_{sc} = 9Isc​=9 A. Find the maximum current per string and total combined maximum current if all three are paralleled.
Model Answer:
Per string maximum current = 9 × 1.25 = 11.25 A
For 3 parallel strings, total = 11.25 × 3 = 33.75 A.
Marks: 3

18. A cable’s rated CCC is 40 A. If temperature derating is 0.9 and grouping derating is 0.8, find final derated CCC.
Model Answer:
Total derating factor = 0.9 × 0.8 = 0.72
Derated CCC = 40 × 0.72 = 28.8 A.
Marks: 3

19. A circuit requires Imax=27I_{max} = 27Imax​=27 A. A cable has rated CCC = 35 A and total derating factor = 0.7. Is the cable acceptable?
Model Answer:
Derated CCC = 35 × 0.7 = 24.5 A
Since 24.5 A < 27 A, the cable is not acceptable.
Marks: 3

20. A project has 5 major milestones. If each milestone is planned 12 days apart, how many days are spanned from the first to the fifth milestone?
Model Answer:
Gap count = 4 intervals
Total days = 4 × 12 = 48 days.
This is a simple milestone planning exercise consistent with project timeline concepts.
Marks: 3

21. A site requires 6 earth pits, each tested at 4 ohms. Does each pit meet the stated resistance requirement?
Model Answer:
Required earth pit resistance is less than 5 ohms.
Given resistance = 4 ohms, so yes, each pit meets the requirement.
Marks: 3

22. A 250 kW commercial case study is to be prepared. If each day it produces 4 kWh per kW, estimate daily energy output.
Model Answer:
Daily energy = 250 × 4 = 1000 kWh/day.
This is an applied estimate consistent with case-based project work.
Marks: 3

23. A 25 kW residential case produces 4 kWh per kW per day. Estimate monthly energy for 30 days.
Model Answer:
Daily energy = 25 × 4 = 100 kWh/day
Monthly energy = 100 × 30 = 3000 kWh/month.
Marks: 3

24. A 5 MW industrial case produces 4,000 kWh per MW per day. Estimate total daily output.
Model Answer:
Daily output = 5 × 4,000 = 20,000 kWh/day.
Marks: 3

25. If a 1 MW plant costs Rs 4.87 crore, estimate the cost of a 5 MW plant assuming linear scaling.
Model Answer:
Cost = 4.87 × 5 = Rs 24.35 crore approximately.
Marks: 3

26. A cable run has Impp = 7 A, R = 0.15 Ω, L = 15 m, total input voltage = 150 V. Calculate voltage drop percentage.
Model Answer:
Vdrop=(7×0.15×15)×100150V_{drop} = \frac{(7 \times 0.15 \times 15)\times100}{150}Vdrop​=150(7×0.15×15)×100​
= 15.75×100150\frac{15.75 \times100}{150}15015.75×100​
= 10.5%
Voltage drop = 10.5%.
Marks:3

27. A cable has reference resistance 0.3 Ω at 25°C. Calculate resistance at 55°C if $\alpha =0.004$.
Model Answer:
$R=0.3[1+0.004(55-25)]$
= $0.3[1+0.12]$
= $0.3\times 1.12$
= 0.336 Ω.
Marks:3

28. A project has a total investment of Rs 20 lakh. If the expected average annual saving is Rs 4 lakh, estimate simple payback period.
Model Answer:
Payback = 20,00,000 / 4,00,000 = 5 years.
This fits within the commercial/industrial payback range mentioned in the notes.
Marks: 3

29. A residential system costs Rs 3.5 lakh and the user saves Rs 50,000 per year. Estimate simple payback.
Model Answer:
Payback = 3,50,000 / 50,000 = 7 years.
This aligns with the residential payback range in the notes.
Marks: 3

30. A plant has expected PR improvement from 92% to 95%. Calculate the improvement in percentage points.
Model Answer:
Improvement = 95 − 92 = 3 percentage points.
Marks: 3