A hot water loop is a common way to transfer heat. Hot water loops are often found in district heating networks and on industrial sites.
But do you know that often these loops are not operated optimally?
Figure 1 – A simple hot water loop
This post will explain how the CAPEX & OPEX costs of many hot water loops are higher than they should be. We explore this by introducing one of the most useful equations in energy engineering.
The equation below shows how to calculate heat transferred in our hot water loop.
The mass flow rate (m, kg/s) is a measurement of the amount of water flowing around the hot water loop.
The specific heat capacity (CP, kJ/kg.°C) is a thermodynamic property specific to the material. We could manipulate the specific heat capacity only by changing the fluid used in the loop. Water is a good fluid choice for cost and safety considerations. The specific heat capacity of water does vary with temperature but for the scope of liquid water heating it can be estimated as constant.
The temperature difference (dT ,°C) is the change in temperature of the hot water. It is the difference in temperature before and after heat transfer.
As engineers we are able to manipulate two of the three variables in our equation – flow rate and temperature difference. We can optimize these through the design and operation of our hot water loops.
We could use two distinct methods to deliver the same amount of heat in our hot water loop:
- High mass flow rate + low temperature difference
- Low mass flow rate + high temperature difference
The second method is optimal and will reduce our scheme CAPEX & OPEX costs.
A low mass flow rate minimizes the amount of electricity required to pump water around the loop.
A high temperature difference means:
- Increase in the maximum capacity of the loop to deliver heat. Pipe size limits the capacity of the loop by limiting the maximum flow rate. More heat can be transferred at the maximum flow rate by using a larger temperature difference.
- Maximises heat recovery from CHP heat sources such as jacket water or exhaust.
- Maximises electric output from steam turbine based systems by allowing a lower condenser pressure.
The CAPEX benefit comes from being able to transfer more heat for the same amount of investment.
OPEX benefits arise from reduced pump electricity consumption and increased CHP system efficiency.
Thanks for reading!
Most people associate thermodynamics with textbooks full of strange symbols and complex equations. It is no surprise that thermodynamics is rarely a topic that gets people excited.
This post aims to highlight what is exciting about thermodynamics. The practical insights that stem from thermodynamics are the foundation of energy engineering.
The First Law
The First Law is about conservation. All that we can do is convert one form of energy to another (most often converting between fuel, power and heat). We can never generate energy out of nothing.
This is all great – but how does it help us?
The First Law allows engineers to use mass & energy balances. Engineers use balances to model energy systems. Unknowns such as flow rates of water or temperatures of gases can be calculated using balances.
A simple example of power generation – we generate 4 units of power from 10 units of fuel. The First Law allows us to calculate the heat generation at 6 units.
What the First Law doesn’t do is place any limit on the amount of power we can generate (except limiting it at 10). This is where the Second Law steps in.
The Second Law
The Second Law has multiple definitions. One form of the Second Law limits how much power we can generate from fuel. No matter how advanced our technology becomes we will never be able to convert 100% of our fuel into power.
There are fundamental thermodynamic limits on the amount of high quality energy we can convert from one form to another. We will never be able to convert 100% of our fuel into power. Even with more advanced technology, we will always end up with some of our fuel converted into low quality energy (i.e. heat).
The Second Law tells us that heat is a necessary by-product of power generation. To maximize the value we get from our fuel requires finding a place for this heat. Combined Heat and Power is a range of technologies that aim to do this.
Essentially the First Law tells us we can only ever break even when converting energy. The Second Law then tells us we will always lose!
Both of these Laws have alternative definitions and interpretations. The insights I have chosen to highlight above are the two I find the most practical.
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Understanding the difference between kilowatts (kW) and kilowatt-hours (kWh) is a basic for energy industry professionals. The key is understanding they are related to time.
An analogy with distance and speed can illuminate the concepts of kW and kWh. We measure distance in fixed amounts – such as 50 kilometers. Energy is measured in the same way using the unit kWh. 50 kWh is a fixed amount of energy.
Speed is the rate at which we are covering distance – such as 50 km/h. The rate of energy generation or consumption is measured using units of kW. This rate is known as power. The kW is a measurement of kJ per second (kJ is a different unit to measure energy. For reference 3,600 kJ = 1 kWh).
Table 1 – Analogy with speed and distance
If you ever see kW/hr – this is almost certainly an error. Technically this is equivalent to acceleration – the rate at which we are changing our power. People often mistakenly use kW/hr when they should be using kWh – don’t be one of these people!
How then to relate kW to kWh? We need one more piece of information – the length of time over which we are producing or consuming energy.
If we were driving a car at a certain speed, to know how far we had driven we need to know how long we had been driving. Likewise if we are consuming energy at a certain rate, to know how much energy we had consumed we need to know how long we had been operating.
