---
title: "Isolated PV sizing: lamps, fan and TV"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:pv-sizing-lamp-fan-tv"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/pv-sizing-lamp-fan-tv
site: BCT Engineering
---

# Isolated PV sizing: lamps, fan and TV

Power Plant Design · Chapter 1: Energy Sources and Electric Power Generation

- Asked in: 2081 Shrawan, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/pv-sizing-lamp-fan-tv
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

A house is using in average the following electrical appliances:
- Two 10 W lamp for 4 hours per day
- One 60 W fan for 2 hours per day
- One 75 W television used for 4 hours per day

The system has to be powered by isolated PV System with 12 V battery. Determine Size of the PV Panel, Inverter and Battery. (Consider wiring losses 20%, assume inverter efficiency 95% & for battery: efficiency is 90%, depth of discharge factor 0.6 and days of autonomy as 3. Make your smart assumption if further required)

## Short answer

Daily load is 500 Wh. Adding inverter, wiring and battery losses gives required PV energy of 702 Wh/day; with 5 peak sun hours and 0.9 derating, panel size is about 156 Wp, so choose 160 Wp. Battery capacity is 292 Ah, choose 12 V, 300 Ah; inverter about 194 W, choose 250 VA.

## Answer

An isolated PV system is sized by first finding the daily energy demand, then adding the losses to get the PV array rating, the battery capacity for the required autonomy, and an inverter that can carry the connected load.

**Assumptions:** all loads are AC (through inverter); wiring losses are taken as 20% extra energy; peak sun hours in Nepal = 5 h/day; panel derating (temperature, dust) = 0.9.

#### 1. Daily load energy

| Load | No. | W | h/day | Wh/day |
|---|---|---|---|---|
| Lamp | 2 | 10 | 4 | 80 |
| Fan | 1 | 60 | 2 | 120 |
| TV | 1 | 75 | 4 | 300 |
| **Total** | | **155** | | **500** |

#### 2. Energy to be supplied by the battery (DC side)

$$
\begin{aligned}
E_{inv,in} &= \frac{500}{0.95} = 526.3\ \text{Wh} \\
E_{DC} &= 1.2 \times 526.3 = 631.6\ \text{Wh/day}
\end{aligned}
$$

#### 3. Battery size

$$
\begin{aligned}
C &= \frac{E_{DC} \times \text{days of autonomy}}{DoD \times \eta_{b} \times V} \\
&= \frac{631.6 \times 3}{0.6 \times 0.9 \times 12} = 292.4\ \text{Ah}
\end{aligned}
$$

Select **12 V, 300 Ah** (e.g. two 12 V, 150 Ah batteries in parallel).

#### 4. PV panel size
Energy the array must give (battery efficiency 90%):

$$
\begin{aligned}
E_{PV} &= \frac{631.6}{0.9} = 701.8\ \text{Wh/day} \\
P_{PV} &= \frac{701.8}{5 \times 0.9} = 155.9\ \text{W}_p
\end{aligned}
$$

Select **160 Wp** (two 80 Wp, 12 V panels in parallel).

#### 5. Inverter size
Connected load = 155 W. With a 25% margin for surge (fan motor, TV) and future load:

$$
P_{inv} = 1.25 \times 155 = 193.75\ \text{W}
$$

Select **12 V DC / 230 V AC, 250 VA pure sine wave inverter**.

#### 6. Charge controller (extra)
$I = 1.25 \times 160/12 = 16.7$ A, so a **12 V, 20 A** charge controller.

**Answer:** PV array ≈ 156 Wp → 160 Wp; battery ≈ 292 Ah → 12 V, 300 Ah; inverter ≈ 194 W → 250 VA.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the daily load energy?
   A. 155 Wh
   B. 500 Wh
   C. 620 Wh
   D. 1000 Wh
   Answer: B. Lamps 2x10x4=80, fan 60x2=120, TV 75x4=300, total 500 Wh. (Key point: Daily energy)

2. How is the DC energy found?
   A. 500 x 0.95 x 1.2
   B. 500 x 1.2 only
   C. 500/0.95, then x 1.2 for wiring loss, giving 631.6 Wh
   D. 500/1.2
   Answer: C. 500/0.95 = 526.3 Wh, times 1.2 = 631.6 Wh/day. (Key point: DC energy)

3. What is the battery capacity formula and result?
   A. C = E x days/(DoD x eta x V) = 292 Ah
   B. C = E/V = 52.6 Ah
   C. C = E x days/V = 158 Ah
   D. C = E x DoD/V = 31.6 Ah
   Answer: A. 631.6 x 3/(0.6 x 0.9 x 12) = 292.4 Ah, so 12 V, 300 Ah. (Key point: Battery size)

4. How is the PV rating found?
   A. 701.8/(5 x 0.9) = 156 Wp, after dividing 631.6 by 0.9 battery efficiency
   B. 631.6/5 = 126 Wp
   C. 500/5 = 100 Wp
   D. 701.8 x 5 = 3509 Wp
   Answer: A. E_PV = 631.6/0.9 = 701.8 Wh, divided by 5 h x 0.9 derating gives 155.9 Wp, so 160 Wp. (Key point: PV array size)

## Related in Power Plant Design

- [Solar panel and battery for a house (6 h sun)](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Apv-sizing-12v-6h-sun) — Power Plant Design, chapter 1, asked 2×
- [Solar panel and battery for a house (7 h sun)](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Apv-sizing-12v-7h-sun) — Power Plant Design, chapter 1, asked 2×
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- [Parameters for planning and sizing a plant](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aplant-sizing-parameters) — Power Plant Design, chapter 2
- [Solar PV versus hydro and their combination](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Asolar-pv-hydro-combination) — Power Plant Design, chapter 1

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