Science 123 Spring 2020
Unit Two (Second Seven-Weeks)
Homework Help


Chapter 8: Divergence, Vorticity and Temperature Advection

1. The wind speeds are in m/sec.

Divergence : $\displaystyle{\frac{\Delta u}{\Delta x} = \frac{8-9}{200,000} = -0.000005}$. Calculate the $\displaystyle{\frac{\Delta v}{\Delta y}}$ term in a similar way. The divergence in this case is zero.

Vorticity: The term $\displaystyle{\frac{\Delta v}{\Delta x} = \frac{3 - -1}{200,000} = 0.00002}$. Calculate the value for the $\displaystyle{\frac{\Delta u}{\Delta v}}$ term in a similar way.
2. The divergence in this case is $-7 \times 10^{-5}$ sec$^{-1}$. The vorticity is $-5 \times 10^{-5}$ sec$^{-1}$.
3. The speed of a Rossby wave is given by $\displaystyle{U_0 - \beta \left(\frac{\lambda}{2\pi} \right)^2}$. The answer is 55 - 14.7 = 40.30 m sec$^{-1}$.
4. The speed of a Rossby wave is given by $\displaystyle{U_0 - \beta \left(\frac{\lambda}{2\pi} \right)^2}$. The answer is 60 - 3.7 = 56.3 m sec$^{-1}$.
7. The speed of the Rossby wave was calculated to be 40.3 m$\cdot$sec$^{-1}$ in Exercise 3. We will assume, correctly, that short wave travel at the speed of the wind, which is 55 m$\cdot$sec$^{-1}$. Therefore, the relative speed of the short wave to the long wave is 55 - 40.3 = 14.7 m$\cdot$sec$^{-1}$. The time required to catch the long wave is (400 $\times$ 1000)/14.7 = 27,137 seconds = 7.5 hours by dividing by 3600 seconds per hour.
9. The temperature gradient at station 1 is $\displaystyle{\frac{\Delta T}{\Delta x} = \frac{35 - 30}{150000}} = 3.33 \times 10^{-5}$ F/m. Be sure to measure $\Delta x$ in meters.
10. The temperature gradient is $2.0 \times 10^{-5}$ F/m.
11. The temperature gradient at station 3 is $\displaystyle{\frac{\Delta T}{\Delta x} = \frac{45 - 40}{100000}} = 5.0 \times 10^{-5}$ F/m. Be sure to measure $\Delta x$ in meters.
13. The velocity vector is $\vec{v} = (10,0)$, so the speed of the wind is $v = \sqrt{10^2 + 0^2} = 10$ m/sec. The angle $\theta$ between the temperature gradient and the velocity vector is zero degrees, so $\cos \theta = \cos 0 = 1$. The wind is blowing in a direction relative to the temperature that will result in a decrease in temperature at Station 1 (cold air advection). Consequently, the temperature advection is $\frac{\Delta T}{\Delta t} = -\cos 0 \times 10 \times 3.3\times 10^{-5} \times 3600 = -1.18$ F/hour. (The factor of 3600 is required to convert the rate to degrees per hour from degrees per second.(
14. The velocity vector is $\vec{v} = (10,10)$, so the speed of the wind is $v = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ m/sec. The angle $\theta$ between the temperature gradient and the velocity vector is 45 degrees, so $\cos \theta = \cos 45 = \sqrt{2}/2$. The wind is blowing in a direction relative to the temperature that will result in a decrease in temperature at Station 1 (cold air advection). Consequently, the temperature advection is $\frac{\Delta T}{\Delta t} = -\cos 45 \times 10\sqrt{2} \times 5.0\times 10^{-5} \times 3600 = -1.8$ F/hour. (The factor of 3600 is required to convert the rate to degrees per hour from degrees per second.)
15. The formula to use is $\displaystyle{-\cos \alpha \|\vec{v} \| \left(\frac{\Delta T}{\Delta x}\right) }$. The angle $\alpha$ is that between the temperature gradient direction (in the positive $x$ direction) and the velocity vector $\vec{v} = (-10, 10)$ direction. This angle is 135 degrees. The norm (or length) of the velocity vector is $\| \vec{v} \| = \sqrt{u^2 + v^2} = \sqrt{(-10)^2 + 10^2} = 2\sqrt{10}$. Therefore

$\displaystyle{\frac{\Delta T}{\Delta t} = -\cos 135 \cdot 10\sqrt{2} \cdot 2.5 \times 10^{-5} \cdot 60 \cdot 60 = 0.9 \mbox{ F}/\mbox{hr}}$
16. The formula to use is $\displaystyle{-\cos \alpha \|\vec{v} \| \left(\frac{\Delta T}{\Delta x}\right) }$. The angle $\alpha$ is that between the temperature gradient direction (in the positive $x$ direction) and the velocity vector $\vec{v} = (-10, 10)$ direction. This angle is 135 degrees. The norm (or length) of the velocity vector is $\| \vec{v} \| = \sqrt{u^2 + v^2} = \sqrt{(-10)^2 + 10^2} = 2\sqrt{10}$. Therefore

$\displaystyle{\frac{\Delta T}{\Delta t} = -\cos 135 \cdot 10\sqrt{2} \cdot 2.5 \times 10^{-5} \cdot 60 \cdot 60 = 0.9 \mbox{ F}/\mbox{hr}}$

Chapter 9: Moisture Measures

1. (a) P = 1000 mb, so 1% water vapor implies that $e = \frac{1}{100} \times 1000 = $10 mb. $MR = \frac{\epsilon \cdot 10}{1000 - 10} = \frac{0.622 \cdot 10}{990} = 0.00626$ gm/gm = 6.26 gm /kg.

