Plot the local solar elevation angle vs. local time for 22 December, 23 March, and 22 June for the following city: g) Montreal, Canada.

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1 ATSC Assignment 3 Answer Key Chapter 2: A5g, A13g, A15g Chapter 10:A1g, A4g, A5g, A6c, A8g, A9g A5g) Plot the local solar elevation angle vs. local time for 22 December, 23 March, and 22 June for the following city: g) Montreal, Canada. Given: The location Montreal, Canada. d (22-Dec) = 356 d (23-Mar) = 82 d (22-Jun) = 173 Find: Ψ (deg) =? Use eq. 2.5: δs = Φr * cos(c*(d - dr)/dy) where: Φr = rad C = rad dr = 172 for 2017 dy = Dec 23-Mar 22-Jun δs (rads) δs (deg) Use eq. 2.6: sin(ψ) = sin(φ) * sin(δs) - cos(φ)*cos(δs)*cos((c*tutc/td)-λe) where: φ = degn for Montreal, Canada λe = degw td = 24.0 h time zone of Montreal, Canada: tutc = t - 5hours Ψ (deg) t(h) 22-Dec 23-Mar 22-Jun

2 Solar Elevation Angle (deg) Solar Elevation Angles for Montreal, Canada Dec 23-Mar 22-Jun Local Standard Time (h) Check: Discussion: Units ok. Physics ok. The above plot shows that the solar elevation angle is much lower in the Northern hemisphere's winter than its summer, as is expected.

3 A13g) Find the kinematic heat fluxes at sea level, given these regular fluxes (W/m^2): g) 400. Given: FH = 400 W/m^2 Find: FH =? K*m/s Use eq. 2.11: FH = FH / rho*cp where rho *Cp ((W/m^2)/(K*m/s))= 1231 for dry air at sea level. FH = K*m/s Check: Discussion: Units ok. Physics ok. This amount of heat flux is less than half as much as we get from the sun (1366 W/m^2). A15g) Plot Planck curves for the following blackbody temperatures (K): g) Given: T = 1500 K Find: Planck curve of blackbody object with temp T. Use eq. 2.13: Eλ* = c1/(λ^5 *(e^(c2/λ*t)-1)) where c1 = 3.74E+08 W*μm^4/m^2 c2 = 1.44E+04 μm*k λ (μm) Eλ* E E

4

5 Eλ* (W/m^2*μm) Planck Radiant Exitance for T = 1500K Wavelength λ (μm) Check: Discussion: Units ok. Physics ok. The temperature of this object (1500K) is just under half of the temperature of the sun, therefore, the emmittance on the y-axis of this plot is significantly less than that of the sun. But the wavelengths are very similar. Chapter 10 A1g) Plot the wind symbol for winds with the following directions and speeds: g) W at 27kt. Given: M = 27 kt direction = West Find: Applicable wind symbol. From Table 10-1: Full barb (full line): 10 speed units Half barb (half line): 5 speed units 27 kt = 2 full barb + 1 half barb Check: direction and symbol ok. Discussion: If only the pressure gradient force was acting here, high pressure would be to the west and low pressure would be to the east.

6 A4g) Find the advective "force" per unit mass given the following wind components (m/s) and horizontal distances (km): g) V = 7, ΔU = -2, Δy = -50. Given: V = 7 m/s ΔV = -2 m/s Δy = -50 km Find: FxAD/m =? m/s^2 FyAD/m =? m/s^2 Use eq. 10.8a: FxAD/m = -U*(ΔU/Δx) -V*(ΔU/Δy) -W*(ΔU/Δz) Use eq. 10.8b: FyAD/m = -U*(ΔV/Δx)-V*(ΔV/Δy) - W*(ΔV/Δz) Convert Δy(km) to Δy(m): Δy = m Since ΔU is not given, we can assume ΔU= 0. Therefore, FyAD/m = 0. Since U and W were not given, we can assume that U = 0 and W = 0. Hence: FyAD/m = -V *(ΔV/Δy) FyAD/m = m/s^2 Check: Discussion: Units ok. Physics ok. The advective force is negative, therefore advection is slowing down the V wind. A5g) Town A is 500km west of town B. The pressure at town A is given below, and the pressure at town B is 100.1kPa. Calculate the pressure-gradient force/mass in between these two towns: g) 99.8 kpa. Given: Δx = 500 km P@A = 99.8 kpa P@B = Kpa Find: FxPG/m =? m/s^2 FyPG/m =? m/s^2

