Chapter 3, Problem 3.1P

A space is at a temperature of 75 F (24 C), and the relative humidity is 45 percent. Find (a) the partial pressures of the air and water vapor, (b) the vapor density, and (c) the humidity ratio of the mixture. Assume standard sea-level pressure.

To determine

The partial pressure of the air.

The partial pressure of water vapor.

Answer to Problem 3.1P

The value of partial pressure of air is $99.57\text{kPa}$ .

The value of partial pressure of vapor is $1.43\text{kPa}$ .

Explanation of Solution

Given:

The temperature of the space is $T=75\text{F}\left(24°\text{C}\right)$ .

The relative humidity is ${\varphi }_r=45\%$ .

Formula used:

The expression for the partial pressure of air is given as,

  $P_a=P_{atm}-P_v$

Here, $P_{atm}$ is the atmospheric pressure and its value is $101\text{kPa}$ and $P_v$ is the vapor pressure.

The expression for the partial pressure of vapor is given as,

  $P_v={\varphi }_r{P}_s$

Here, $P_s$ is the saturation temperature at $24°\text{C}$ .

Calculation:

The partial pressure of vapor can be calculated as,

  $\begin{array}{l}{P}_v={\varphi }_r{P}_s\\ P_v=0.45\times \left(3.17\text{kPa}\right)\\ P_v=1.43\text{kPa}\end{array}$

The partial pressure of air can be calculated as,

  $\begin{array}{l}{P}_a=P_{atm}-P_v\\ P_a=\left(101-1.43\right)\text{kPa}\\ P_a=99.57\text{kPa}\end{array}$

Conclusion:

Therefore, the value of partial pressure of air is $99.57\text{kPa}$ .

Therefore, the value of partial pressure of vapor is $1.43\text{kPa}$ .

To determine

The vapor density.

Answer to Problem 3.1P

The value of vapor density is $0.0104\text{kg}/{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

The temperature of the space is $T=75\text{F}\left(24°\text{C}\right)$ .

The relative humidity is ${\varphi }_r=45\%$ .

Formula used:

The expression for the vapor density is given as,

  ${\rho }_v=\frac{P_v}{R_v\times T}$

Here, $R_v$ is the gas constant of water vapor and its value is $462.5\text{J}/\text{kg}\cdot \text{K}$ .

Calculation:

The vapor density can be calculated as,

  $\begin{array}{l}{\rho }_v=\frac{P_v}{R_v\times T}\\ {\rho }_v=\frac{1430\text{Pa}}{462.5\text{J}/\text{kg}\cdot \text{K}\times \left(297\text{K}\right)}\\ {\rho }_v=0.0104\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the value of vapor density is $0.0104\text{kg}/{\text{m}}^{\text{3}}$ .

To determine

The humidity ratio of the mixture.

Answer to Problem 3.1P

The value of humidity ratio is $0.00893\text{kgv}/\text{kga}$ .

Explanation of Solution

Given:

The temperature of the space is $T=75\text{F}\left(24°\text{C}\right)$ .

The relative humidity is ${\varphi }_r=45\%$ .

Formula used:

The expression for the humidity ratio is given as,

  $W=\frac{0.6219\times P_v}{\left(P_a+P_v\right)-P_v}$

Calculation:

The humidity ratio can be calculated as,

  $\begin{array}{l}W=\frac{0.6219\times P_v}{\left(P_a+P_v\right)-P_v}\\ W=\frac{0.6219\times 1.43\text{kPa}}{\left(99.57+1.43\right)\text{kPa}-1.43\text{kPa}}\\ W=0.00893\text{kgv}/\text{kga}\end{array}$

Conclusion:

Therefore, the value of humidity ratio is $0.00893\text{kgv}/\text{kga}$ .