KNOW YOUR ENGINE DYNAMICS How many KGs of air do you think your engine consumes? EQUATION USED (for a 4-stoke engine): AIRFLOW = 0.5 x Engine Size x RPM x % V/Efficiency
 The chart below shows the quantity of air(kgs/min.) passes into the engine (1-6.5 litre) under different engine speeds. We assumed the engine's Volumetric Efficiency is 80% and Air Density = 1.2g/M³ (25C). The engine is a naturally aspirated type and operating at sea level. A typical two litre engine at 6000 RPM (red dot) consumes 5.64 Kg of air per minute, if water was injected at a rate of 250cc/minute (0.25kg), the water/air ratio is only 2.9% ! The percentage will drop further if fuel is included as a total mass in the calculation. * * * * * * * CALCULATING THE TEMPERATURE RISE Pity that temperature rise is proportional to pressure increase EQUATION USED: T2={T1[P2/P1]^0.283} x n(65%) T1=inlet T2=outlet n=compressor efficiency
 Doubling the boost pressure ratio doesn't yield twice the amount of power, unfortunately. Law of Physics rules, when a gas is compressed, temperature rise will follow. We will examine how the theory can be applied to an automotive engine, step by step we will try to unveil the mystery behind the true workings of an engine. Since we are dealing with force induced engines, we will first look at the amount of adiabatic temperature rise when pressure is applied to a gas. Once the theory is established, the mass airflow quantity can then be calculated. We use the van der Walls equation. The following chart shows the compressor exit temperature operating under various compressor efficiencies, ranging from adiabatic (100% compressor efficiency 'n') down to 60% efficiency. Notice the massive temperature rise (as much as 180C !) when the turbo efficiency dropped down to 60%. * * * * * * * CALCULATING THE DENSITY RATIO We have to obtain density ratio first before calculating Mass Airflow EQUATION USED: Density=(T1*P2)/(T2*P1) T1=ambient+273 T2=outlet-temp.+273. P1=atmos P2=boost
 Now that we know that the temperature rise, we will calculate the air density ratio next and then we can obtain the mass airflow. Chart below shows the difference in temperature rise due to pressure. As the compressor efficiency drops, the density ratio also drops. Even if you have a turbocharger that has 100% efficiency, there is still a density ratio drop of 15%. At 60% efficiency the density ratio falls by 30% ! * * * * * * * THE DENSITY INCREASE AS TEMPERATURE DROPS We have to obtain density ratio first before calculating Mass Airflow EQUATION USED: % DENSITY INCREASE={(T1*P2)/(T2*P1)*(P2/P1)-1}*100% T1=ambient+273 T2={T1*(P2/P1)^0.283)-Td} Td=Tdrop P1=atmos P2=boost
 The range selected should be reasonably accurate up to 300°C, taking into account that the compressor is working close to its full pumping capacity. * * * * * * *

 MASS AIRFLOW FOR A 2-litre TURBO ENGINE Due to ease of calculation, air density is kept constant at 25C through out the calculation. EQUATION USED: Mass = Density x [flow x density ratio] Density of air (at 25C) = 1.2 (un-corrected) At last, we can plot the mass airflow chart of a typical 2-litre turbo charged engine, notice that the mass airflow doesn't double up when the pressure ratio is 2. The blue dot indicates the non-turbo engine. * * * * * * *
 CALCULATING THE EFFECT OF WATER INJECTION Water flow=250g/min (25C), Airflow =8.64 kg/min (124C) EQUATION USED: MaCpT1 + MwHf = MaCpT3 + MwHg T1=124C, T2=25C, T3=final Cp=Specific heat of air=1.005 Ma=mass of air/s, Mw=mass of water/s, Hf=Enthalpy of sat liquid, Hg=Enthalpy of sat vapour (obtained from steam tables) Air mass=8.64/60 = 0.144Kg/s. We need to guess the final temperature of the mixture to look up tables: Let T3=(T1+T2)/2 =74.5°C. say 75°C. From the steam tables: Hg @75°C = 2635.3 kJ/kg Saturated vapour Hf @25°C= 105kJ/Kg Saturated liquid re-arrange the equation: MaCp(T1-T3)= Mw(Hg-Hf) to: T3= 124-(2635.3-105)/(0.144*1.005)Mw = 17547Mw Substitute Mw to obtain final temperature: For Mw=0.00333Kg/s (200ml/min): T3=124-57.91=66.09°C For Mw=0.00417kg/s (250ml/min): T3=124-73.12=50.88°C For Mw=0.00500kg/s (300ml/min): T3=124-87.74=36.27°C Now that we have obtained the results for three different flow rate, a chart can be plotted to predict the intercooling effect of water. * * * * * * *
 COMPARISONS INTERCOOLER WATER INJECTION
 EFFICIENCY COST SIZE FITTING MAINTENANCE 25% - 75% HIGH BIG COMPLEX NONE 0 - 110%+ MEDIUM SMALL 2-3 HRS (1s) WATER TOP UP

 Despite the comparisons, both systems gave good results. It is in our opinion that intercooler should be your first choice unless it is physically impossible. Water injection should be your preferred choice to upgrading your existing stock intercooler. Our reason for suggesting the above preference is purely based on common sense, it is best to dump the unwanted heat into the atmosphere via the intercooler rather than spending some of the cooling properties of water which is essential for in-cylinder cooling and detonation control. The best of both world can be combined- if affordable, imagine injecting super-chilled water immediately after the intercooler, the density gain is un-imaginable ... It is worth noting that water injection is originally designed for in-cylinder cooling, we are delighted with the theoretical results.
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file updated on 1st May 2000