Chapter 15: Problem 20
The temperature \(\theta_{2}\) of a winding which is being heated electrically at time \(t\) is given by: \(\theta_{2}=\theta_{1}\left(1-\mathrm{e}^{-t / \tau}\right)\) where \(\theta_{1}\) is the temperature (in degrees Celsius) at time \(t=0\) and \(\tau\) is a constant. Calculate. (a) \(\theta_{1}\), correct to the nearest degree, when \(\theta_{2}\) is \(50^{\circ} \mathrm{C}, t\) is \(30 \mathrm{~s}\) and \(\tau\) is \(60 \mathrm{~s}\) (b) the time \(t\), correct to 1 decimal place, for \(\theta_{2}\) to be half the value of \(\theta_{1}\)
Short Answer
Step by step solution
Substitute Given Values and Rearrange Equation
Solve for \(\theta_{1}\)
Set Up Equation for Part (b) and Simplify
Solve for \(t\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Temperature Calculation
- \( \theta_{1} \) is the initial temperature at time \( t = 0 \).
- \( \theta_{2} \) is the temperature at a later time \( t \).
- \( \tau \) is a constant representing the system's response time.
Exponential Function
- The base \( e \) is a mathematical constant approximately equal to 2.718.
- The exponent \( -t/\tau \) indicates a rapid change at the start which gradually slows down.
Natural Logarithm
- When solving for time \( t \), the logarithm allows us to "undo" the exponent, turning complex exponential equations into simpler linear forms.
- For example, to solve \( e^{-t/\tau} = 0.5 \), you compute the natural log of both sides: \( \ln(0.5) \).
- This makes solving equations exponentially simpler.
Time Constant
- In many physical systems, \( \tau \) relates to the time it takes for the system to stabilize or reach a significant fraction of its final value.
- A small time constant means the system responds quickly, while a larger one indicates a slower response.
Equation Rearrangement
In the given exercise, equation rearrangement was key:
- For part (a), after substituting known values, it's necessary to rearrange the equation to solve for \( \theta_{1} \).
- In part (b), the equation is rearranged to solve for \( t \) by setting \( \theta_{2} \) as half of \( \theta_{1} \).