Heat Loss

This polyethylene heat storage tank occupies a surface area of about 6 ft^2. Can you estimate the conductivity of polyethylene based on the heat loss from 5 gallons of water over the period of one hour at an ambient temp of 52*F?

There are a number of ways to measure or estimate heat losses. The most direct method involves measuring the drop in temperature over the period of one hour and multiplying this temperature drop by the mass of water in the tank. Of course the rate of heat loss will depend on the surface area of the container, the conductance of the container, the thickness of the container and also the ambient temperature surrounding the container. Other factors will involve the R factor of a film of air around the container and the level to which the container is filled. When a storage tank is filled and surrounded with insulation the R factor is easy to calculate since it is large compared to the container heat losses but I'd like to estimate the heat losses per sq ft of container surface and calculate the conductivity of polyethylene from this information, CAN YOU HELP?

Q = A x u x (T1-T2)

**The heat loss from the container between 11PM and 12AM was
8*F x 40lbs = 320 BTUs/hr with and ambient temperature of 52*F. What is the
conductivity of the polyethylene tank.. Given a tank surface area of 6ft ^{2}
half filled with water?**

**Thank You John Canivan **

COLLECTOR HEAT GAIN

It would be nice to know the conductivity of
polyethylene to calculate the heat loss of the storage tank, but this is not
needed to calculate the total solar collector heat gain and the collector
efficiency. As long as we know how much heat is lost through the tank under the
same differential temp as conditions we should be able to estimate the total
heat gain of the collector with this graph of input and output temp at a flow
rate of 1.2 gal/min. and a collector surface area of 8 ft^{2}

Total Heat gain of this serpentine drainback system is 15*F x 72 gal/hr = 1080 BTU/hr

Collector Heat Gain = Total Heat Gain + tank heat loss

Qc = 1,080 BTU/hr + 320 BTU/hr = 1,400 BTU/hr

Solar Heat Availability = 280 BTU/ft^{2}
x 8ft^{2 } = 2,240
BTU/hr

Efficiency of serpentine collector = 1,400/2,240 = 62 %