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electrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [2023/02/12 06:00] – mexleadmin | electrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1 | ||
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- | A thermistor is used as a temperature sensor in a refrigeration system. The thermistor has a resistance of $10 k\Omega$ at $+25 °C$. \\ | ||
- | Its temperature coefficients are: $\alpha=0.01 {{1}\over{K}}$ and $\beta=71 \cdot 10^{-6}{{1}\over{K^2}}$ | ||
- | The temperature inside the refrigeration system can reach down to $-40 °C$. | ||
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- | 1. Calculate the resistance of the thermistor at $-40 °C$. | ||
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- | \begin{align*} | ||
- | R &= R_0 \cdot (1 + \alpha \cdot \Delta T + \beta \cdot \Delta T^2) && | \text{with | ||
- | R &= 10 k\Omega \cdot (1 + 0.01 {{1}\over{K}} \cdot (-40°C - 25°C) + 71 \cdot 10^{-6}{{1}\over{K^2}} \cdot (-40°C - 25°C)^2) \\ | ||
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- | \begin{align*} | ||
- | R &= 6.5 k\Omega \\ | ||
- | \end{align*} | ||
- | \\ | ||
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- | 2. Additionally, | ||
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- | Resistors transfer electrical energy out of the circuit and generate heat. Therefore, a resistive sensor might heat up the refrigeration system. \\ \\ | ||
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- | 3. Regarding question 2.: Given a constant sensor voltage, would a sensor with tenfold the resistance be better or worse? Give an explanation for you answer. | ||
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- | The power of the resistor $P = U \cdot I = R \cdot I^2 = {{U^2}\over{R}}$ is equivalent with the heat flow. \\ | ||
- | Therefore, with constant $U$ and increasing $R$ the power decreases. Ten times more resistance decreases the heat flow to one tenth. | ||
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- | </ | ||