Chapter 5, Problem 2PA

(a)

To determine

The height of the LCL if the air parcel has a temperature of $25°\text{C}$ and dew point temperature of $14°\text{C}$ and the temperature of the air parcel if it rises to 4,000 meters.

Answer to Problem 2PA

The height of the Lifting Condensation Level (LCL) is $1375\text{meters,}$ and the temperature of air parcel that is at $\text{4000 meters}$ is $\text{- 1}\text{.9}°\text{C}$.

Explanation of Solution

The height of an air parcel at which the condensation begins is termed as Lifting Condensation Level (LCL). The formula to determine the LCL is as follows:

$\text{Lifting Condensation Level}=125\text{meters}\times \left(\text{Celcius temperature}-\text{Celcius dew point}\right)$

Substituting the values of Celsius temperature as $25°\text{C}$ and Celsius dew point as $14°\text{C}$,

$\begin{array}{l}\text{Lifting}\text{Condensation}\text{Level}=125\text{meters}\times \left(\text{Celcius temperature}-\text{Celcius dew point}\right)\\ =125\text{meters}\left(25-14\right)\\ =125\text{meters}\left(11\right)\\ =1375\text{meters}\end{array}$

To identify the temperature of air parcel at $4000\text{meters}$,

Given info:

The temperature of the air parcel = $25°\text{C}$

The height of the Lifting Condensation Level (LCL) $=1375\text{meters}$

To identify the distance between the starting point and the point of condensation, subtract the point of condensation with the starting point.

$\begin{array}{l}\text{Distance between the starting point }\\ \text{and the point of condensation}\end{array}\}\text{ = The point of condensation - The starting point}$

Substituting the values $1375\text{meters}$ in the point of condensation and 0 meter in the starting point in the above expression,

$\begin{array}{l}\begin{array}{l}\text{Distance between the starting point}\\ \text{and the point of condensation}\end{array}\}\text{ = 1375m - 0m}\\ \text{ = 1375m}\end{array}$

Now, divide $1375\text{meters}$ by $100\text{meters}$ and multiply by $1°\text{C}$. The reason for dividing by 100m and multiplying by $1°\text{C}$ is because the air parcel rises for every $100\text{meters}$ and the air temperature drops by $1°\text{C}$.

$\frac{1375\text{meters}}{100\text{meters}}\times \text{ 1}°\text{C = 13}\text{.75}°\text{C}$ ----------------- (1)

Now, subtract the starting temperature ($25°\text{C}$) with the above temperature ($\text{13}\text{.75}°\text{C}$), as the above temperature is not the actual temperature at the point of condensation as it is in fact the total drop in temperature from the starting point to the point of the condensation level.

$25°\text{C - 13}\text{.75}°\text{C = 11}\text{.25}°\text{C}$

Therefore, the temperature at the point of condensation is $\text{11}\text{.25}°\text{C}$.

Now, to determine the distance between the highest point ($\text{4000meters}$) and the LCL ($1375\text{meters}$),

$4000\text{meters - 1375meters = 2625meters}$

Now, divide $2625\text{meters}$ by $100\text{meters}$ and multiply by $0.5°\text{C}$. The reason for dividing by 100m and multiplying by $0.5°\text{C}$ is because the air parcel rises for every $100\text{meters}$ after the point of condensation and the wet adiabatic lapse rate of $5°\text{C for every 1000meters}$ becomes operational, and the air temperature drops by $0.5°\text{C for every100meters}$.

$\frac{2625\text{meters}}{100\text{meters}}\times \text{ 0}\text{.5}°\text{C = 13}\text{.125}°\text{C}$ -------------- (2)

Now, to identify the temperature of air parcel at 4000 meters,

$\begin{array}{l}\text{The temperature of air }\\ \text{parcel at 4000 meters}\end{array}\}=\text{ Starting temperature - }\left[\left(\text{Value at 1}\right)+\left(\text{Value at 2}\right)\right]$

Substituting the value $25°\text{C}$ in the starting temperature and (1) and (2) values in the above expression,

$\begin{array}{l}\begin{array}{l}\text{The temperature of air }\\ \text{parcel at 4000 meters}\end{array}\}=\text{25}°\text{C - }\left[\left(\text{13}\text{.75}°\text{C}\right)+\left(\text{13}\text{.125}°\text{C}\right)\right]\\ \text{ = 25}°\text{C - 26}\text{.895}°\text{C}\\ \text{ = - 1}\text{.895 }\simeq \text{ - 1}\text{.9}°\text{C}\end{array}$

Therefore, the temperature of air parcel that is at $\text{4000 meters}$ is $\text{- 1}\text{.9}°\text{C}$.

(b)

To determine

The calculation of temperature and dew point temperature of an air parcel that has descended to 2,000 meters, which was previously at 6,000 meters having a temperature of –5°C and a dew point of –10°C.

Answer to Problem 2PA

The temperature of the descending air parcel is $35°\text{C}$, and the dew point temperature is $-2°\text{C}$.

Explanation of Solution

Given info:

$\begin{array}{l}\text{The height of air parcel = 6000meters}\hfill \\ \text{The temperature of air parcel = -5}°\text{C}\hfill \\ \text{The dew point temperature = -10}°\text{C}\hfill \end{array}$

The temperature of air parcel at 2000meters = ?

The general rule is that the temperature increases by $10°\text{C for every 1000meters}$ for a descending air parcel, and the dew point temperature increases by $2°\text{C for every 1000meters}$.

Height of air parcel (in meters)	Temperature (celsius)	Dew point temperature (celsius)
$6000$	$-5$	$-10$
$5000$	$5$	$-8$
$4000$	$15$	$-6$
$3000$	$25$	$-4$
$2000$	$35$	$-2$

Therefore, the temperature of the descending air parcel is $35°\text{C}$, and the dew point temperature is $-2°\text{C}$.