---
title: "Which motor is more economical (two-part tariff)"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:economical-motor-two-part-tariff"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/economical-motor-two-part-tariff
site: BCT Engineering
---

# Which motor is more economical (two-part tariff)

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2079 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/economical-motor-two-part-tariff
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

A consumer requires an induction motor of 35 kW. He is offered two motors of the following specifications.
Motor A: Efficiency = 85%; power factor 0.9
Motor B: Efficiency = 90%; power factor 0.8
The consumer is being charged on a two-part tariff of Rs. 250 per kVA of the maximum demand plus Rs. 8 per unit. The power factor of the motor B is to be raised to 0.85 by installing condensers. The motor costs Rs. 1500 less than motor A.
The cost of condenser is Rs. 400 per kVAR. Determine which motor is more economical. Assume rate of interest and depreciation as 10% and working hours of motors as 3000 hours in a year.

## Short answer

Annual cost is demand charge (Rs 250 per kVA) plus energy charge (Rs 8 per kWh) plus capital charges. Both motors draw 45.75 kVA. Motor A: 41.18 kW input, total Rs 999,823. Motor B: 38.89 kW input with condenser, total Rs 944,974. Motor B is more economical by Rs 54,849 per year.

## Answer

Compare the **annual cost** of each motor: demand charge + energy charge + interest and depreciation on the extra capital. The tariff of Rs 250 per kVA of maximum demand is taken as an annual charge, and "the motor costs Rs 1500 less" is read as **motor B costs Rs 1500 less than motor A**.

#### Motor A ($\eta$ = 85%, pf = 0.9)
$$
\begin{aligned}
P_{in,A} &= \frac{35}{0.85} = 41.18\ \text{kW} \\
S_A &= \frac{41.18}{0.9} = 45.75\ \text{kVA} \\
\text{Demand charge} &= 250 \times 45.75 = \text{Rs } 11{,}438 \\
\text{Energy charge} &= 8 \times 41.18 \times 3000 = \text{Rs } 988{,}235 \\
\text{Extra capital cost charge} &= 0.10 \times 1500 = \text{Rs } 150
\end{aligned}
$$
Annual cost of A = 11,438 + 988,235 + 150 = **Rs 999,823**

#### Motor B ($\eta$ = 90%, pf raised from 0.8 to 0.85)
$$
\begin{aligned}
P_{in,B} &= \frac{35}{0.9} = 38.89\ \text{kW} \\
\tan\phi_1 &= \tan(\cos^{-1}0.8) = 0.75,\quad \tan\phi_2 = \tan(\cos^{-1}0.85) = 0.6197 \\
Q_C &= 38.89\,(0.75 - 0.6197) = 5.07\ \text{kVAR} \\
\text{Cost of condenser} &= 400 \times 5.07 = \text{Rs } 2026 \\
\text{Interest and depreciation} &= 0.10 \times 2026 = \text{Rs } 202.6 \\
S_B &= \frac{38.89}{0.85} = 45.75\ \text{kVA} \\
\text{Demand charge} &= 250 \times 45.75 = \text{Rs } 11{,}438 \\
\text{Energy charge} &= 8 \times 38.89 \times 3000 = \text{Rs } 933{,}333
\end{aligned}
$$
Annual cost of B = 11,438 + 933,333 + 202.6 = **Rs 944,974**

#### Comparison
| Item | Motor A | Motor B |
|---|---|---|
| Input (kW) | 41.18 | 38.89 |
| Max demand (kVA) | 45.75 | 45.75 |
| Demand charge (Rs) | 11,438 | 11,438 |
| Energy charge (Rs) | 988,235 | 933,333 |
| Capital charge (Rs) | 150 | 202.6 |
| Total (Rs/year) | 999,823 | 944,974 |

Both motors draw the same kVA, so the higher efficiency of B decides the result.

**Answer: Motor B (with condensers) is more economical, saving about Rs 54,849 per year.**

## Check yourself

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

1. Which items are added to get the annual cost of each motor in this two-part tariff comparison?
   A. Demand charge + energy charge + interest and depreciation on the extra capital
   B. Energy charge only
   C. Demand charge + energy charge, ignoring capital
   D. Capital cost of the motor + condenser cost only
   Answer: A. The answer compares demand charge, energy charge and interest and depreciation on the extra capital (motor cost difference or condenser). (Key point: Annual cost components)

2. How is the input power of Motor A found from the 35 kW output?
   A. 35 x 0.85 = 29.75 kW
   B. 35 / 0.85 = 41.18 kW
   C. 35 / 0.9 = 38.89 kW
   D. 35 x 0.9 = 31.5 kW
   Answer: B. Input = output / efficiency, so 35/0.85 = 41.18 kW. (Key point: Input power from efficiency)

3. How is the kVAR of the condenser for Motor B calculated?
   A. Q = P(tan(phi1) - tan(phi2)) with P = 38.89 kW input
   B. Q = 35 x 0.85
   C. Q = S x 0.05
   D. Q = P(cos phi1 - cos phi2)
   Answer: A. Q_C = P_in(tan phi1 - tan phi2) = 38.89(0.75 - 0.6197) = 5.07 kVAR. (Key point: Condenser kVAR)

4. Both motors give 45.75 kVA. Why is Motor B cheaper overall?
   A. Its demand charge is much lower
   B. Its higher efficiency lowers the input kW and so the energy charge
   C. Condensers cost nothing
   D. Its power factor stays at 0.8
   Answer: B. Demand charges are equal, so B's lower input (38.89 kW vs 41.18 kW) cuts the energy charge; B costs about Rs 944,974 vs Rs 999,823. (Key point: Deciding factor)

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