Comment: Solve all questions with full details of how you obtained the final result.
In answering the questions take the density of air to be 1.2 kg m-3, the specific heat of air at constant pressure cp = 1005 J kg-1 K-1, the latent heat of vapourization λ = 2.5 × 106 J kg-1
Part A : The Reynolds Averaged Temperature Equation
Derive the Reynolds averaged equation for the potential temperature of the boundary layer over a horizontally homogeneous surface. Take the equation for potential temperature to be,
∂θ/∂t + ∂Uθ/∂x + ∂Vθ/∂y + ∂Wθ/∂z = 0 (1)
where θ is potential temperature and (U, V, W ) is the velocity vector. (Note: this is known as the flux form of the thermodynamic equation).
State what properties of the Reynolds average you use in the derivation and any assumptions you make.
Part B : Heating of the Boundary Layer in the Afternoon
|
Time
|
Heat Flux Wm-2
|
Temperature oC
|
∂θ/∂t
oC hr-1
|
θscrn
oC
|
|
12:00
|
150.0
|
25.0
|
-
|
+
|
|
13:00
|
147.0
|
-
|
+
|
-
|
|
14:00
|
135.0
|
+
|
-
|
+
|
|
15:00
|
120.0
|
-
|
+
|
-
|
|
16:00
|
93.0
|
+
|
-
|
+
|
|
17:00
|
65.0
|
-
|
+
|
-
|
|
18:00
|
33.0
|
+
|
-
|
+
|
|
19:00
|
0.0
|
27.4
|
-
|
27.4
|
In the table the positions with a plus sign are where you should put the results of your calculations. The positions with minus signs should remain blank.
What is the heating rate of the mixed layer in terms of the surface heat flux and boundary layer depth (assume that the potential temperature of the mixed layer below the capping inversion, is constant with height). Give the assumptions that you use to obtain your answer.
Use the values in the table to calculate the evolution of the mixed layer temperature during the afternoon. You can assume that the depth of the boundary layer is 1000m and the ratio of the entrainment heat flux and the surface flux is -0.2 (note: you should not give details of the numerical calculations).
Part C : Forecasting the Screen Level Temperature.
Near the surface there are strong gradients of potential temperature which were neglected in Part B. The difference in temperature between heights z1 and z2 is,
θ(z2) - θ(z1) = θ∗/k(ln(z2/z1) - ψh(z2/L) + ψh(z1/L)) (2)
where θ∗ = -ω'θ'0/u∗, u∗ is the friction velocity, κ is the von Karman constant = 0.4 and L is the Obukhov length. The function ψh is
ψh = 2ln[(1+x2)/2]
X = (1.0 - 16 z/L)1/4
Assume for the calculations that the top of the surface layer is at 0.2h, where h is the depth of the boundary layer, screen level is 1.25m and u∗ = 0.4ms-1.
What difference is there between the evolution of the screen level temperature and the boundary layer temperature. Give a qualitative description of why this difference occurs.
Part D: The Depth of the Boundary Layer
Changes in the depth of the boundary layer were ignored in the calculations of the temperature evolution in Part B. Calculate the rate at which the boundary layer deepens at 14:00 assuming that the temperature immediately above the boundary layer is 29.0oC, stating the equation that you are using.
Give your answer in metres per hour. Would taking account of the variation in boundary layer depth have a significant effect on your calculations of the evolution of the temperature of the boundary layer (mixed layer or screen level)?
|
Time
|
Heat Flux
Wm-2
|
Temperature oC
|
∂θ/∂t
oC hr-1
|
Screen Temp
oC
|
|
12:00
|
150.0
|
25.0
|
|
|
|
13:00
|
147.0
|
|
|
|
|
14:00
|
135.0
|
|
|
|
|
15:00
|
120.0
|
|
|
|
|
16:00
|
93.0
|
|
|
|
|
17:00
|
65.0
|
|
|
|
|
18:00
|
33.0
|
|
|
|
|
19:00
|
0.0
|
27.4
|
|
27.4
|