Measuring Energy and Power

It is important that the plumber has a sound knowledge of the of power and energy to understand the relative sizes of the gains and losses in solar systems and the auxiliray systems to which they are connected.  We have already drawn the disinction between the power of the sun at any instant and the energy that it can provide over a period of time, either day or year.  The calculations below will help you to calculate the amount of energy required to heat a cylinder, the heat losses from pipe work and the equivalent amount of fossil fuel required to heat domestic hot water.  The concepts will help somebody understand domestic hot water systems and will be invaluable in the assessment of the solar contribution to the space heating requirement of a house.

Units of Energy

In 1840 the English scientist James Prescott Joule discovered that heat is a type of energy, he determined the numerical relation between heat and mechanical energy, or the mechanical equivalent of heat by measuring the heat developed through the friction of water stirred by means of a paddle wheel attached to a falling weight.  He found that 772 lbs raised by one foot would raise the temperature of one pound (1 lb) of water a single degree Fahrenheit.  

A Joule is defined as the whe work done by a force of one newton traveling through a distance of one meter;  or roughly 0.1 kg dropping (or being raised) a distance of 1 meter. 

But energy can also be measured in other units;

• kWhrs (units of gas or electricity)
• BOE (Barrels of Oil Equivalent)
• kCal (Calories)

Energy itself exists in many forms.

• Chemical (e.g. Fuel Oil)
• Electrical
• Mechanical

Units of Power

The watt (symbol: W) is equal to one joule of energy per second. It measures a rate of energy conversion. So 1 watt equates to lifting a 0.1 kg weight at 1 meter per second, (or a 100g apple being dropped from 1 meter)   

Power and Energy

Power and energy are frequently confused. Power is the rate at which energy is used (or generated). Power is measured in a unit called a watt which is defined as one joule of energy per second. For example, if a 100 watt light bulb is turned on for one hour, the energy used is 100 watt-hours (W·h) or 0.1 kilowatt-hour.

In other words, 100 watts is the power of the lightbulb, but if left on of one hour it uses 0.1 kilowatt-hours. So kilowatt-hours (kWh) is the measure of energy used. Energy depends on power as well as time.

This same quantity of energy would light a 40-watt bulb for 2.5 hours.
Energy used is the amount of power acting multiplied by the amount of time.

Energy = Power x Time (in seconds)

Using this formula
1000 Watts or 1 kW is the equivalent of converting 1000 Joules every second. 
1 kWh is defined as 1000 Watts acting for 1 hour (3600 seconds), or in other words
1 kWh = 1000 W x 3600 seconds = 3,600,000 Joules

Energy and Water

About 4200 Joules are required to heat 1 litre of water by 1° Centigrade. 

Example 1

1 litre of water is heated with a 2kW kettle from 10°C to 100°C. 

2kW means that 2000J are supplied every second. 

To raise 1 litre by 90C 378,000J are needed, (4200 x 1 x 90).

This means that a 2kW kettle needs 189 seconds to boil the litre of water.

Example 2

A 30 gallon (130 litre) cylinder is heated from 10C to 60C. 

The energy required is;

50C x 4200 x 130 = 27,300,000 J

This is a very large number that doesn’t mean an awful lot, so to convert this to kWhr, simply divide by 3,600,000

= 7.58 kWhrs 

a 3kW immersion heater would therefore require 2½ hours to heat the cylinder. 7.58 divided by 3. 


Shortcut to quickly work out energy required to heat water.

If we divide 4200 by 3,600,000J we can get a factor to work out how much energy is required in kWhrs.

0.001166

A simple formula now exists
 

Example 3

How much energy is used to heat a 200 litre cylinder from 10C to 57C? Change in Temperature = 47°C
Volume = 200 litres
Answer= 200 x 47 x 0.001166
= 10.96 kWhr
So a 18kW boiler (60,000 BTU) will be capable of heating this water in
36½ minutes
 
or a 3kW immersion would take
3 hours 39 minutes

Example 4

A typical daily shower uses 30 litres of water at 39C. If the incoming mains water is at 10C, how much energy is used over the whole year?

  
Change in temperature = 29C
Volume = 365 x 30 = 10950

Energy in kWhrs = 29 x 10950 x 0.001166 = 370 kWh

Example 5

An 130 litre (30 gallon) uninsulated cylinder in a “hot press” is heated by the central heating every evening from 20C to 60C. It cools overnight to 20C, so that the central heating is required in the morning to re-heat this water. How much energy is wasted by overnight cooling when the heating is used for 300 days per year?

Change in Temperature = 40°C
Volume = 130 x 300
 
Energy = 40 x 130 x 300 x 0.001166
 
= 1819 kWhr. 
(note that this is almost the annual output of a solar panel). 
Note: Replacing a cylinder with a properly insulated cylinder (with lagged pipework) can achieve the similar energy savings as the installation of a solar panel. 

