Jug Test
Solar Collectors and solar panels should be installed in locations that have unobstructed sunlight. They should be oriented to face the sun. Unfortunately ideal installation locations are not always possible so a person must decide if the location they have is practical. Suppose you have a south-facing roof that receives sunlight but you also have a tree that casts a shadow on your roof. Should you remove the tree. Well I would, but you’ll have to decide how important the removal of a tree will be to your system. You’ll have to decide if a few shadows will render your application impractical. A data logger could help you decide if tree removal is necessary, but what if you don’t have a data logger?. is there a simple test that can approximate the energy available from sunlight. This is an idea that Gary Reysa had to estimate the amount of sunlight energy available to a given location. I will be exploring the concept with milk jugs.
After the blackened jug is filled with water and placed in the sun it’s temperature will rise in proportion to the amount of solar energy available. You’ll also have to take into account heat losses due to ambient temperature. The more data reading you take the better, but the lag time associated with heating the jug allows it to be used as a sort of low cost data logger. it will get hot gradually and actually record the amount of heat available over a given period of time. We’ll use a data logger with the jug collectors to see how the pattern of sunlight and ambient temperature affect the temperature of water in the jug. From this information we should be able to estimate the actual solar heat harvest and confirm the accuracy of this method of collecting solar thermal data.
Milk Jug heat Loss
The flat surface of a half gallon milk jug provides a nice surface area for solar heat collection.
· Fill two half-gallon milk containers with hot tap water.
· Place one milk container in an insulated box.
· Place the other container on a counter
· Log the temperature of each container each hour.
· Also log the average ambient temperature
· Calculate the heat loss of each container.
Q= ( T1-T2 ) x mass
Q= heat loss
(T1 –T2 ) = Starting temp. – Ending temp…. Fahrenheit
mass = weight of water in pounds.
This
graph illustrates the temperature drop of insulated and un-insulated jugs over a
period of 7 hours. We will only be calculating the heat losses that occur during
the first hour although the heat loss per degree differential will be about the
same at any place on the curve.
Mass of water = 4 pounds
Initial temp T1 = 107*F
Ending temp T2 = 102.5 F
Q = ( 107 – 102.5 ) x 4lbs.
Q = 18 BTU
Mass of water = 4 pounds
Initial temp T1 = 107*F
Ending temp T2 = 95* F
Q = ( 107 – 95 ) x 4 = 48 BTU
Heat loss per degree differential
Average temperature of water during first hour = 105*F
Average ambient temperature during first hour = 71*F
Differential temperature 105* – 71* = 34* F
Heat Loss during first hour = Q = 18 BTU
Heat loss per degree differential = 18/34 = .5 BTU
Average temperature of water during first hour = 101*F
Average ambient temperature during first hour = 71*F
Differential temperature 105* – 71* = 30* F
Heat Loss during first hour = Q = 48 BTU
Heat loss per degree differential = 48/34 =1.4 BTU
So the heat loss from the un-insulated jug is almost 3x greater than the heat loss from the insulated jug. When we calculate the heat gain from the sun we’ll need to factor in this heat loss per degree differential.
We could just measure the temp change of the jug collector at the end of one hour in bright sunlight if we have some idea how intense the sun is. This will of course depend on your location and your weather conditions and time of year. A good first approximation of sunlight energy on a clear sunny day would be about 200 BTU/ft^2.
OK If you're still with me I ran the Jug Test from Jan. 9 till Jan. 10. Unfortunately i forgot to connect the temp. probe to the insulated jug so the only data available is the ambient temp, the un-insulated Jug temp and the solar FLUX... Also the solar flux density should be less than that illustrated on the graph.
Since the ambient temperature dropped below 24* F overnight ice formed in the jug so that heat gain and heat loss calculations are difficult to approximate on the second day so we will confine our solar heat gain calculation to the first day of data collection between 10 AM and 11 AM.
For our un-insulated jug I am assuming a surface area exposed to sunlight of .3 ft.^2 and a solar flux density of 200 BTU/ft.^2. From the above graph you can see a 7* jug temp rise between 10 AM and 11 AM. Raising 4 lbs. of water 7* requires 7* x 4lbs. or 28 BTUs of energy... HOWEVER we must also take into account the heat losses that take place due to the difference in temp between the jug and the ambient temp.
SOLAR HEAT GAIN = OBSERVED
HEAT GAIN + AMBIENT HEAT LOSS OF 1.4 BTU/degree differential
SOLAR HEAT GAIN = 7* X 4
lbs.
+ 13* X 1.4 BTU/degree
SOLAR HEAT GAIN = 28 BTU + 18 BTU = 46 BTU
SOLAR HEAT AVAILABLE FROM SUN
= 200 BTU x 0.3ft^2 = 60 BTU's
Efficiency of Heat
Collection =
28/60 = 47%
Solar Heat Transfer
Efficiency = 46/60 = 77%
These experimental results are a bit crude but I will refine the process when freezing water is no longer an experimental complication. With a little work I do believe a jug of water could be used as an experimental tool used to estimate solar energy availability. As a matter of fact I was thinking that a brick inside an insulated box might work just as well if the brick is calibrated... If you have some ideas please sent them to:
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