Chapter 18, Problem 35P

To determine

Estimate the air temperatures and corresponding speeds of sound at altitudes of $1700\text{m}$ and $11,000\text{m}$.

Answer to Problem 35P

The air temperatures and corresponding speeds of sound at altitudes of $1700\text{m}$ is $277.15\text{K}$ and $333\frac{\text{m}}{\text{s}}$, the air temperatures and corresponding speeds of sound at altitudes of $11000\text{m}$ is $222\text{K}$ and $298\frac{\text{m}}{\text{s}}$.

Explanation of Solution

Given data:

Refer to Table given in problem 18.35 in textbook,

The air temperature at altitude $1000\text{m}$ is $281.7\text{K}$.

The air temperature at altitude $2000\text{m}$ is $275.2\text{K}$.

The speed of sound at altitude $1000\text{m}$ is $336\frac{\text{m}}{\text{s}}$.

The speed of sound at altitude $2000\text{m}$ is $332\frac{\text{m}}{\text{s}}$.

The air temperature at altitude $10000\text{m}$ is $223.3\text{K}$.

The air temperature at altitude $15000\text{m}$ is $216.7\text{K}$.

The speed of sound at altitude $1000\text{m}$ is $299\frac{\text{m}}{\text{s}}$.

The speed of sound at altitude $15000\text{m}$ is $295\frac{\text{m}}{\text{s}}$.

Formula used:

Formula for the linear interpolation is,

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ (1)

Calculation:

To find the air temperature at altitude $1700\text{m}$ use the neighboring values of temperature at $1000\text{m}$ and $2000\text{m}$.

The diagrammatic representation for the given value is drawn below,

Substitute $1000\text{m}$ for $x_1$, $1700\text{m}$ for $x$, $2000\text{m}$ for $x_2$, $281.7\text{K}$ for $y_1$,$T_{1700}$ for $y$ and $275.2\text{K}$ for $y_2$ in equation (1),

$\begin{array}{l}\frac{T_{1700}-281.7\text{K}}{1700\text{m}-1000\text{m}}=\frac{275.2\text{K}-281.7\text{K}}{2000\text{m}-1000\text{m}}\\ \frac{T_{1700}-281.7\text{K}}{700\text{m}}=\frac{275.2\text{K}-281.7\text{K}}{1000\text{m}}\\ \frac{T_{1700}-281.7\text{K}}{700\text{m}}=\frac{-6.5\text{K}}{1000\text{m}}\end{array}$ (2)

Equation (2) can be reduced as follows,

$\begin{array}{l}\frac{T_{1700}-281.7\text{K}}{700\text{m}}=\frac{-6.5\text{K}}{1000\text{m}}\\ T_{1700}-281.7\text{K}=\left(\frac{-6.5\text{K}}{1000\text{m}}\right)\left(700\text{m}\right)\\ T_{1700}-281.7\text{K}=-4.55\text{K}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{T}_{1700}=-4.55\text{K+}281.7\text{K}\\ T_{1700}=277.15\text{K}\end{array}$

Therefore, the air temperature at an altitude $1700\text{m}$ is $277.15\text{K}$.

To find the speed of sound at an altitude $1700\text{m}$ use the neighboring values of speed of sound at $1000\text{m}$ and $2000\text{m}$.

The diagrammatic representation for the given value is drawn below,

Substitute $1000\text{m}$ for $x_1$, $1700\text{m}$ for $x$, $2000\text{m}$ for $x_2$, $336\frac{\text{m}}{\text{s}}$ for $y_1$,$v_{1700}$ for $y$ and $332\frac{\text{m}}{\text{s}}$ for $y_2$ in equation (1),

$\begin{array}{l}\frac{v_{1700}-336\frac{\text{m}}{\text{s}}}{1700\text{m}-1000\text{m}}=\frac{332\frac{\text{m}}{\text{s}}-336\frac{\text{m}}{\text{s}}}{2000\text{m}-1000\text{m}}\\ \frac{v_{1700}-336\frac{\text{m}}{\text{s}}}{700\text{m}}=\frac{332\frac{\text{m}}{\text{s}}-336\frac{\text{m}}{\text{s}}}{1000\text{m}}\\ \frac{v_{1700}-336\frac{\text{m}}{\text{s}}}{700\text{m}}=\frac{-4\frac{\text{m}}{\text{s}}}{1000\text{m}}\end{array}$ (3)

Equation (3) can be reduced as follows,

$\begin{array}{l}\frac{v_{1700}-336\frac{\text{m}}{\text{s}}}{700\text{m}}=\frac{-4\frac{\text{m}}{\text{s}}}{1000\text{m}}\\ v_{1700}-336\frac{\text{m}}{\text{s}}=\left(\frac{-4\frac{\text{m}}{\text{s}}}{1000\text{m}}\right)\left(700\text{m}\right)\end{array}$

Reduce the equation as,

$\begin{array}{l}{v}_{1700}-336\frac{\text{m}}{\text{s}}=-2.8\frac{\text{m}}{\text{s}}\\ v_{1700}=-2.8\frac{\text{m}}{\text{s}}+336\frac{\text{m}}{\text{s}}\\ v_{1700}=333.2\frac{\text{m}}{\text{s}}\\ v_{1700}\approx 333\frac{\text{m}}{\text{s}}\end{array}$

Therefore, the approximate value of speed of sound at an altitude $1700\text{m}$ is $333\frac{\text{m}}{\text{s}}$

To find the air temperature at altitude $11000\text{m}$ use the neighboring values of temperature at $10000\text{m}$ and $15000\text{m}$.

