Exercise EE1-04.R3 (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-12-07)

Relates to EE1-04 Resistance

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Task

A photovoltaic system on a flat roof generates a direct current of $I=32\mathrm{A}$ at full load and a voltage of $U_{\mathrm{DC}}=400\mathrm{V}$. The copper electric line from the PV generator to the inverter in the basement has a one-way length of $l=35\mathrm{m}$.

The installer plans to use a cable with a cross-section of $A=6\mathrm{m}{\mathrm{m}}^2$.

Boundary conditions:

• The resistivity of the copper electric line at $20°\mathrm{C}$ is ${\rho }_{20}=0.0178\Omega \mathrm{m}{\mathrm{m}}^2/m$
• The temperature coefficient of copper: ${\alpha }_{20}=3.93\cdot 10^{-3}{\mathrm{K}}^{-1}$
• In summer, the cable temperature on the roof can reach $T=70°\mathrm{C}$
• The maximum permissible voltage drop on the DC side is $2\%$ of the system voltage

a) Calculate the conductor resistance at $20°\mathrm{C}$ (take the outgoing and return conductors into account).

b) Calculate the conductor resistance at the maximum operating temperature of $70°\mathrm{C}$.

c) Determine the voltage drop and the power loss in the electric line at $70°\mathrm{C}$.

d) Check whether the planned cable cross-section complies with the $2\%$ limit. If not, calculate the minimum required cross-section.

e) What efficiency results for the transmission of electricity via the electric line?

Solution

a) $R_{20}=0.208\Omega$
b) $R_{70}=0.249\Omega$
c) $\Delta U=8.0\mathrm{V}$ ($2.0\%$), $P_{\mathrm{loss}}=255\mathrm{W}$
d) Limit just met; with reserve: $A_{\min }=10\mathrm{m}{\mathrm{m}}^2$
e) $\eta =98.0\%$

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Solution

Given

Given explicitly:

• Operating current: $I=32\mathrm{A}$
• DC voltage: $U_{\mathrm{DC}}=400\mathrm{V}$
• One-way cable length: $l=35\mathrm{m}$ → total length: $\ell =2l=70\mathrm{m}$
• Planned cross-section: $A=6\mathrm{m}{\mathrm{m}}^2$
• Resistivity of copper: ${\rho }_{20}=0.0178\Omega \mathrm{m}{\mathrm{m}}^2/m$
• Temperature coefficient of copper: ${\alpha }_{20}=3.93\cdot 10^{-3}{\mathrm{K}}^{-1}$
• Maximum operating temperature: $T=70°\mathrm{C}$
• Maximum permissible voltage drop: $\Delta U_{\max }=2\%$ of $U_{\mathrm{DC}}$

Known:

• Conductor resistance: $R=\rho \cdot \frac{\ell }{A}$
• Temperature dependence: $R(T)=R_{20}\cdot (1+{\alpha }_{20}\cdot \Delta T)$
• Voltage drop: $\Delta U=R\cdot I$
• Power loss: $P_{\mathrm{loss}}=R\cdot I^2=\Delta U\cdot I$
• Efficiency: $\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{tot}}}=1-\frac{P_{\mathrm{loss}}}{P_{\mathrm{tot}}}$

where:

• $R$ is the resistance in ohms ($\Omega$)
• $\rho$ is the resistivity in $\Omega \mathrm{m}{\mathrm{m}}^2/m$
• $\ell$ is the total length (outgoing and return conductors) in metres ($\mathrm{m}$)
• $A$ is the cross-sectional area in square millimetres ($\mathrm{m}{\mathrm{m}}^2$)
• ${\alpha }_{20}$ is the temperature coefficient at $20°\mathrm{C}$ in ${\mathrm{K}}^{-1}$
• $\Delta T$ is the temperature difference in kelvins ($\mathrm{K}$)
• $\Delta U$ is the voltage drop in volts ($\mathrm{V}$)
• $P_{\mathrm{loss}}$ is the power loss in watts ($\mathrm{W}$)
• $\eta$ is the efficiency (dimensionless)

Required

a) Conductor resistance $R_{20}$ at $20°\mathrm{C}$
b) Conductor resistance $R_{70}$ at $70°\mathrm{C}$
c) Voltage drop $\Delta U$ and power loss $P_{\mathrm{loss}}$ at $70°\mathrm{C}$
d) Check of the $2\%$ limit and, if necessary, minimum cross-section
e) Efficiency $\eta$ of the transmission of electricity

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a) Conductor resistance at 20 °C

The total length (outgoing and return conductors):

$$\ell =2\cdot l=2\cdot 35\mathrm{m}=70\mathrm{m}$$

The conductor resistance at $20°\mathrm{C}$:

$$R_{20}={\rho }_{20}\cdot \frac{\ell }{A}=0.0178\frac{\Omega \mathrm{m}{\mathrm{m}}^2}{\mathrm{m}}\cdot \frac{70\mathrm{m}}{6\mathrm{m}{\mathrm{m}}^2}$$ $$R_{20}=0.0178\cdot 11.67\Omega =\underline{0.208\Omega }$$

b) Conductor resistance at 70 °C

The temperature difference is:

$$\Delta T=70°\mathrm{C}-20°\mathrm{C}=50\mathrm{K}$$

The temperature-corrected resistance:

$$R_{70}=R_{20}\cdot (1+{\alpha }_{20}\cdot \Delta T)$$ $$R_{70}=0.208\Omega \cdot \left(1+3.93\cdot 10^{-3}{\mathrm{K}}^{-1}\cdot 50\mathrm{K}\right)$$ $$R_{70}=0.208\Omega \cdot (1+0.1965)=0.208\Omega \cdot 1.1965=\underline{0.249\Omega }$$

