Chapter 3.8, Problem 131RP

To determine

The average temperature of the air in the balloon for the atmospheric temperature of $15°\text{C}$ and $30°\text{C}$.

Answer to Problem 131RP

The average temperature of the air in the balloon for the atmospheric temperature of $15°\text{C}$ is $306.5\text{K}$.

The average temperature of the air in the balloon for the atmospheric temperature of $30°\text{C}$ is found to be $323.6\text{K}$.

Explanation of Solution

Write the expression to obtain the buoyancy force acting on the hot-air balloon $\left(F_B\right)$.

$F_B={\rho }_{\text{cool air}}g{V}_{\text{balloon}}$ (I)

Here, density of the cool air is ${\rho }_{\text{cool air}}$, acceleration due to gravity is $g$, and volume of the hot-air balloon is $V_{\text{balloon}}$.

Write the expression to obtain the volume of the hot-air balloon $\left(V_{\text{balloon}}\right)$.

$V_{\text{balloon}}=\frac{4\pi r^3}{3}$ (II)

Here, radius of the hot-air balloon is $r$.

Write the expression to obtain the density of the cool air $\left({\rho }_{\text{cool air}}\right)$.

${\rho }_{\text{cool air}}=\frac{P}{RT}$ (III)

Here, atmospheric pressure is P, and atmospheric air temperature is T.

Write the expression to obtain the vertical force balance acting on the hot-air balloon.

$\begin{array}{c}{F}_B=W_{\text{hot air}}+W_{\text{cage}}+W_{\text{people}}\\ =m_{\text{hot air}}g+m_{\text{cage}}g+m_{\text{people}}g\\ =\left(m_{\text{hot air}}+m_{\text{cage}}+m_{\text{people}}\right)g\end{array}$ (IV)

Here, weight of the hot air is $W_{\text{hot air}}$, weight of the cage is $W_{\text{cage}}$, weight of the people is $W_{\text{people}}$, mass of the hot air is $m_{\text{hot air}}$, mass of the cage is $m_{\text{cage}}$, and mass of the people is $m_{\text{people}}$.

Write the expression to obtain the average temperature of the air in the balloon $\left(T\right)$ .

$T=\frac{PV}{m_{\text{hot air}}R}$ (V)

Conclusion:

Consider that the air behaves as an ideal gas.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”, obtain the gas constant of air $\left(R\right)$ as $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

Substitute $\text{10}\text{m}$ for r in Equation (II).

$\begin{array}{c}{V}_{\text{balloon}}=\frac{4\pi {\left(\text{10}\text{m}\right)}^3}{3}\\ =4,189{\text{m}}^{\text{3}}\end{array}$

Substitute $\text{90}\text{kPa}$ for P, $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for R, and $15°\text{C}$ for T in Equation (III).

${\rho }_{\text{cool air}}=\frac{\text{90}\text{kPa}}{\begin{array}{l}\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(15°\text{C}\right)\\ =\frac{\text{90}\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(273+15\text{K}\right)}\\ =\frac{\text{90}\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(288\text{K}\right)}\\ =1.089\text{kg}/{\text{m}}^{\text{3}}\end{array}}$

Substitute $1.089\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{cool air}}$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for g, and $4,189{\text{m}}^{\text{3}}$ for $V_{\text{balloon}}$ in Equation (I).

$\begin{array}{c}{F}_B=\left(1.089\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(4,189{\text{m}}^{\text{3}}\right)\\ =44,700\text{kg}\text{m}/{\text{s}}^{\text{2}}\\ =44,700\text{kg}\text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\\ =44,700\text{N}\end{array}$

Substitute $44,700\text{N}$ for $F_B$, $80\text{kg}$ for $m_{\text{cage}}$, $195\text{kg}$ for $m_{\text{people}}$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for g in Equation (IV).

$\begin{array}{c}44,700\text{N}=\left(m_{\text{hot air}}+80\text{kg}+195\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ 44,700\text{N}=\left(m_{\text{hot air}}+275\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\\ 44,700\text{N}=\left(m_{\text{hot air}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\right)+275\text{kg}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\\ 44,700\text{N}=\left(m_{\text{hot air}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\right)+2,695\text{N}\end{array}$

$\begin{array}{c}{m}_{\text{hot air}}=\frac{42,005\text{N}}{9.8\text{m}/{\text{s}}^{\text{2}}}\left(\frac{\text{1}\text{kg}\text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ =4,287\text{kg}\end{array}$

Substitute $\text{90}\text{kPa}$ for P, $4,189{\text{m}}^{\text{3}}$ for $V_{\text{balloon}}$, $4,287\text{kg}$ for $m_{\text{hot air}}$, and $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for R in Equation (V).

$\begin{array}{c}T=\frac{\left(\text{90}\text{kPa}\right)\left(4,189{\text{m}}^{\text{3}}\right)}{\left(4,287\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)}\\ =306.5\text{K}\end{array}$

Thus, the average temperature of the air in the balloon for the atmospheric temperature of $15°\text{C}$ is $306.5\text{K}$.

Similarly the average temperature of the air in the balloon for the atmospheric temperature of $30°\text{C}$ is calculated in the same manner.

Thus, the average temperature of the air in the balloon for the atmospheric temperature of $30°\text{C}$ is found to be $323.6\text{K}$.