How much power are we getting from the Sun though the panel?

I’ll use three separate ways to calculate this:

These results are all in the same ball park. (!)


The sun is shining in a cloudless sky.
The Latitude is 52°N, Longitude 0°.  ( I've got a bit carried away with mapping here.)
The panel area is 2 m2.
The time is noon.
We’re at an equinox.


The manifold capacity is 3.4L
Heat exchanger coil is 6 metre long of "Integron" copper finned tube. External diameter 22mm, internal diameter 17mm.


The panel is at 30° to the horizontal.
The panel faces SE (135°)
The transit time is 2 min 45 seconds (165s). The time for the water to circulate once round the loop.
Coil transit time is 35 seconds.
Pipe run length (15mm) 22m
Internal diameter of 22mm pipe is 20 mm, 15mm is 13mm.

The power saved by not firing the gas boiler.

This is the lamest calculation but is also the easiest! 

Our average hot water demand requires the gas boiler (29.3kW Output rating) to burn for 10 minutes. This is 4.88 kWh.
8hrs duration of sunshine satisfies the same hot water  demand.
4.8/ 8 = 610 W 

Measuring flow rate and the temperature difference.

Need to find the capacity of the various components in the circulating system

Manifold,  pipe work, heat exchanger coil.
The coil

Coil is 6 metres long of "Integron" copper finned tube. External diameter 22mm, internal diameter 17mm

Capacity 1.6 L.

The pipe-work
21.5m of 15mm pipe-work in circulation.
Pipe capacity (13mm mean internal diameter.)
3.14 * 6.5 * 6.5 * 22,000 mm3 = 2.8L
Capacity of circulating system
(CapTot) = Manifold + pipework + coil.
(CapTot) = 3.4 + 2.9 + 1.6 = 7.8L
Validation of the coil capacity against total pipework capacity can be checked with the transit times. (The time a "packet" of water takes to flow around the system.)
flow rate in coil 1.6/35 = 46 mL s-1
flow rate in circuit 7.8/165 = 47 mL s-1
47mL s-1 ( 0.17 m3h-1)
The Grundfos pump manual quotes a range of flows from 0 to 0.7 m3h-1. At 0.19 m3h-1 the pump produces a head of 1.2m. (.32mA at 243V) P = 77W not the rated 25W !!!!
Heat calculation
7.2°C is measured across the coil.
SH of water = 4.19
KWs (kJ) = dt * SH/ flow = 30.2 / 0.047 = 642W (This is useful heat – it warms the water!)  

1.0°C is measured from the manifold to the top coil.
SH of water = 4.19
KWs (kJ) = dt * SH/ flow = 4.19 / 0.047 = 89.1W 
Assume the return pipe drop is the same = 178W (This is not useful heat – it warms the attics!)  

From the manufacturer’s data sheet.  

Thermomax TMO500 is 60% efficient at 60°C.
Full sun power 1 kW per m2
At noon the angle of the sun (altitude angle) is 90 – 52 = 38°.  
Insolation value at surface = sin (38°) = 615W (The extra path length through the atmosphere.)
Panel area is 2 m2.
The roof tilt effectively increases this by 1/ cos (tilt) = 1.154 * 2 m2 = 2.3 m2
Manifold Power = 2.3 * 615 * 60% = 851W
Heat loss in pipe run. 
Water 60°C
Ambient 20°C  
15mm diameter pipe.

Insulation calculations

The material is cheap unbranded rubbish from a "home care" DIY shed. Outside diameter 40mm.

The material is urethane foam: R6 per inch

R6 is equivalent to = .042 °C W  m2  1 mm thick

I shall ignore any heat transfer coefficients within the water and the copper of the pipe. Also I shall ignore transfer coefficients from the surface of the insulation. (Convection and Radiation effects.)

Using Rcyl = ln(Outer radius / Inner radius)/ (2 *  PI * length * conductivity)

 = ln(20 / 7.5) / 2 * PI * 21.5 * .042

 = 0.172 C/W

Q = dt / R =  40 / 0.172 = 231W

(Using 65mm outside diameter insulation would reduce this to 156W)

Power across coil
851 – 231 = 620W   

Some Notes

SI Thermal Conductivity

The SI unit of thermal conductivity is W/m °C.

More sense might be to rewrite as (W*m/m2)°C. This now hints that conductivity is the heat transfer (W) through a thickness (m) of a sheet area (m2) with a temperature difference of (°C).

Perhaps another rewrite...  W m2 °C / m. This suggests heat transfer (W) per area (m2) for a temperature gradient.

1 W/m °C = 0.5778 Btu/h * ft * °F.


Insulation is measured in R-value, the capacity of an insulating material to resist heat flow. The higher the R-value, the greater the insulating power.
When you're purchasing insulation, buy or specify R-value, not inches, as R-values of materials vary.
To achieve higher total insulating power, R-values can be added together. For example, R-38 added to an R-11 results in R-49.
Achieve a higher total R-value by "combining" two different R-values.
R Value (°F·hr·ft² / Btu)
From: Nick Pine <>
Sent: Tuesday, September 21, 1999 7:59 AM
Subject: Re: What does the R mean for insulation?
Mike Swift <> wrote:
>...From my ASHRAE (American Society of Heating, Refrigerating and
>Air-Conditioning Engineers) Handbook of Fundamentals the term R is
>defined as "the reciprocal of thermal conductance". Conductance
>is the quantity of heat that will flow through a material when a
>temperature difference exists between too surfaces.
Not exactly. Take a look at these units.
> The units are W/m^2 *°C .
You must have the metric version of the HOF. Metric quantities of heat
are measured in joules or watt-hours, not "W/m^2-C," and W measures
heatflow over time, ie heat power in watts (joules of heat energy per
second) vs "a quantity of heat." One might say "a metric thermal
conductance IS a ratio of heatflow in watts to a certain Celsius
temperature difference for heat that flows through a square meter
of some material."
>As an example if a one square meter wall were made and a one watt
>heater was placed on one side, and that caused a one degree C rise
>in temperature then the wall would have a [metric] R value of one.
If 1 watt (3.41 Btu/h) flows through 1 m^2 (10.76 ft^2) with a 1 C
(1.8F) temperature difference, the US R-value is 10.76ft^2x1.8F/3.41Btu/h
= 5.68 ft^2-F-h/Btu [yuck], and you'd see "R5.68" stamped in big letters
on insulation (eg 1 inch of Styrofoam) with that resistance in a US store.
Nick <Pine>