# Heat storage - PowerPoint PPT Presentation

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Heat storage. Conservation of energy requires that incoming energy balances outgoing energy plus a change in storage. To relate changes in heat content with temperature, we use: Δ Q S / Δ z = C s Δ T / Δ t.

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Heat storage

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### Heat storage

• Conservation of energy requires that incoming energy balances outgoing energy plus a change in storage.

• To relate changes in heat content with temperature, we use:

• ΔQS / Δz = CsΔT/ Δt.

• where the term on the lhs denotes the heat flux density change in layer Δz, and the term on the rhs represents the heat capacity times the heating rate.

• If we use as an example Qin= 100 W m-2 and Qout= 10 W m-2, and a layer thickness Δz = 0.5 m of dry clay, we then obtain:

• ΔT/ Δt = 90 J m-2 s-1 / {(0.5 m)(1.42 × 106 J m-3 K-1)}

• then, ΔT/ Δt = 1.27 × 106 K s-1 = 0.46 K h-1

Oke (1987)

### Layers in the Lower Atmosphere

• Laminar boundary layer

### Laminar Boundary Layer

• This skin is only a few mm thick and adheres to all surfaces.

• In this layer, the motion is laminar, i.e. streamlines are continuously parallel to the surface.

• Thus adjacent layers of the fluid remain distinct and do not intermix.

• In addition, there is no convection such that transfers of heat, water, etc. are by conduction.

Oke (1987)

• As an example, take a laminar boundary layer that is 3 mm thick, a sensible heat flux QH = 100 W m-2, and an air temperature Ta = 10oC.

• Then what is the gradient in temperature between the surface and the top of the laminar boundary layer?

• Use QH = -Ka CaΔT/ Δz (= -k ΔT/ Δz) .

• 100 W m-2 = (20.5 × 10-6 m2 s-1) ×

(0.0012 × 106 J m-3 K-1) ×ΔT/0.003 m

• solving for ΔT yields:

• ΔT = {(100 W m-2)(0.003 m )}/ {(20.5

m2 s-1)(0.0012 J m-3 K-1)} = 0.3/0.0246 K = 12.2 K

• Thus very large temperature gradients exist in the laminar boundary layer.

• Equations for water vapour and momentum transfer are similar:

• E = -ρaKva∂ρv/∂z and

• τ = ρaKma∂u/∂z

• Since molecular diffusivities (“K”-values) are small, gradients are large in the laminar boundary layer.

### Roughness Layer

• The surface roughness causes complex 3D flows, including eddies and vortices, that are dependent on the details of the surface.

• Exchanges of heat, mass and momentum and related climatic characteristics are difficult to express in this zone, but generalized features can be established.

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### Turbulent Surface Layer

• The TSL is above the roughness layer where small scale turbulence dominates and vertical fluxes are approximately constant (“constant flux layer”) - about 10% of the PBL depth.

• Processes of transfer are turbulent, not molecular, in this layer.

• However, we can write a flux gradient transfer equation that is analogous to conduction, by replacing the K's with “eddy diffusivities”.

• These are not simple constants, but vary with time and space (if they were constant, turbulent would be a solved problem and weather forecasting would be nearly perfect!).

• The eddy diffusivities vary with the size of the eddies, that tend to increase with height above the surface.

• Values of K increase from about 10-5 m2

s-1 near the laminar boundary layer to as large as 102 m2 s-1 higher up in the PBL (that equates to 7 orders of magnitude!).

• Since the flux is approximately constant but that the diffusivities increase with height, the related climatic property (wind, temperature, humidity) has a curved (logarithmic) shape with a decreasing gradient away from the surface.

Oke (1987)

• In an analogy with the soil, the greatest temperature range is near the surface and decreases away from it and there is a time lag between surface and air temperatures.

• However it is less than for soil because turbulent transfers are more efficient than conduction at moving heat around.

• [see Oke, p. 51].

### Stability

• A dominant process in the lower atmosphere is convection, and a major control on the amount of convection is the vertical temperature structure (stability).

• To look at stability, consider a discrete “parcel” of air that does not exchange any heat with the air around it as it moves (“adiabatic motion”).

• If you move the parcel up it will encounter lower pressure because the mass of air above it becomes progressively less dense.

• As it encounters lower pressure it will tend to expand to make its internal pressure match that of its environment, but the expansion requires both work and energy.

• Since the only available energy is in the form of heat, the rising parcel will cool.

• In unsaturated air the parcel cools at the constant rate of 9.8 × 10-3 oC m-1 called the “Dry Adiabatic Lapse Rate” (DALR).

• On the other hand, a parcel moving downward will warm at the DALR.

• If a parcel is saturated, some water vapour will condense as it rises, thus releasing latent heat and reducing the rate of cooling.

• In this case, the parcel of air will cool at the “Saturated Adiabatic Lapse Rate” (SALR) that has an approximate value of 6.0 × 10-3 oC m-1.

• The actual temperature profile of the atmosphere (not the DALR!) is called the Environmental Lapse Rate (ELR).

• When considering stability, it is useful to use “potential temperature” (θ) instead of temperature.

• Potential temperature is the temperature that a parcel would have if it were moved adiabatically to 1000 hPa.

• This is like correcting the observed temperature to allow for Γ (DALR) and effectively rotates T curves by Γ.

• If θ is used rather than T, analysis of stability is simplified

θ = T + Γ z

Oke (1987)