If the predicted regression output equation for a solar still is: Production of
ID: 1836842 • Letter: I
Question
If the predicted regression output equation for a solar still is:
Production of water (l/m2) = 0.92 I (kW-h/m2) where I is the solar insolation.
What is the overall efficiency of the system? Assume a total radiation for a typical day to be 6.2kW-h/m2 or 6.2 sun hours.
Given:
Production equation above and 1 liter(l) of water = 1 kilogram
1 hour = 3600 seconds; 24 hours in a day
The latent heat of vaporization for water is 539.6 cal/gram; this is the amount of energy required to convert liquid water to a vapor (exactly what the still is doing for us).
Solution:
Determine production for the day (per m2 of still)
Determine the energy going into the still over a full day per m2.
(efficiency) = stuff out/stuff in and is unit-less. Therefore, we must get these values into the same units (this is the trick – the old comparing apples and oranges problem). Remember that All units must cancel in % values. The output of the still is in liters (l) of water, while input to the still (solar energy) has units of kW-h
Explanation / Answer
Production of water (L/sq.m) = 0.92*I =0.92*6.2 (Kwh-sq.m) = 5.7 Ltrs
The still produces 5.7 Ltr of water by evaporation and energy conusmed = 6.2e+3*3600 = 2.232 J
latent heat of vapourisation for water = 539.6 cal/gm
energy required
for 5.7 Ltr = 5.7*1000*539.6 cal
5.7*1000*539.6*4.186 J ( 1cal = 4.186 J)
= 1.288e+7 J
efficiency of the still = 1.288e+7/2.232e+7 = 0.58 or 58%
Prodction of water per day = 5.7 Lts