Exercise EE1-02.03 – Plant lighting, power and efficiency (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-21)

Relates to EE1-02 Electrical quantities and units, terms, symbols and models

Task

A passively cooled LED lamp for lighting plants with a rated power of $P=300\mathrm{W}$ is connected to a standard household socket-outlet. Its efficiency is specified as $\eta =0.35$.

a) What is the power that the lamp emits as heat?

b) What current flows through the supply cable of the socket-outlet when two lamps are operated on a multiple socket-outlet with supply cable?

c) What is the cost of operating two of these lamps for one year with a lighting cycle of 18/6 (18 hours of light, 6 hours of darkness)? Assume an energy price of $40ct/kWh$.

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Solution

Given

Given explicitly:

• Rated power (electrical power input) $P=300\mathrm{W}$
• Efficiency $\eta =0.35$
• Lighting cycle: $18\mathrm{h}$ per day
• Energy price $k=40ct/kWh=0.40EUR/kWh$

Given implicitly from ‘standard household socket-outlet’:

• Mains voltage $U=230\mathrm{V}$

Known:

• The efficiency is defined as: $\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}$

Required

a) Thermal power $P_{\text{heat}}$ in watts ($\mathrm{W}$)
b) Current $I$ with two lamps in amperes ($\mathrm{A}$)
c) Annual cost for two lamps in euros ($\mathrm{EUR}$)

a) Thermal power

The rated power of a lamp is the electrical power input: $P=300\mathrm{W}$.

The useful output power (light) is:

$$P_{\mathrm{light}}=\eta \cdot P=0.35\cdot 300\mathrm{W}=105\mathrm{W}$$

The power loss (heat) is obtained from:

$$P_{\text{heat}}=(1-\eta )\cdot P=0.65\cdot 300\mathrm{W}=\underline{195\mathrm{W}}$$

Alternatively via the difference:

$$P_{\text{heat}}=P-P_{\mathrm{light}}=300\mathrm{W}-105\mathrm{W}=\underline{195\mathrm{W}}$$

The lamp emits a thermal power of $195\mathrm{W}$. It therefore heats the room considerably.

b) Current with two lamps

Total power of two lamps:

$$P_{\mathrm{tot}}=2\cdot P=2\cdot 300\mathrm{W}=600\mathrm{W}$$

Current from the power formula:

$$P=U\cdot I\Rightarrow I=\frac{P_{\mathrm{tot}}}{U}$$ $$I=\frac{600\mathrm{W}}{230\mathrm{V}}=2.608\mathrm{A}\approx \underline{2.61\mathrm{A}}$$

A current of $2.61\mathrm{A}$ flows through the supply cable.

c) Annual cost

Operating time per year:

$$t=18\frac{\mathrm{h}}{\mathrm{d}}\cdot 365\mathrm{d}=6570\mathrm{h}$$

Energy consumption per year:

$$W=P_{\mathrm{tot}}\cdot t=600\mathrm{W}\cdot 6570\mathrm{h}=3942000\mathrm{Wh}=3942\mathrm{kWh}$$

Cost:

$$K=W\cdot k=3942\mathrm{kWh}\cdot 0.40\frac{\mathrm{EUR}}{\mathrm{kWh}}=1576.8\mathrm{EUR}\approx \underline{1577\mathrm{EUR}}$$

Operating two lamps costs about $1577\mathrm{EUR}$ per year.

Energy efficiency of plant lamps with light emitting diodes

With an efficiency of $35\%$, this LED lamp is relatively efficient. $65\%$ of the electrical energy input ($195\mathrm{W}$ of $300\mathrm{W}$) is converted into heat. For comparison: conventional incandescent bulbs have an efficiency of only about $5\%$ and convert $95\%$ into heat. Compact fluorescent lamps have losses of about $75\%$.