Question

Air at \(320 \mathrm{K}\) is flowing in a duct at a velocity of \((a) 1\) (b) \(10,(c)\) 100, and \((d) 1000 \mathrm{m} / \mathrm{s}\). Determine the temperature that a stationary probe inserted into the duct will read for each case.

Short answer

Based on the given initial temperature and velocities in a duct, calculate the temperature that a stationary probe inside the duct will read for each case. Case (a) - velocity: 1 m/s Case (b) - velocity: 10 m/s Case (c) - velocity: 100 m/s Case (d) - velocity: 1000 m/s Initial air temperature: 320 K Specific heat of air at constant pressure: 1005 J/kg·K

Step-by-step solution

Calculate the probe temperature for case (a) - 1 m/s

To calculate the temperature that the probe will read for case (a), we will use the provided initial temperature (\(T_{initial} = 320 \mathrm{K}\)), the velocity of air flow (\(v = 1 \mathrm{m} / \mathrm{s}\)), and the specific heat of air at constant pressure (\(c_p = 1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\)): \(T_{probe} = 320 \mathrm{K} + \frac{(1 \mathrm{m} / \mathrm{s})^2}{2(1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K})} \approx 320.0005 \mathrm{K}\)

Calculate the probe temperature for case (b) - 10 m/s

To calculate the temperature that the probe will read for case (b), we will use the provided initial temperature (\(T_{initial} = 320 \mathrm{K}\)), the velocity of air flow (\(v = 10 \mathrm{m} / \mathrm{s}\)), and the specific heat of air at constant pressure (\(c_p = 1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\)): \(T_{probe} = 320 \mathrm{K} + \frac{(10 \mathrm{m} / \mathrm{s})^2}{2(1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K})} \approx 320.0498 \mathrm{K}\)

Calculate the probe temperature for case (c) - 100 m/s

To calculate the temperature that the probe will read for case (c), we will use the provided initial temperature (\(T_{initial} = 320 \mathrm{K}\)), the velocity of air flow (\(v = 100 \mathrm{m} / \mathrm{s}\)), and the specific heat of air at constant pressure (\(c_p = 1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\)): \(T_{probe} = 320 \mathrm{K} + \frac{(100 \mathrm{m} / \mathrm{s})^2}{2(1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K})} \approx 324.9751 \mathrm{K}\)

Calculate the probe temperature for case (d) - 1000 m/s

To calculate the temperature that the probe will read for case (d), we will use the provided initial temperature (\(T_{initial} = 320 \mathrm{K}\)), the velocity of air flow (\(v = 1000 \mathrm{m} / \mathrm{s}\)), and the specific heat of air at constant pressure (\(c_p = 1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\)): \(T_{probe} = 320 \mathrm{K} + \frac{(1000 \mathrm{m} / \mathrm{s})^2}{2(1005 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K})} \approx 819.9005 \mathrm{K}\) In conclusion, the temperature that the stationary probe will read in the duct for each case is approximately: 1. \(320.0005 \mathrm{K}\) for case (a) with a velocity of \(1 \mathrm{m} / \mathrm{s}\) 2. \(320.0498 \mathrm{K}\) for case (b) with a velocity of \(10 \mathrm{m} / \mathrm{s}\) 3. \(324.9751 \mathrm{K}\) for case (c) with a velocity of \(100 \mathrm{m} / \mathrm{s}\) 4. \(819.9005 \mathrm{K}\) for case (d) with a velocity of \(1000 \mathrm{m} / \mathrm{s}\)