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In the second instalment of this two-part blog series, we take a closer look at how hydraulic systems can be fine-tuned for greater efficiency and long-term performance. With insights from Stefan Lay, UK Cooling and Heating Product Manager at Swegon, this post explores how temperature differences (TDs), flow rates and load conditions interact – and what that means for design choices on site.
Optimisation
Machines have a minimum and maximum flow field – this means the widest acceptable TD at the slowest flow rate and the narrowest TD at the fastest flow rate. Data published against EN14511 is given at the standard 5°C TD with subsequent flow rates detailed.
Be aware that in a fixed flow situation, your expected 5°C TD will only be achievable at full load. For the rest of the year, this TD will depend on how much of the machine needs to operate i.e. if only 50% of the available compressors need to work, expect a 2.5°C TD. You must also consider that as the outdoor temperature increases for heat pumps or decreases for chillers, it is easier for the system to manage the energy requirements and therefore capacities increase meaning the TD varies.
Let’s assume a machine with 4 equal sized compressors was selected for 100kW at 45/40 A-5 (OAT) and we want to work out the flow rate needed to achieve a 5°C TD:
One parameter required regards the Specific Heat Capacity of the fluid being used. This example includes for pure water, but you would have to adapt accordingly if using a glycol mix.
Flow rate = 100kW ÷ 4.2 (Water SHC) ÷ 5°C TD
Fixed Flow rate = 4.76l/s
Expected TDs based on no. of comps in operation using fixed water flow:
4 comps = 100kW = 4.76l/s x 4.2SHC x 5°C
3 comps = 75kW = 4.76l/s x 4.2SHC x 3.75°C
2 comps = 50kW = 4.76l/s x 4.2SHC x 2.5°C
1 comp = 25kW = 4.76l/s x 4.2SHC x 1.25°C
The same machine but at A7 (OAT) can produce 135kW heat so, using the fixed flow rate from before, our TD changes:
TD = 135kW ÷ 4.2 (Water SHC) ÷ 4.76l/s
TD = 6.4°C
4 comps = 135kW = 4.76l/s x 4.2SHC x 6.42°C
3 comps = 101.25kW = 4.76l/s x 4.2SHC x 4.82°C
2 comps = 67.5kW = 4.76l/s x 4.2SHC x 3.21°C
1 comp = 33.75kW = 4.76l/s x 4.2SHC x 1.60°C
If the flow rate is fixed for 5°C TD at A7 (OAT) operation, the TD at A-5 (OAT) would be reduced:
Flow Rate at A7 (OAT) = 135kW ÷ 4.2 ÷ 5°C TD
Fixed Flow Rate = 6.42l/s
Expected TDs based on no. of comps in operation using fixed water flow:
4 comps = 100kW = 6.42l/s x 4.2SHC x 3.70°C
3 comps = 75kW = 6.42l/s x 4.2SHC x 2.78°C
2 comps = 50kW = 6.42l/s x 4.2SHC x 1.85°C
1 comp = 25kW = 6.42l/s x 4.2SHC x 0.92°C
To fix the TD at 5°C, the water flow needs to become variable to match the capacity of the machine.
Fixed TD at 5°C for A-5(OAT) = Variable water flow:
100kW = 4.76l/s x 4.2SHC x 5°C
75kW = 3.57l/s x 4.2SHC x 5°C
50kW = 2.38l/s x 4.2SHC x 5°C
25kW = 1.19l/s x 4.2SHC x 5°C
Our built-in functionality can help simplify this approach using onboard logic which can manage flow either by using system pressures or by directly measuring the TD. Be aware that variable flow systems need to account for the minimum flow requirements of the connected machine to ensure this is not exceeded.
The machine should be selected and set to the more favoured outdoor condition which can be considered accurately if “bin” hours for local ambient temperatures are understood. Most selections requested are at design conditions which, if they are ever reached, are never sustained. A more realistic understanding of the actual conditions and any diversification in loads is crucial to avoid oversizing equipment which can lead to operational issues.
Efficiency
To be able to calculate the efficiency of a machine you need to know its capacity output and electrical energy input in that moment i.e. COP (Heating) or EER (Cooling) = kW Out ÷ kW In.
To calculate the kW Out, the following are needed:
Water Flow Rate (l/s) x Fluid type (SHC) x Inlet/Outlet Temperature Difference (°C)
Operating at 45/40 A7 (OAT) with a fixed 6.42l/s flow rate using pure water:
kW Out = 6.42l/s (Measured) x 4.2 (SHC Water) x 5°C (Measured)
kW Out = 135kW
Let’s assume we have measured 35.8kW Electrical Energy Input to the machine at full load…
COP = kW Out ÷ Kw In
COP = 135kW at A7 (OAT) ÷ 35.8kW (Measured)
COP = 3.77
The efficiency of a heat pump will reduce as the outside air temperature decreases because the machine has to “work harder” to produce the desired output:
COP = 100kW at A-5 (OAT) ÷ 35.8kW (Measured)
COP = 2.79
Conclusion
Designing, implementing and operating an efficient and reliable system needs the proper expertise and in Swegon we have the right tools and the right people to support.
To better understand system operation / building performance and to find opportunities for optimisation and increased efficiencies, the job shouldn’t stop after installation. In most new build cases, the building itself is not at full use when equipment is commissioned and therefore nowhere near the selected design condition, so it’s best to include remote monitoring which allows you to review operation both historically and in real time, facilitating the possibility to make relevant cost saving adjustments.