7 Use eq. 10.9a: FxPG/m = -(1/ρ)*(ΔP/Δx) Use eq. 10.9b: FyPG/m = -(1/ρ)*(ΔP/Δy) where ρ = where ΔP = P@B - P@A 1.1 kg/m^3 Convert Δx(km) to Δx(m): Δx = m Convert P@A(kPa) to P@A(Pa) and P@B(kPa) to P@B(Pa): P@A = Pa P@B = Pa Since town A is 500km to the west of town B, there is no pressure change in the North - South direction. Δy = 0, FyPG/m = 0. FxPG/m = m/s^2 Check: Units ok. Physics ok. Discussion: The pressure gradient force is acting only in the x-direction, and is negative because low pressure is west at town A and high pressure is east at town B. A6c) Suppose that U = 8m/s and V = -3 m/s, and latitude = 45 deg. Calculate centrifugal force components around a: c) 400km radius low in S. hemisphere Given: U = 8 m/s V = -3 m/s lat = 45 degn R = 400 km Find: FxCN/m =? m/s^2 FyCN/m =? m/s^2 Use eq a: FxCN/m = +s *(V*M)/R Use eq b: FyCN/m = -s *(U*M)/R where M = (U^2 + V^2)^1/2 and s = -1 in the Southern hemisphere for low pressure center: s= -1

8 Convert R(km) to R(m): R = M is equal to: M = m m/s FxCN/m = FyCN/m = 6.408E-05 m/s^ m/s^2 Check: Units ok. Physics ok. Discussion: Both the x and y components of the centrifugal force are positive, which would correspond to eastern and northern directions respectively. A8g) What is the magnitude and direction of Coriolis force/mass in Los Angeles, USA, given: g) U(m/s) = -5, V(m/s) = -5. Given: U = -5 m/s V = -5 m/s lat = degn of Los Angeles, USA Find: FCF/m =? m/s^2 Direction of FCF/m. Use eq a: FCF/m = 2*Ω* sin(φ)*m or use eq b: FCF/m = fc*m and eq : fc = 2*Ω*sin(φ) where M = (U^2 + V^2)^1/2 and FCF/m = M = Ω = 5.77E-04 m/s^ m/s 7.29E-05 s^-1 Direction is towards the northwest. Check: Units ok. Physics ok.

9 Discussion: Nonzero wind in the horizontal means the coriolis force will act only in the vertical due to it acting 90 deg to the right of the wind in the N.H. A9f) Same wind components as exercise A8, but find the magnitude and direction of turbulent drag force/mass in a statically neutral atmospheric boundary layer over an extensive forested region. Given: U = -5 m/s V = -5 m/s lat = degn Find: FTD/m =? m/s^2 Direction of FTD/m. Use eq : FTD/m = wt *(M/zi) where M = (U^2 + V^2)^1/2 and wt = CD*M for statically neutral conditions Hence: CD = zi = 0.02 over forests. 1.5 km Convert zi(km) to zi(m): zi = 1500 m (values of CD and zi given in Sample Application pg. 300; or could be estimated given the latitude of LA) M = wt = m/s 0.14 m/s FTD/m = m/s^2 Direction: to the northeast. Check: Units ok. Physics ok. Discussion: The drag force will act exactly opposite of the x wind component, therefore acting towards the northeast.

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