Heat storage

1 / 20

Heat storage - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Heat storage

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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
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.
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.
• 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.

Why? free pizza, meet colleagues & faculty, hear about program changes, etc…

Where? Bentley Garden

When? Thursday,

September 25th

from 11:00 – 12:30

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.
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