Chapter 12, Problem 5P

To determine

(a)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

Answer to Problem 5P

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

Explanation of Solution

Given information:

The temperature of air is $320\text{K}$ and the velocity is $1\text{m}/\text{s}$.

Expression for the stagnation temperature

   $T_0=T+\frac{V^2}{2c_p}$     ...... (I)

Here, the stagnation temperature of air is $T_0$, static temperature is $T$, air velocity is $V$ and the specific heat at constant pressure is $c_p$.

Calculation:

The specific heat for air is $1.005\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $320\text{K}$ for $T_0$, $1\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(1\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(1\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{1{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.999\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

(b)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

Given information:

The temperature of air is $320\text{K}$ and the velocity is $10\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $10\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{10{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.999\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

(c)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.950\text{K}$.

Given information:

The temperature of air is $320\text{K}$ and the velocity is $100\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $100\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(100\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(100\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{100{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.950\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.950\text{K}$.

(d)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.5025\text{K}$.

Given information:

The temperature of air is $320\text{K}$ and the velocity is $1000\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $1000\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(1000\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(1000\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{1000{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.5025\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.5025\text{K}$.

To determine

(b)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

Answer to Problem 5P

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

Explanation of Solution

Given information:

The temperature of air is $320\text{K}$ and the velocity is $10\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $10\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(10\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{10{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.999\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.999\text{K}$.

To determine

(c)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

Answer to Problem 5P

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.950\text{K}$.

Explanation of Solution

Given information:

The temperature of air is $320\text{K}$ and the velocity is $100\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $100\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(100\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(100\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{100{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.950\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.950\text{K}$.

To determine

(d)

The temperature that a stationary probe inserted into the duct will read for the given velocity.

Answer to Problem 5P

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.5025\text{K}$.

Explanation of Solution

Given information:

The temperature of air is $320\text{K}$ and the velocity is $1000\text{m}/\text{s}$.

Calculation:

Substitute $320\text{K}$ for $T_0$, $1000\text{m}/\text{s}$ for $V$ and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in the Equation (I).

   $\begin{array}{c}T=320\text{K}-\frac{{\left(1000\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{{\left(1000\text{m}/\text{s}\right)}^2}{2\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =320\text{K}-\frac{1000{\left(\text{m}/\text{s}\right)}^2}{2.01\text{kJ}/\text{kg}\cdot \text{K}\times \frac{1000\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{kJ}}}\\ =319.5025\text{K}\end{array}$

Conclusion:

The temperature that a stationary probe inserted into the duct will read for the given velocity is $319.5025\text{K}$.