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Intake Pressure Calculation

Below is the thermodynamic proof with a concrete comparison:

  • Case A (cool/dry): 15 °C, 20% RH
  • Case B (hot/humid): 45 °C, 80% RH
  • Assume barometric pressure (P = 101.325\ \text{kPa})

1) Same power ⇒ approximately same oxygen mass flow requirement

Brake power is approximately:

Pb≈ηb,m˙f,LHVP_b \approx \eta_b ,\dot m_f, LHV


For a given operating point (same RPM, similar efficiency), holding (P_b) constant implies similar fuel flow (\dot m_f). If we hold lambda (and therefore AFR) constant:

m˙air≈AFR⋅m˙f\dot m_{air} \approx AFR \cdot \dot m_f

and because only oxygen supports combustion, “same power” implies similar required oxygen mass flow ( \dot m_{O2} ) (first order).

2) Humidity reduces the dry-air (and oxygen) partial pressure

Ambient air is a mixture of dry air + water vapour. Water vapour takes up part of the total pressure, reducing the dry-air partial pressure:

Pd=P−e, where (e) is water vapour partial pressureP_d = P – e,\ where\ (e)\ is\ water\ vapour\ partial\ pressure

Water vapour partial pressure comes from relative humidity:

e=RH⋅es(T)e = RH \cdot e_s(T)

with saturation vapour pressure (e_s(T)) (engineering approximation, Tetens):

es(T)≈0.61094,exp⁡(17.625TT+243.04) kPa ((T) in °C)e_s(T)\approx 0.61094, \exp\left(\frac{17.625T}{T+243.04}\right)\ \text{kPa} \ ((T)\ in\ °C)

Compute (e) for both cases:

At 15°C:(es≈1.70 kPa)[e15≈0.20⋅1.70=0.34 kPa][Pd,15=101.325−0.34=100.985 kPaAt\ 15 °C: (e_s \approx 1.70\ \text{kPa}) [ e_{15} \approx 0.20 \cdot 1.70 = 0.34\ \text{kPa} ] [ P_{d,15} = 101.325 – 0.34 = 100.985\ \text{kPa}
At 45°C:(es≈9.59 kPa)[e45≈0.80⋅9.59=7.67 kPa][Pd,45=101.325−7.67=93.653 kPaAt\ 45 °C: (e_s \approx 9.59\ \text{kPa}) [ e_{45} \approx 0.80 \cdot 9.59 = 7.67\ \text{kPa} ] [ P_{d,45} = 101.325 – 7.67 = 93.653\ \text{kPa}

Oxygen is a fixed fraction of dry air (about (x_{O2}\approx 0.2095)), so:

PO2=xO2,PdP_{O2} = x_{O2} , P_d
PO2,15≈0.2095⋅100.985=21.16 kPaP_{O2,15} \approx 0.2095 \cdot 100.985 = 21.16\ \text{kPa}
PO2,45≈0.2095⋅93.653=19.62 kPaP_{O2,45} \approx 0.2095 \cdot 93.653 = 19.62\ \text{kPa}

So humidity alone drops oxygen partial pressure by about:

19.6221.16≈0.927(∼7.3%)\frac{19.62}{21.16} \approx 0.927 \quad (\sim 7.3\%)

3) Temperature further reduces oxygen density (oxygen per m2)

For oxygen density in the intake air, use ideal gas:

ρO2=PO2MO2RuT\rho_{O2}=\frac{P_{O2} M_{O2}}{R_u T}

where

MO2=0.031998 kg/mol,Ru=8.314 J/(mol·K),(T)inKelvin,(T15=288.15 K),(T45=318.15 K)M_{O2}=0.031998\ \text{kg/mol}, R_u=8.314\ \text{J/(mol·K)}, (T) in Kelvin, (T_{15} = 288.15\ \text{K}), (T_{45} = 318.15\ \text{K})

Plugging in gives:

ρO2,15≈0.2826 kg/m3ρO2,45≈0.2373 kg/m3\rho_{O2,15}\approx 0.2826\ \text{kg/m}^3 \rho_{O2,45}\approx 0.2373\ \text{kg/m}^3

Ratio:

ρO2,45ρO2,15≈0.23730.2826=0.84\frac{\rho_{O2,45}}{\rho_{O2,15}}\approx \frac{0.2373}{0.2826}=0.84

Interpretation: each cubic metre of intake air at 45 °C & humid contains about 16% less oxygen than at 15 °C & dry-ish.

This 16% reduction is the combined effect of:

  • hotter air (higher (T) → lower density), and
  • higher humidity (higher (e) → lower (P_d) → lower (P_{O2})).

4) Same oxygen mass flow ⇒ higher volumetric flow (what creates suction load)

Oxygen mass flow is:

m˙O2=ρO2,V˙\dot m_{O2}=\rho_{O2},\dot V

Holding (\dot m_{O2}) approximately constant:

V˙∝1ρO2\dot V \propto \frac{1}{\rho_{O2}}

So the volumetric flow needed at 45 °C humid vs 15 °C dry is:

V˙45V˙15=ρO2,15ρO2,45≈0.28260.2373=1.19\frac{\dot V_{45}}{\dot V_{15}}=\frac{\rho_{O2,15}}{\rho_{O2,45}} \approx \frac{0.2826}{0.2373}=1.19

Result: for the same power output, the intake system may need about 19% more volumetric airflow in hot/humid conditions.

5) Intake/filter “vacuum” upstream of throttle increases ~with flow²

Most intake restrictions behave approximately like:

ΔP∝V˙2\Delta P \propto \dot V^2

Therefore:

ΔP45ΔP15≈(V˙45V˙15)2≈(1.19)2≈1.42\frac{\Delta P_{45}}{\Delta P_{15}} \approx \left(\frac{\dot V_{45}}{\dot V_{15}}\right)^2 \approx (1.19)^2 \approx 1.42