Chapter 20, Problem 52PQ

(a)

To determine

The high relative humidity and the low relative humidity of the places.

Answer to Problem 52PQ

The high relative humidity of the place is $\underset{\_}{108\%}$, and the low relative humidity is $\underset{\_}{60\%}$.

Explanation of Solution

Given that the high temperature is $21°\text{F}$, low temperature is $3°\text{F}$, and the dew point is $5°\text{F}$.

Relative humidity is the ratio of the partial pressure of water to the saturated vapor pressure of water. The value of partial pressure of water is not given directly, but the value can be derived using the information given in the question. The dew point of the place is given, which is the temperature at which the partial pressure of water equals to the saturated vapor pressure.

Write the expression to convert temperature from Fahrenheit scale to Celsius scale.

$T\left(\text{°C}\right)=\frac{5}{9}\left(T\left(\text{°F}\right)-32\right)$ (I)

Here, $T\left(\text{°C}\right)$ is the temperature in Celsius scale, and $T\left(\text{°F}\right)$ is the temperature in Fahrenheit scale.

Refer table 20.5 which gives the saturated vapor pressure values for various temperatures.

From the table look for the values for the high temperature and the low temperature. But the values of vapor pressure for both the temperatures is not available. So use the values for temperatures $-25°\text{C}$ and $0°\text{C}$, interpolate between these points to get the corresponding value for the given high temperature and low temperature of the place.

Write the expression for the rate of change of saturated vapor pressure with respect to the change in temperature between $0°\text{C}$ and $-25°\text{C}$.

$\frac{\Delta P_{\text{svp}}}{\Delta T}=\frac{P_{\text{svp0}}-P_{\text{svp25}}}{T_0-T_{25}}$ (IV)

Here, $\frac{\Delta P_{\text{svp}}}{\Delta T}$ is the rate of change of saturated vapor pressure with respect to temperature, $P_{\text{svp0}}$ is the saturated vapor pressure at temperature $T_0$, $P_{\text{svp25}}$ is the saturated vapor pressure at temperature $T_{25}$, $T_0$ and $T_{25}$ are the upper and lower limits of temperatures considered for interpolation.

Now the rate of change of vapor pressure with respect to the change of temperature is found. Now using this value, the vapor pressure of water at $5°\text{F}$ can be found using interpolation formula.

Write the interpolation expression for vapor pressure at the dew point.

$P_D=P_{\text{spv0}}+\left(T_D-T_0\right)\left(\frac{P_{\text{spv25}}-P_{\text{spv0}}}{T_{25}-T_0}\right)$ (V)

Here, $P_D$ is the vapor pressure in Celsius scale at the dew point, and $T_D$ is the dew temperature.

The vapor pressure at the dew point is equal to the vapor pressure of water.

$P_{\text{water}}=P_D$ (VI)

Use expression (V) in (VI) for $P_{\text{water}}$.

$P_{\text{water}}=P_{\text{spv0}}+\left(T_D-T_0\right)\left(\frac{P_{\text{spv25}}-P_{\text{spv0}}}{T_{25}-T_0}\right)$ (VII)

Similarly write expression for finding vapor pressure for high temperature using the values of $0°\text{C}$ and $25°\text{C}$.

$P_H=P_{\text{spv0}}+\left(T_H-T_0\right)\left(\frac{P_{\text{spv25}}-P_{\text{spv0}}}{T_{25}-T_0}\right)$ (VIII)

Here, $P_H$ is the vapor pressure corresponding to the high temperature, and $T_H$ is the high temperature in Celsius scale.

Similarly write expression for finding vapor pressure for low temperature using the values of $0°\text{C}$ and $25°\text{C}$.

$P_L=P_{\text{spv0}}+\left(T_L-T_0\right)\left(\frac{P_{\text{spv25}}-P_{\text{spv0}}}{T_{25}-T_0}\right)$ (IX)

Here, $P_L$ is the vapor pressure for low temperature, and $T_L$ is the low temperature in Celsius scale.

Write the expression to find the relative humidity at high temperature.

$R{H}_H=\frac{P_{\text{water}}}{P_H}\times 100\%$ (X)

Here, $R{H}_H$ is the relative humidity at high temperature.

Write the expression to find the relative humidity at low temperature.

$R{H}_L=\frac{P_{\text{water}}}{P_L}\times 100\%$ (XI)

Here, $R{H}_L$ is the relative humidity at low temperature.

Conclusion:

Substitute $21°\text{F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$ corresponding to the high temperature.

