Heat transfer indirect hot water

[webtecmark]

I have designed, built & installed a home made solar hot water heating system and know roughly the energy gained. I have done some research and come up with a basic calculation for energy transfered in to the hot water storage but would like opinions as to how accurate it is.

I have made my own micro-controller and have many temeperature sensors around the system. I have also measured the flow with an accurate meter. Data I have is:

1. Heat exchanger coil is 22mm copper pipe formed in to 5 coils with a centerline diameter of 0.33m.
2. Heat drop across the heat exchanger coil in degrees C, ie. input and output temperature.
3. Flow rate of 4.0LPM.

I calculated the volume of the heat exchanger coil:
V = (0.011^2 x pi) x (0.33 x pi) x 5 = 1.97 x10-3 m^3
as there are 1000Cu liters per cu meter the volume is
V = 1.97 liters (1.97Kg)

assuming there are no losses/reductions through the copper wall (scale will not form) then the heat lost from the heat exchanger coil:
Q = 4.19 x DT x volume
Q in Mj
4.19 is specific heat capacity of water
DT is differential temp in K or C
volume in Kg.

Considering the flow rate of 4Kg/min
volume = 4
Q = 4.19 x 2 x 4 = 33.52Mj/min
per hour
Q = 33.52 x 60 = 2011.2Mj/hr

Convert to Kw/h multiply by 0.27777

Kw/h = Q x 0.2777 = 2011.2 x 0.2777 = 0.558Kw/h

This figure seems believable but I am not sure if I have made the right assumptions regarding the flow rate and heat exchanger capacity! Knowing the flow rate does the heat exchanger capacity matter?

How much difference would taking the thickness of the copper wall of the heat exchanger in to account affect the outcome?

 

[FMJalink]

Postwebtecmark,
I assume that the 5 coils heat up the water and you want to calculate the heat transferred from the coils to the water.

In formula would this stationary heat transfer becomes:
Q = ? x ?vol x Cp x ?T, where

Q = Heat transferred in kW or (kWh/h)
?vol = Volumetric flow in m3/s; 4 LPM = 4 /(1000 x 60) m3/s
Cp = Heat capacity at constant pressure in kJ/kg.K (4.19 kJ/kg.K)
?T = Temperature loss over the coils (entrance minus exit) in °C or K (2 °K)

This leads to a Q of 0.559 kW or 0.559 kWh/h.

From this formula you can see:
(1) You do not have to calculate the water mass in the coils, as you have already the flow; and
(2) The thickness of the tube is not important for this stationary heat transfer calculation.

Kind regards,
Frank Jalink

 

[IRstuff]

I concur with FMJalink. You made a conversion error somewhere; can't tell where since units aren't shown in your calculations.

Is this actual data or just a test of the calculation? 559W is about 50% of the worldwide peak insolation on a 1 m^2 absorber, so that's extraordinarily good, if those are the conditions.

Note, however, that the total insolation during a day is only about 5 times the peak energy hour, i.e., if the peak power is about 1 kW/m^2, the total energy is about 5 kWh/m^2.

Note also, a 2°C rise is too small to be of practical use. You really need to get the DT to be at least 35°C if you're starting with city tap water; you may still need to supplement with your regular water heater.

Note also, 4 L/min is barely enough for 1 shower since most showerheads are designed to flow around 8 L/min. This also means that you need to have a storage capability to deal with these transients.

TTFN

FAQ731-376

 

[webtecmark]

Thank you for the comments, Franks reply seems to correspond with my calculation (0.559KwH), all well and good.
A little more information about the setup I have made:
1. There are three flat panel collectors just >1m^2 each; 3m^2 total area.
2. The panels are connected in series and all seem to have the same temperature gain, input to output.
3. The solar panels are connected in a pressurised sealed system that has a total of approximately 9liters of water.
4. The heat is transfered in to my hot water storage tanks via the heating coil mentioned in the previous post.
5. Differential heat at the hot water storage tank coil varies from 2 DegC to 4 or 5 DegC so the overall heat extracted from the sun equates to around 0.5Kwh to 1.4Kwh leading me to assume a system efficiency of around 0.45Kwh peak per sq meter.

I know from how it runs that the efficiency drops back when the heat transfer fluid gets hotter (50 DegC +) as there is noticeable heat leakage. I am planning to install better insulation!

How would I go about calculating the dynamic temperature transfer in to the storage tank considering there is a constant 4LPM flow rate?

Do I need to monitor the storage tank water temperature at the top and bottom of the coil?

Thank for your interest.
Regards
Mark

 

[zekeman]

If you know the capacity of the tank, then the electrical analogy which you may be more familiar with, is

dV/dt=i/C

V= temperature
i = your input heat rate
C= tank capacity= water weight in tank*specific heat

 

[IRstuff]

If it's a steady state flow, that means that the water is actually recirculating through the tank and heater. If that's the case, then the steady state condition is when the heat loss of the stoarge tank is exactly balanced by the incoming heat from from the heater. Since you've calculated your inflow heat flux, you then determine what storage tank temperature will result in a loss of that same heat flux. The resultant tank water temperature will be the steady state condition.


TTFN

FAQ731-376