(b) P = 1000 mb, so 4% water vapor implies that $e = \frac{4}{100} \times 1000 = $40 mb. $MR = \frac{\epsilon \cdot 40}{1000 - 40} = \frac{0.622 \cdot 40}{960} = 0.0259$ gm/gm = 25.9 gm /kg.
2. (a) 6.28 g/kg     (b) 25.9 g/kg
3. (a) Begin by solving the mixing ratio formula for $e$. $MR = \frac{\epsilon \cdot e}{P - e} \Rightarrow MR(P - e) = \epsilon \cdot e \Rightarrow MR\cdot P - MR \cdot e = \epsilon \cdot e \Rightarrow MR \cdot P = (MR + \epsilon)e \Rightarrow e = \frac{MR \cdot P}{MR + \epsilon}$

Then $ e = \frac{(10/1000) \cdot 1020}{10/1000 + 0.622} = 16.1$ mb

(b) $ e = \frac{(10/1000) \cdot 700}{10/1000 + 0.622} = 11.1$ mb
4. (a) 31.1 mb     (b) 25.6 mb
5. From the SVP table: $T$ = 15 C $\rightarrow$ $e_s$ = 17.2 mb
6. T = 13$^{\circ}$ F
7. From the SVP table: $e_s$ = 12 mb $\rightarrow$ $T$ = 10.0 C
8. T = 6.7$^{\circ}$ F
9. (a) The parcel temperature determines the SVP, and SVP(25 C) = 32.3 mb (from the SVP table). $RH = \frac{e}{e_s} \times 100 = \frac{10}{32.3} \times 100 \approx$ 31%.

(b) The parcel dew point is determined using its vapor pressure $e$. From the SVP table, $T$ = 7 C.

10. 7.02 g/kg
11. The parcel temperature determines its SMR. Reading from the SMR column for $P$ = 1000 mb, the SMR for 90 F is 32.1 gm/kg. Use the RH to determine the parcel mixing ratio: $RH = \frac{MR}{SMR} \times 100 \Rightarrow MR = \frac{RH}{100} \times SMR = 0.7 \times 32.1 = 22.5$ gm/kg
12. The relative humidity of the parcels is given by $RH = 100 \times \frac{e}{e_s}$. Saturation vapor pressure $e_s$ is found from the tables using a temperature of 60.; $e_s \approx 17.5$ mb. The actual vapor pressure $e$ of the parcel is determined using its dew point. The dew point tells us the temperature at which the parcel would be saturated with the vapor pressure it has. So, use the table to find the saturation vapor pressure for a temperature of 45; $e_s \approx 10.3$ mb. So $e$ is 10.3 mb for this parcel. Consequently RH = 100(10.3/17.5) = 59%.
13. The relative humidity of the parcels is given by $RH = 100 \times \frac{e}{e_s}$. Saturation vapor pressure $e_s$ is found from the tables using a temperature of 80 F; $e_s \approx 35.6$ mb. The actual vapor pressure $e$ of the parcel is determined using its dew point. Use the table to find the saturation vapor pressure for a temperature of 35 F; $e_s \approx 7.1$ mb. So $e$ is 7.1 mb for this parcel. Consequently RH = 100$\times$(7.1/35.6) $\approx$ 20%.
14. 55.4$^{\circ}$ F
15. The parcel temperature of 45 F determines its SVP, and from the table we get $e_s$ = 10.5 mb. Use the fact that the parcel RH is 80% to give $\frac{e}{e_s} = \frac{80}{100} = 0.8 \Rightarrow e = 0.8 \times 10.5 \approx 8.4$ mb. The vapor pressure $e$ of the parcel determines its dew point (and vice versa), so from the table we get $T_d$ = 40 F.
16. 11.4$^{\circ}$ C or 53$^{\circ}$ F
17. $T_d$ = 20 C, so $e$ = 23.7 mb (from the table). RH = 70%, so $\frac{e}{e_s} = \frac{70}{100} \Rightarrow e_s = \frac{23.7}{0.7} = 33.8$ mb. The parcel temperature is given by its SVP, so T $\approx$ 25.5 C (from the table).

Chapter 12: Stability

1. (a) at 400 mb and T = -40 C, (using the skew T log P chart) the potential temperature is approximately $\theta$ = 30$^{\circ}$ C
2. (a) A parcel with P = 1000 mb, T = 10 C and T$_d$ = 5 C (using the skew T log P chart) the equivalent potential temperature is approximately $\theta_e$ = 28$^{\circ}$ C
3. (a) The parcel at the 800 mb level (using the skew T log P chart) will have T$_d$ = -3 C (approximately) and a mixing ration W = 3.8 g/kg (approximately)

(b) The parcel at the 500 mb level (using the skew T log P chart) will have T$_d$ = -22 C (approximately) and a mixing ration W = 1.3 g/kg (approximately)