Thermal Physics: Solutions to Practical Problems - Prof. Graig A. Spolek, Study notes of Heat and Mass Transfer

Solutions to practical problems related to thermal physics, including greenhouses, insulation, heat transfer, thermos bottles, human body heat loss, thermal touch, car roof frost, window pane frost, warm water ice cubes, cooling effect of open refrigerators, microwave cooking, potato baking, hot tub operation, and adding cream to coffee. It explains the underlying principles and provides insights into the behavior of various systems.

Typology: Study notes

Pre 2010

Uploaded on 08/16/2009

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Solutions to Practical Problems
1. Greenhouse – The greenhouse works due to selective transmissivity of glass.
Short wave radiation from sun is transmitted through the glass; long wave
radiation from plants is blocked. Internal temperature rises until heat gain by
radiation balances shell heat loss.
2. Insulation – Insulation inhibits air movement, which eliminates convection.
Hence, heat transfer occurs by conduction through air/glass fiber blend and by
radiation between fibers. For radiation, color matters. But pink and yellow have
about the same emissivity, so little difference could be detected.
3. Parallel plates on heaters – Plates act as fins, which increase the area for
convection from the heater to the surrounding air. Q = h A T. For a Q set by
heater power, as A the T for the same h value. Reducing T lowers the
operating temperature of the heating element (increases useful life) and reduces
surface temperature of all parts (safety).
4. Thermos bottle – The thermos bottle flask is a sealed, double wall container. Air
is evacuated between the two walls, so only radiation can transmit thermal energy,
which greatly inhibits heat loss. Coating the surfaces with a shiny layer (silver)
reduces the emissivity to reduce radiation losses further.
5. Chill factor – The human body internally generates thermal energy which is lost
continuously by convection at the surface (as well as respiration and perspiration).
The heat loss in still air can be calculated by Q = h A T, where T = Tsurface - T.
When the wind blows, the convective heat transfer coefficient hwind is much
larger, so the heat loss is the same as it would be for still air with the original h
but a much lower temperature Twind chill.
Q = h A (Tsurface - Twind chill) = hwind A (Tsurface - T)
6. Thermal touch – When objects initially at a temperature lower than that of the
skin surface is touched, heat will be lost to that object. The initial rate of heat lost
affects the sensation of “warm” or “cool”. That rate is predicted by factor
p
Ck
ρ
for the material being touched: the larger this factor, the cooler an object
will feel. (See section 5.7 of Incropera and DeWitt).
7. Car roof frost – A car’s roof will reach an equilibrium temperature which
balances the convection from surrounding air and radiation loss to the clear sky.
Since deep space is approximately -40ºF, the roof’s surface can cool to below the
freezing point. During day, indirect radiation changes the balance. The side of
the car does not freeze because of reduced shape factor with the cold sky.
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Solutions to Practical Problems

  1. Greenhouse – The greenhouse works due to selective transmissivity of glass. Short wave radiation from sun is transmitted through the glass; long wave radiation from plants is blocked. Internal temperature rises until heat gain by radiation balances shell heat loss.
  2. Insulation – Insulation inhibits air movement, which eliminates convection. Hence, heat transfer occurs by conduction through air/glass fiber blend and by radiation between fibers. For radiation, color matters. But pink and yellow have about the same emissivity, so little difference could be detected.
  3. Parallel plates on heaters – Plates act as fins, which increase the area for convection from the heater to the surrounding air. Q = h A ∆T. For a Q set by heater power, as A↑ the ∆T↓ for the same h value. Reducing ∆T lowers the operating temperature of the heating element (increases useful life) and reduces surface temperature of all parts (safety).
  4. Thermos bottle – The thermos bottle flask is a sealed, double wall container. Air is evacuated between the two walls, so only radiation can transmit thermal energy, which greatly inhibits heat loss. Coating the surfaces with a shiny layer (silver) reduces the emissivity to reduce radiation losses further.
  5. Chill factor – The human body internally generates thermal energy which is lost continuously by convection at the surface (as well as respiration and perspiration). The heat loss in still air can be calculated by Q = h A ∆T, where ∆T = Tsurface - T∞. When the wind blows, the convective heat transfer coefficient hwind is much larger, so the heat loss is the same as it would be for still air with the original h but a much lower temperature Twind chill.

Q = h A (Tsurface - Twind chill) (^) = hwind A (Tsurface - T∞)

  1. Thermal touch – When objects initially at a temperature lower than that of the skin surface is touched, heat will be lost to that object. The initial rate of heat lost affects the sensation of “warm” or “cool”. That rate is predicted by factor

k ρ C p for the material being touched: the larger this factor, the cooler an object

will feel. (See section 5.7 of Incropera and DeWitt).

  1. Car roof frost – A car’s roof will reach an equilibrium temperature which balances the convection from surrounding air and radiation loss to the clear sky. Since deep space is approximately -40ºF, the roof’s surface can cool to below the freezing point. During day, indirect radiation changes the balance. The side of the car does not freeze because of reduced shape factor with the cold sky.
  1. Window pane frost – As air is cooled, its density increases and it drops. Air cooled by contact with a cold window surfaces cools to the ice point near the bottom of the window pane on its drop from the top down. Frost forms after adequate cooling.
  2. Warm water ice cubes – In an all things equal experiment, warm water will not freeze faster than cold water, because the amount of heat to be removed is greater. However, open trays of warm water will often freeze faster than cold ones because warm water evaporates faster (especially into a frost-free freezer compartment). The more water that evaporates, the less remains to freeze. Since latent heat rather than sensible hear dominates this entire cooling/freezing process, freezing less water will appear to produce ice cubes faster.
  3. Cooling effect of open refrigerator – As you stand before an open refrigerator, your skin will begin to lose heat faster due to radiation exchange with the cold interior refrigerator surfaces. Greater heat loss is sensed as feeling cold.
  4. Microwave cooking – Yes, a microwave oven does, indeed, cook from the inside out. Microwaves cause water molecules in the food to vibrate, which is equivalent to higher thermal energy. Microwaves are absorbed more-or-less uniformly by food, representing uniform internal generation of heat. Thus, the core of the food is warmer than the surface.
  5. Potato baking. This can be solved using transient heat transfer, assuming the same dimensionless temperature for both potatoes. Only the Bi -1^ changes due to the size change, so the Fo number for the larger potato can be found. Typical numbers reveals about 32% longer cooking for a potato twice the weight.
  6. Hot tub operation – The energy required to heat the hot tub is equal to the energy loss from the hot tub, integrated over time. Hot tub heat loss is dictated by ∆T between the water and the surroundings. A reduction in ∆T reduces energy costs. Allowing the hot tub to cool down between uses reduces ∆T and, therefore, energy use.
  7. When to add cream to coffee – The cream will cool down the coffee when added, and by approximately the same amount whether added when the coffee is hotter or cooler. So by adding the cream when the coffee is first served, the coffee/cream mixture has a reduced temperature for longer prior to drinking, the heat loss is smaller, and the drinking temperature is higher. So add the cream initially to keep the coffee warmer for drinking.
  8. Beverage cooling – The cooling rate is driven by ∆T, but also by the convective coefficient h: Q = h A ∆T. The h for water is likely to be 10-100 times as great as that for air, while the temperature difference ∆T for the two cases is not that large. So leave the beverage in the air, at least until the beverage reaches water temperature.