The location Boston, MA, USA. d (22-Dec) = 356 d (23-Mar) = 82 d (22-Jun) = 173

Transcription

1 ATSC Assignment 3 Answer Key Total mark out of 50 Chapter 2: A5h, A13h, A15h Chapter 10:A1h, A4h, A5h, A6d, A8h, A9h Chapter 2 A5h) Plot the local solar elevation angle vs. local time for 22 December, 23 March, and 22 June for the following city: h) Boston, MA, USA. Given: The location Boston, MA, USA. 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 2018 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 Boston, MA, USA λe = degw td = 24.0 h time zone of Boston, MA, USA: tutc = t - 5hours (EST) Ψ (deg) t(h) 22-Dec 23-Mar 22-Jun

2 Solar Elevation Angles for Boston, MA, USA 80 Solar Elevation Angle (deg) Dec 23-Mar 22-Jun Local Standard Time (h)

3 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. A13h) Find the kinematic heat fluxes at sea level, given these regular fluxes (W/m^2): h) 300. Given: FH = 300 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). A15h) Plot Planck curves for the following blackbody temperatures (K): f) Given: T = 1000 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 = c2 = 3.74E+08 W*μm^4/m^2 1.44E+04 μm*k λ (μm) Eλ* E E E E

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5 Planck Radiant Exitance for T = 1000K Eλ* (W/m^2*μm) Wavelength λ (μm) Check: Discussion: Units ok. Physics ok. The temperature of this object (1000K) is about 1/5th 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 A1h) Plot the wind symbol for winds with the following directions and speeds: h) NW at 50kt. Given: M = 50 kt direction = Northwest Find: Applicable wind symbol. From Table 10-1: Penant (triangle): 50 speed units

6 Half barb (half line): 5 speed units 50 kt = 1 pennant Check: direction and symbol ok. Discussion: If only the pressure gradient force was acting here, high pressure would be to the northwest and low pressure would be to the southeast. A4h) Find the advective "force" per unit mass given the following wind components (m/s) and horizontal distances (km): h) V = -9, ΔV = -10, Δy = -6. Given: V = -9 m/s ΔV = -10 m/s Δy = -6 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, FxAD/m = 0. Since U and W were not given, we can assume that U = 0 and W = 0. Hence: FyAD/m = -V *(ΔV/Δy)

7 FyAD/m = Check: Discussion: m/s^2 Units ok. Physics ok. The advective force is positive, therefore advection is accelerating the V wind. A5h) 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: h) kpa. Given: Δx = 500 km P@A = 100 kpa P@B = kpa Find: FxPG/m =? m/s^2 FyPG/m =? m/s^2 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. The force is very small because the pressure difference is very small

8 A6d) Suppose that U = 8m/s and V = -3 m/s, and latitude = 45 deg. Calculate centrifugal force components around a: d) 500km radius high in S. hemisphere Given: U = 8 m/s V = -3 m/s lat = 45 degs R = 500 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 high pressure center: s= 1 Convert R(km) to R(m): R = M is equal to: M = m m/s FxCN/m = FyCN/m = E-05 m/s^ m/s^2 Check: Units ok. Physics ok. Discussion: Both the x and y components of the centrifugal force are negative, which would correspond to western and southern directions respectively. These wind 'coordinates' about a high in the Southern hemisphere would point to a location away from the high pressure center. A8h) What is the magnitude and direction of Coriolis force/mass in Los Angeles, USA, given: f) U(m/s) = -5, V(m/s) = 5. Given: U = -5 m/s V = 5 m/s lat = degn of Los Angeles, USA

9 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 M = Ω = m/s 7.29E-05 s^-1 FCF/m = 5.77E-04 m/s^2 Direction is towards the northeast or 45deg. Check: Units ok. Physics ok. Discussion: The wind is southeast and so the force is 90 to it's right in the N.H so the coriolis force will be towards the northeast. A9h) 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

10 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: towards the southeast or 135deg. Check: Discussion: Units ok. Physics ok. The wind is coming from the southeast and so the force is directly opposite the wind and will be towards the southeast.