1 / 21

Energy Transfer

Energy Transfer. Energy is “the ability to do work”. Heat and the Earth’s Atmosphere: Radiation. Nearly all the Earth’s energy comes from the Sun Sun: nuclear reactor Earth receives less than one billionth of the Sun’s energy

lacey-bridges
Download Presentation

Energy Transfer

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Energy Transfer Energy is “the ability to do work”

  2. Heat and the Earth’s Atmosphere: Radiation • Nearly all the Earth’s energy comes from the Sun • Sun: nuclear reactor • Earth receives less than one billionth of the Sun’s energy • If the Earth could be viewed from the Sun, it would appear as a speck in the solar system, with the equivalent diameter of a 5p coin observed from a distance of 265m

  3. Thermal conduction • All materials, and in particular solids, transfer heat by contact of individual atoms or molecules. • Faster molecules hit slower ones and produce an exchange of energy in which the speed is averaged out • For example, if a bar of iron is hotter on the left side, the atoms on the left side are vibrating faster than those on the right side. If the bar is isolated, the atoms of the left side will transfer heat to the atoms on the right side and, after a while, both sides will have the same temperature

  4. The basic law of thermal conduction is straightforward: the rate of heat conduction for a given area is proportional to the difference between two sides of a material. • The law of thermal conduction can be expressed as: 1/A*[dQ/dt]=-kdT/dx Where k is the coefficient of thermal conductivity

  5. Values of thermal conductivity for various substances are: Silver  418.6 (w*/(m*deg)) Clay  2.9 Quartz  8.8 Air  0.025 Water  0.545 Ice  7 to 21 (depends on bubbles) Snow  0.8 (for fresh snow)

  6. Let us make a practical example: What is the heat flux from a silver cup of boiling coffee (100C) to your skin (37C) if the cup is 1mm thick ? 1/A*[dQ/dt]=418.6*[100-37]/.001=2.6367*10^7 w/m^2 Which is 20000 times larger than the solar constant

  7. Why do we use Styrofoam ? Because it is full of air and air does not conduct heat well and Styrofoam is full of air and, therefore, it does not conduct heat well. The same for new snow: new snow is made for 90 % by air and therefore it can insulate plants and animals from very low temperatures Heat conduction is, therefore, seldom mentioned when dealing with heat moving through atmosphere .Conduction of heat by air is only important in the lowest millimeter or so of the atmosphere

  8. Turbulent transfer – Convection • Heat conduction is important for ground and ice, for example, but it is insignificant in the atmosphere. • More heat is transferred by motion due to turbulent convection. As the air is stirred, eddies or blobs of air swirl downward while others swirl upward to replace them. • If the two blobs have different amount of heat, then heat will be exchanged as well.

  9. The transport rate due to turbulent convection depends on the size of the gradient and on the intensity of the turbulence. • Turbulence is more intense for stronger winds and also when the lapse rate is unstable • Near the ground the wind speed drops to zero as a result of friction ,reducing the effectiveness of turbulence. Therefore, the gradient must be enormous to transport quantity, to compensate the lack of turbulence. • Thus, the largest gradient of all quantities occur near the ground

  10. The Turbulent Heat Transfer Equation is dQ/(A*dt) = C*r*c*V*[T(s)-T(a)] Where V is the wind speed, T(s) is surface temperature, T(a) is the temperature at the shelter, or a temperature near the ground, C is the turbulent exchange coefficient and must be determined by experiment.

  11. Typical daytime values for C are : C = 10^-3  over the ocean C = 3*10^-3  over rough land and C = 10 ^ -3 over land during the night

  12. Example: During the day the surface temperature is 50C and air temperature is 25C. Surface pressure is 1000mb and the wind speed is 5 m/s. What is the rate of heat transfer ? dQ/(A*dt) = 3*10^-3*1.17*1004*5*25 = 439 W/m2

  13. Let us compare, now, the heat transferred with turbulent flux with the intensity heat transferred by conduction into the ground when the surface temperature is 50 C, the temperature 20 cm lower is 25 C and the soil conductivity is k = 1.0. dQ/(A*dt) = k*dT/dz=1*25/.2=125 W/m2 Thus a significant amount of heat can be conducted into the ground during the day

  14. Evaporation • The final way to transport heat in the atmosphere is through evaporation and condensation. • When the ground is wet, most of the solar radiation is used to evaporate water and the temperature does not rise so high as when the ground is dry • Vapor is then transported upward by turbulent convection and, therefore, the equation is similar to the turbulent heat transfer equation: dQ/(A*dt) = C*L*V/[Rw*T(a)]*[e(s)-e(a)] Where Rw = 461.7 is the gas constant for water vapor, and e is the vapor pressure

  15. The equation dQ/(A*dt) = C*L*V/[Rw*T(a)]*[es-ea] can be transformed into an equation for the rate of evaporation: dz/dt = C*V/[Rw*T(a)*r(H2O)]*[es-ea]

  16. The above equations need an expression for the vapor pressure e, in order to be solved. • The saturated vapor pressure, es, is given by: es(T) = 610.8*exp[Lavg/Rw*(1/273.15-1/T)] Where Lavg is the average latent heat or Lavg = 2.5*10^6-1250*T(C) Note: to get the actual vapor pressure simply take the saturated vapor pressure and multiply it by relative humidity e (T) = es(T)*RH

  17. Finally, the mixing ratio is related to the vapor pressure by the formula: w = Rd/Rw*[e/(p-e)] Or approximately: w = .622*e/p

  18. Example: What is the saturated mixing ratio of air with T = 25 C ? • Lavg = 2.5*10^6-1250*25 = 2.46875*10^6 • es(298) = 610.8*exp[Lavg/461.7*(1/273-1/298)] = 3158 p • w = .622*e/p = .0196

  19. Note that es increases exponentially with temperature and consequently evaporation rates increase rapidly with temperature • One consequence is that a warm region needs more precipitation to avoid being arid than a cold place . • For example, in the tropics, places with less than 28’’ of rain/year are semiarid while London, with ~ 20’’ rain/year is known for its humid climate. This is because London is much cooler. • Another effect is that the moister the soil the more of sun’s heat will be used to evaporate water and thus the smaller the temperature rise during the day

  20. Example: Find the heat flux due to evaporation when T(s) = 50C, T(a) = 25C, RH = 50 % and V = 5 m/s. • es (298) = 3158 p and es(323) = 12190 p • dQ/(A*dt) = 3*10^-3*2.5*10^6/(4761.7*298)*5*(.5*12190-.5*3158) = 1231 W/m2

  21. More example: If this evaporation keeps up for 8 hours a day for a month, what will be the depth of evaporated water ? dz = C*V/[Rw*T(a)*r(H20)]*[e(s) – e(a)]*dt=3*10^-3*5/[461.7*298*1000]*4516*30*8*3600 = .425 m • This would require .425*12 = 5.1 m of rain per year to keep from getting drier, Few places get so much rain

More Related