The diagrammatic representation for the given value is drawn below,

Substitute $10000\text{m}$ for $x_1$, $11000\text{m}$ for $x$, $15000\text{m}$ for $x_2$, $223.3\text{K}$ for $y_1$,$T_{11000}$ for $y$ and $216.7\text{K}$ for $y_2$ in equation (1)

$\begin{array}{l}\frac{T_{11000}-223.3\text{K}}{11000\text{m}-10000\text{m}}=\frac{216.7\text{K}-223.3\text{K}}{15000\text{m}-10000\text{m}}\\ \frac{T_{11000}-223.3\text{K}}{1000\text{m}}=\frac{216.7\text{K}-223.3\text{K}}{5000\text{m}}\\ \frac{T_{11000}-223.3\text{K}}{1000\text{m}}=\frac{-6.6\text{K}}{5000\text{m}}\end{array}$ (4)

Equation (4) can be reduced as follows

$\begin{array}{l}\frac{T_{11000}-223.3\text{K}}{1000\text{m}}=\frac{-6.6\text{K}}{5000\text{m}}\\ T_{11000}-223.3\text{K}=\left(\frac{-6.6\text{K}}{5000\text{m}}\right)\left(1000\text{m}\right)\\ T_{11000}-223.3\text{K}=-1.32\text{K}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{T}_{11000}=-1.32\text{K+}223.3\text{K}\\ T_{11000}=221.98\text{K}\\ \approx \text{222}\text{K}\end{array}$

Therefore, the air temperature at an altitude $11000\text{m}$ is $222\text{K}$

To find the speed of sound at an altitude $11000\text{m}$ use the neighboring values of speed of sound at $10000\text{m}$ and $15000\text{m}$.

The diagrammatic representation for the given value is drawn below,

Substitute $10000\text{m}$ for $x_1$, $11000\text{m}$ for $x$, $15000\text{m}$ for $x_2$, $299\frac{\text{m}}{\text{s}}$ for $y_1$,$v_{11000}$ for $y$ and $295\frac{\text{m}}{\text{s}}$ for $y_2$ in equation (1)

$\begin{array}{l}\frac{v_{11000}-299\frac{\text{m}}{\text{s}}}{11000\text{m}-10000\text{m}}=\frac{295\frac{\text{m}}{\text{s}}-299\frac{\text{m}}{\text{s}}}{15000\text{m}-10000\text{m}}\\ \frac{v_{11000}-299\frac{\text{m}}{\text{s}}}{1000\text{m}}=\frac{295\frac{\text{m}}{\text{s}}-299\frac{\text{m}}{\text{s}}}{5000\text{m}}\\ \frac{v_{11000}-299\frac{\text{m}}{\text{s}}}{1000\text{m}}=\frac{-4\frac{\text{m}}{\text{s}}}{5000\text{m}}\end{array}$ (5)

Equation (5) can be reduced as follows

$\begin{array}{c}\frac{v_{11000}-299\frac{\text{m}}{\text{s}}}{1000\text{m}}=\frac{-4\frac{\text{m}}{\text{s}}}{5000\text{m}}\\ v_{11000}-299\frac{\text{m}}{\text{s}}=\left(\frac{-4\frac{\text{m}}{\text{s}}}{5000\text{m}}\right)\left(1000\text{m}\right)\\ v_{11000}-299\frac{\text{m}}{\text{s}}=-0.8\frac{\text{m}}{\text{s}}\end{array}$

Reduce the equation as follows,

$\begin{array}{l}{v}_{11000}=-0.8\frac{\text{m}}{\text{s}}+299\frac{\text{m}}{\text{s}}\\ v_{11000}=298.2\frac{\text{m}}{\text{s}}\\ v_{11000}\approx 298\frac{\text{m}}{\text{s}}\end{array}$

Therefore, the approximate value of speed of sound at an altitude $11000\text{m}$ is $298\frac{\text{m}}{\text{s}}$.

Conclusion:

Thus, the air temperatures and corresponding speeds of sound at altitudes of $1700\text{m}$ is $277.15\text{K}$ and $333\frac{\text{m}}{\text{s}}$, the air temperatures and corresponding speeds of sound at altitudes of $11000\text{m}$ is $222\text{K}$ and $298\frac{\text{m}}{\text{s}}$.