Example 6

What size boiler would be required to heat a 1000 litre buffer tank by 10°C in one hour?

Answer

Using the formula, the energy required is
Vx Delta T x 0.001166 = Energy in kWhrs
1000 x 10 x 0.001166 = 11.66 kWhrs
 
So the boiler would need to be 11.66 kW
 
If the Boiler was 23.3kW, then the buffer could be heated by 20°C in one hour or 10°C in a half hour. 

Example 7

How much energy is required to heat the following;
 
200 litres from 10°C to 60°C
1000 litres from 20°C to 30°C
 
300°C from 10°C to 60°C
 
Answer
 
Vx Delta T x 0.001166 = Energy in kWhrs
 
200 x 50x 0.001166 = 11.66 kWhrs
1000 x 10 x 0.001166 = 11.66 kWhrs
300 x 50 x 0.001166 = 17.49 kWhrs
 
These are very important answers and should be remembered
 
A 200 litre cylinder heated by 50 C needs about 12kWhrs 
A 1000 litres of water heated by 10 C needs about 12kWhrs 
 
A 300 litre cylinder heated by 50C needs about 18kWhrs

Water Content of Copper Tubes

Example 8

A solar panel is installed with a total of 30 meters of pipework (15m flow and 15m return). Calculate the volume of heat transfer fluid contained in the pipework. Calculate for 10mm, 15mm and 22mm.

Secondly, if the pipework cools from 50°C to 20°C at night, calculate the energy lost each night the solar panel is running. 

Thirdly, assume that the solar panel runs 300 days per year and produces 2000kWh. What is the percentage efficiency drop caused by the volume of liquid in the pipework cooling. Calculate for 10mm, 15mm and 22mm.

Answer

Nominal Size (mm)	Water Content per Meter (Litres)	Content in 30 Meters
10	0.055	1.65
15	0.145	4.35
22	0.320	9.60

 
Water in pipework cools by 30°C
 
Energy in kWhrs = Volume (V) x Change in Temp (Δt) x 0.001166
 
(This figure means 1000Joules every second for 1 hour, i.e. 3600 seconds)
 
10mm pipe = 30 x 1.65 x 0.001166= 0.0577 kWhrs per day
15mm pipe = 30 x 4.35 x 0.001166 = 0.1521 kWhrs per day
22mm pipe = 30 x 9.60 x 0.001166 = 0.3358 kWhrs per day
 
If the panel runs for 300 days per year and creates 2000 kWhrs of energy, then 
 
for
10mm pipe: 0.0577 kWhrs per day x 300 = 17.31 kWhrs
15mm pipe: 0.1521 kWhrs per day x 300 = 45.63 kWhrs
22mm pipe: 0.3358 kWhrs per day x 300 = 100.74 kWhrs
 

This equates to an efficiency drop as follows
17.31  x 100% = 0.87% (10mm pipe)
 2000
 
 
45.63  x 100% = 2.28% (15mm pipe)
 2000
 
 
100.74  x 100% = 5.03% (22mm pipe)
 2000

Units of Energy and Oil.

As can be seen from the index page above Heating oil contains about 10.6 kWhrs per litre and produces 3.02 kg of CO2 when burned. 

A lot of people are surprised to find that a litre of oil weighing about 900g is transformed into over 3kg of CO2, this is because as oil is burned the chemical reaction takes Oxygen from the air (a heavy molecule) and combines with the carbon (a light molecule) so the net result is the high weight of CO2.

Example 5 revisited

In Example 5 above, the wasted 1819 kWhrs would equate to 171.4 litres of oil.  But only if burned in a 100% efficient boiler. 

An typical efficiency might be 80%, so in actual fact the litre of oil is only supplying about 8.48 kWhrs of energy. So the cylinder in example 5 is actually wasting 214.5 litres of oil per year.

In CO2 terms this equates to 647.8 kg

In fact, the energy wastage would possibily be much higher, in many cases heating zones are not installed or working correctly and the complete house is heated to provide hot water generated for showers. 
Another overlooked important energy wasteage is the pipe run between the boiler and the cylinder.  This is often poorly lagged and very wasteful of energy. 
So poor is this performance, is that some solar simulation packages use a default boiler efficiency of just 55% when calculating water heating compared with 80% for the heating system.  

Other units of Energy

Example 9

Calculate the kWhr and Oil equivalent of a 3000 kCal a day diet. 

Answer

If there are 4200 joules in one kcalorie.
 
4,200 x 3000 = 12,600,000 Joules
to convert to kWhrs divide by 3,600,000
 
= 3.5 kWhrs 
or in terms of oil divide by 10.08 (See table)
= 0.35 litres of Oil

The importance of these calculations will be seen in the analysis of the losses in the system, as a means to optimise the solar yield from the collectors.