Increase in resistance due to temperature

The resistance increases by $19.7\%$ as a result of the heating. This is significant and must be taken into account when dimensioning the cable. Many installers forget the temperature correction and therefore underestimate the voltage drop during summer operation.

c) Voltage drop and power loss at 70 °C

Voltage drop (German: Spannungsfall):

$$\Delta U=R_{70}\cdot I=0.249\Omega \cdot 32\mathrm{A}=\underline{7.97\mathrm{V}}$$

Relative voltage drop:

$$\frac{\Delta U}{U_{\mathrm{DC}}}=\frac{7.97\mathrm{V}}{400\mathrm{V}}=0.0199=\underline{1.99\%}$$

Power loss (German: Verlustleistung):

$$P_{\mathrm{loss}}=R_{70}\cdot I^2=0.249\Omega \cdot (32\mathrm{A}{)}^2=0.249\Omega \cdot 1024{\mathrm{A}}^2=\underline{255\mathrm{W}}$$

Check via $P_{\mathrm{loss}}=\Delta U\cdot I$:

$$P_{\mathrm{loss}}=7.97\mathrm{V}\cdot 32\mathrm{A}=255\mathrm{W}$$

✔

d) Checking the 2 % limit

Maximum permissible voltage drop:

$$\Delta U_{\max }=2\%\cdot U_{\mathrm{DC}}=0.02\cdot 400\mathrm{V}=8.0\mathrm{V}$$

Comparison:

$$\Delta U=7.97\mathrm{V}<8.0\mathrm{V}=\Delta U_{\max }$$

At $1.99\%$, the voltage drop lies just within the $2\%$ limit. However, there is practically no reserve.

Recommendation for the minimum cross-section with reserve:

For a target value of $\Delta U_{\mathrm{target}}=1.5\%$ (common in practice):

$$\Delta U_{\mathrm{target}}=0.015\cdot 400\mathrm{V}=6.0\mathrm{V}$$

Maximum permissible resistance at $70°\mathrm{C}$:

$$R_{\max ,70}=\frac{\Delta U_{\mathrm{target}}}{I}=\frac{6.0\mathrm{V}}{32\mathrm{A}}=0.1875\Omega$$

Conversion back to $20°\mathrm{C}$:

$$R_{\max ,20}=\frac{R_{\max ,70}}{1.1965}=\frac{0.1875\Omega }{1.1965}=0.157\Omega$$

Required minimum cross-section:

$$A_{\min }={\rho }_{20}\cdot \frac{\ell }{R_{\max ,20}}=0.0178\frac{\Omega \mathrm{m}{\mathrm{m}}^2}{\mathrm{m}}\cdot \frac{70\mathrm{m}}{0.157\Omega }$$ $$A_{\min }=0.0178\cdot 445.9\mathrm{m}{\mathrm{m}}^2=7.94\mathrm{m}{\mathrm{m}}^2$$

From the standard series, you choose the next larger cross-section: $\underline{A=10\mathrm{m}{\mathrm{m}}^2}$

Practical recommendation

Although the $6\mathrm{m}{\mathrm{m}}^2$ cross-section is sufficient by calculation, $10\mathrm{m}{\mathrm{m}}^2$ is advisable in practice:

• Reserve for ageing and contact resistances
• Safety on particularly hot summer days ($>70°\mathrm{C}$)
• Reduced losses increase the annual yield of the system
• The additional costs for the thicker cable pay for themselves over the operating time

e) Efficiency of the transmission of electricity

Generated power (at the PV generator):

$$P_{\mathrm{tot}}=U_{\mathrm{DC}}\cdot I=400\mathrm{V}\cdot 32\mathrm{A}=12800\mathrm{W}$$

Usable power (at the inverter):

$$P_{\mathrm{out}}=P_{\mathrm{tot}}-P_{\mathrm{loss}}=12800\mathrm{W}-255\mathrm{W}=12545\mathrm{W}$$

Efficiency (German: Wirkungsgrad):

$$\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{tot}}}=\frac{12545\mathrm{W}}{12800\mathrm{W}}=0.980=\underline{98.0\%}$$

Alternative calculation:

$$\eta =1-\frac{P_{\mathrm{loss}}}{P_{\mathrm{tot}}}=1-\frac{255\mathrm{W}}{12800\mathrm{W}}=1-0.0199=98.0\%$$

✔

Estimating annual losses

For a typical PV system with $1000$ full-load hours per year, the energy loss in the electric line is:

$$W_{\mathrm{loss}}=P_{\mathrm{loss}}\cdot t=255\mathrm{W}\cdot 1000\mathrm{h}=255\mathrm{kWh}$$

At an assumed electricity price of $0.30EUR/kWh$, this corresponds to about $77\mathrm{EUR}$ per year. Investing in a larger cable cross-section can therefore pay off over the 20- to 25-year operating time of the system.