For example, if we had been consuming at 50 kW for 1 hour, we would have consumed 50 kWh. Two hours would lead to a consumption of 100 kWh, half an hour 25 kWh.
Table 2 – Relationship between rate, length of time and consumption
|How varying the rate (speed or power) affects consumption
|How varying the length of time affects consumption
These concepts need to be grasped forwards, backwards and side to side. You must be comfortable with moving from:
- kW * hr = kWh
- kWh / hr = kW
- kWh / kW = hr
We can now use these concepts to analyze energy data.
Suppose you have the following data for a CHP scheme. This is a CHP scheme with the facility to dump heat (not all heat generated is necessarily recovered).
Table 3 – Sample data
|Engine electric size||kWe||400
|Engine thermal size||kW||400
|Annual heat recovered||kWh||1,809,798
|Annual power generated||MWh||2,571
|Annual operating hours||hr||5,638
What insights can we gain from this? I would look at the following:
Annual heat recovered
- We can calculate the Maximum heat generation = Engine thermal size * Annual operating hours (This is assuming full load operation for each hour).
- This gives us 22,55,200 kWh of heat.
- This validates the annual heat recovery as reasonable at 80% of the maximum available heat.
- We can calculate the Average Heat Recovery = Annual heat recovery / Annual operating hours. This gives us an average of 321 kW.
- We do not know whether or not this number is low because of part load operation of the CHP or due to heat dumping.
Annual power generated
- The Maximum power generated = Engine electric size * Annual operating hours.
- This gives us a maximum generation of 2,255 MWh (note I have divided by 1,000 to convert from kWh to MWh).
- This flags up an error with the Annual power generated value of 2,571 MWh as it is greater than our maximum!
- We could also spot this error by calculating Average Electric Output = Annual power generated / Annual operating hours. This gives us an average engine electric output of 456 kWe (greater than the size of our engine).
The two concepts of an amount of energy (kWh) and the rate of energy consumption/generation (kW) are related to each other by a length of time.
Understanding this allows energy industry professionals to check the validity of data, as well as calculate new data to evaluate performance. Both skills are valuable to engineers and non-engineers alike.
Two European and British energy engineers are arguing about to calculate the gas consumption of a boiler. The European engineer wants to assume an efficiency of 89% – the British engineer wants to assume 80%.
Which one is correct?
Higher Heating Value vs. Lower Heating Value
In the same way that two different currencies can value the same thing with a different amount of the currency, two conventions exist for quantifying the amount of heat produced in fuel combustion. These two conventions are the:
– higher heating value (HHV) – also known as gross calorific value (GCV)
– lower heating value (LHV) – also known as net calorific value (NCV)
Figure 1 – A fire-tube shell boiler
These conventions arise from practical engineering realities. Specifically it’s about what happens to the water vapour produced during combustion. Condensing water vapour releases a lot of energy.
The gross calorific value (HHV) includes this energy. The net calorific value (LHV) doesn’t include it. This is why a gross calorific value is higher than a net calorific value.
– HHV = water vapour is condensed = more heat is recovered
– LHV = water vapour remains as vapour – less heat is recovered
The reason for the distinction is that water vapour in combustion products is not often condensed in practice:
– Steam or hot water boilers – condensing water requires reducing the flue gas temperature low enough where acids present in the flue gas will also condense out and cause stack corrosion and potential failure.
– Power generation – water usually remains as a vapour as the temperature within the power turbine or pistons of the engine is too high.
The two engineers
We return now to the argument between our European and British engineers. Which efficiency (89% or 80%) is the best assumption?
Table 1 – Typical HHV and LHV efficiencies
|Typical efficiencies||% HHV||% LHV
|Gas engine (2 MWe)||38%||42%
|Gas turbine (5 MWe)||28%||31%
The correct assumption depends on how the calculated gas consumption will be used. To calculate the cost using a UK gas price we would want to have assumed an efficiency of 80 % HHV. This is because UK gas prices are given on an HHV basis.
Either convention can be used as long as all of our fuel consumptions, efficiencies and energy prices are given on the same basis. Consistency is crucial.
Most data sheets will specify gas consumption or efficiency on an LHV basis. If you are working in a country that prices fuel on an HHV basis (such as the UK) that you convert this gas consumption to an HHV basis.
Using a fuel consumption on an LHV basis with a HHV gas price can lead to a significant underestimation of fuel costs.
Using an efficiency or gas consumption straight off a data sheet can easily wipe out the typical margins expected on energy sale contracts. It can also tip a project IRR below the hurdle rate.
Best practice is to always be specific HHV or LHV when working with fuel consumptions, efficiencies and prices. Be the engineer who always writes ‘kWh HHV’!
Thanks for reading!