  $\begin{array}{c}{T}_H\left(\text{°C}\right)=\frac{5}{9}\left(21-32\right)\\ =-6°\text{C}\end{array}$

Substitute $3°\text{F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$ corresponding to the low temperature.

  $\begin{array}{c}{T}_L\left(\text{°C}\right)=\frac{5}{9}\left(3-32\right)\\ =-16°\text{C}\end{array}$

Substitute $5°\text{F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$ corresponding to the dew point.

  $\begin{array}{c}{T}_D\left(\text{°C}\right)=\frac{5}{9}\left(5-32\right)\\ =-15°\text{C}\end{array}$

From the table 20.5, vapor pressure at $0°\text{C}$ is $610.3\text{Pa}$, and at $-25°\text{C}$ is $70.11\text{Pa}$.

Substitute $610.3\text{Pa}$ for $P_{\text{spv0}}$, $70.11\text{Pa}$ for $P_{\text{spv25}}$, $0°\text{C}$ for $T_0$, and $-25°\text{C}$ for $T_{25}$ in equation (IV) to find $\frac{\Delta P_{\text{svp}}}{\Delta T}$.

  $\begin{array}{c}\frac{\Delta P_{\text{svp}}}{\Delta T}=\frac{\left(610.3\text{Pa}-70.11\text{Pa}\right)}{0°\text{C}-\left(-25°\text{C}\right)}\\ =21.6\text{Pa}/\text{°C}\end{array}$

Substitute $610.3\text{Pa}$ for $P_{\text{spv0}}$, $70.11\text{Pa}$ for $P_{\text{spv25}}$, $0°\text{C}$ for $T_0$, $-15°\text{C}$ for $T_D$ in equation (VII) to find $P_{\text{water}}$.

  $\begin{array}{c}{P}_{\text{water}}=610.3\text{Pa}+\left(-15°\text{C}-0°\text{C}\right)\left(\frac{70.11\text{Pa}-610.3\text{Pa}}{-25°\text{C}-\left(-0°\text{C}\right)}\right)\\ =286.3\text{Pa}\end{array}$

Substitute $610.3\text{Pa}$ for $P_{\text{spv0}}$, $70.11\text{Pa}$ for $P_{\text{spv25}}$, $0°\text{C}$ for $T_0$, $-6°\text{C}$ for $T_H$ in equation (VIII) to find $P_H$.

  $\begin{array}{c}{P}_H=610.3\text{Pa}+\left(-6°\text{C}-0°\text{C}\right)\left(\frac{70.11\text{Pa}-610.3\text{Pa}}{-25°\text{C}--0°\text{C}}\right)\\ =480.7\text{Pa}\end{array}$

Substitute $610.3\text{Pa}$ for $P_{\text{spv0}}$, $70.11\text{Pa}$ for $P_{\text{spv25}}$, $0°\text{C}$ for $T_0$, $-16°\text{C}$ for $T_L$ in equation (IX) to find $P_L$.

  $\begin{array}{c}{P}_L=610.3\text{Pa}+\left(-16°\text{C}-0°\text{C}\right)\left(\frac{70.11\text{Pa}-610.3\text{Pa}}{-25°\text{C}-\left(-0°\text{C}\right)}\right)\\ =264.7\text{Pa}\end{array}$

Substitute $286.3\text{Pa}$ for $P_{\text{water}}$, and $480.7\text{Pa}$ for $P_H$ in equation (X) to find $R{H}_H$.

  $\begin{array}{c}R{H}_H=\frac{286.3\text{Pa}}{480.7\text{Pa}}\times 100\%\\ =60\%\end{array}$

Substitute $286.3\text{Pa}$ for $P_{\text{water}}$, and $264.7\text{Pa}$ for $P_L$ in equation (XI) to find $R{H}_L$.

  $\begin{array}{c}R{H}_L=\frac{286.3\text{Pa}}{264.7\text{Pa}}\times 100\%\\ =108\%\end{array}$

Therefore, the high relative humidity of the place is $\underset{\_}{108\%}$, and the low relative humidity of the place is $\underset{\_}{60\%}$.

(b)

To determine

Whether dews form or not in the places.

Answer to Problem 52PQ

Yes, dews are formed in the atmosphere of the places.

Explanation of Solution

Dews are formed by the condensation of water. Dew point of Minneapolis and Minnesota is $5°\text{F}$ which is equal to $-15°\text{C}$. Dews are formed at a temperature of $-16°\text{C}$. This is below the dew point of these places. The relativity humidity at this temperature is more than hundred percentage. This indicates that the amount of water vapor in the air is very high and they can be condensed to form dews at this place.

Conclusion:

Therefore, dews can be formed at Minneapolis and Minnesota.