Trying to design a system to store ammonia at -33°C

1 Comment

1. 

   Dr. Randall Barron

   June 3, 2011

   Generally, the solar radiation is maximum at solar noon, and the rate of solar insolation at noon [I(noon)] can be found from the daily solar insulation [H]:

   I(noon){MJ/m2-hour} = 0.205 H{MJ/m2-day}

   (To convert MJ/m2-hour to kW/m2, multiply by (1000)/(3600 sec/hr)

   Values of the daily solar insolation may be found from meteorological data for the specific location. Another source is: J.A. Duffie and W.A.
   Beckman, SOLAR ENGINEERING OF THERMAL PROCESSES, 2nd ed., John Wiley & Sons, New York, (1991), pp. 843-881. This book has solar data for locations in Africa, Asia, Europe, South America, Australia, Canada, Mexico, and about 230 cities in the US. For example, for Shreveport, LA, during the month of June, the daily solar insolation is H = 23.44 MJ/m2-day, so I(noon) = 4.805 MJ/m2-hour = 1.335 kW/m2.

   The maximum energy absorbed by the surface from the sun is:

   Q(sun) = a A I(noon)

   where: a = surface absorptance
   A = area of the surface projected on a horizontal plane

   For solar radiation, the absorptance values are [see the ASHRAE Guide]:

   a ~ 0.90 for building materials
   a ~ 0.25 for unpainted steel
   a ~ 0.15 for a surface painted with white (ZnO) paint

   For the example of Shreveport, La, if the surface is painted with a white paint, the heat flux absorbed from the sun would be:

   q(sun) = Q(sun)/A = (0.15)(1.335 kW/m2) = 0.200 kW/m2

   According to the ASHRAE Guide, a design value for the convective heat transfer coefficient between the surface and ambient air is:

   h ~ 1.63 Btu/hr-ft2-°F = 9.26 W/m2-°C

   One of the more simple methods of reducing (or actually
   eliminating) the solar radiation from the sun on a surface, such as the ammonia storage vessel, is to cover the vessel with a roof (shed) covering with no closed walls, so that ventilation is good. For a vessel covered with a roof, the only heat transfer to the vessel would be convection from ambient air at Th to the vessel contents at Tc. The overall heat transfer coefficient would be

   1 1 Do ln(Do/Di)
   –– = – + ––––––––––––
   Uo h 2 k

   where: Do = outside diameter of the vessel,
   Di = inside diameter of the vessel, and
   k = thermal conductivity of the insulation on the vessel.

   Q = Uo Ao (Th – Tc)

   where: Ao = total surface area of the vessel
   Tc = temperature of the inner surface of the vessel (essentially, the temperature of the fluid inside the vessel).

   If the vessel is exposed to the sun, the outside surface temperature of the vessel Ts will be increased, due to the solar insolation. In this case,

   Q = U* Ao (Ts – Tc)

   where: 2 k
   U* = ––––––––––––
   Do ln(Do/Di)

   q(sun) + h Th + U* Tc
   Ts = ––––––––––––––––––––– – vessel surface temperature
   h + U*

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