Six Air Compressor Calculation Formulas ~ Examples Demonstration

1. Unit conversion inference between “Nm³/min” and “m³/min”

We know that 1Nm³/min≈1.07m³/min. So how did this conversion come about? Firstly, let’s understand the concept of Nm³/min. The definition of the standard state of air is: air pressure 0.1MPa. The temperature is 0°C. In addition, humidity is 0%. In the standard state, the density of air is 1.185kg/m³. If use it as the unit of air capacity of air compressor and dryer, it should be written as Nm³/min. This is so-called “standard cube”. The equation of state for an ideal gas: pV=nRT. This formula describes the change law of ideal gas state. It’s resented by Clapeyron. Among them: p is the pressure (Pa); V is the volume (m³); n is the amount of substance (mol); T is the absolute temperature (K Kelvin); R is the universal gas constant (about 8.31); In both states, the pressure, the amount of matter, and the constant are the same. The only difference is the temperature (thermodynamic temperature K). Derive: V₁/T₁=V₂/T₂ (that is, Guy Lussac’s law). Assume: V₁, T₁ are standard cubic, V₂, T₂ are cubic Then: V₁: V₂=T₁: T₂ That is: V₁: V₂=273: 293 Therefore: V₁≈1.07V₂ Result: 1Nm³/min≈1.07m³/min Common conversion of flow units of air compressor: ① 1m3/min = 1000L/min; ② 1Nm3/min = 1.07 m3/min; ③ 1CFM =0.02832 M3/min; ④ 1M3/min=35.311CFM; ⑤ CFM — ft3/min, cubic feet per minute;

2. Calculate the oil fuel consumption of the air compressor

Suppose there is an 8bar air compressor. Power is 250kW. The discharge air flow is 40m³/min. In addition, the oil content is 3PPM. If the air compressor runs for 1000 hours, how many liters of oil fuel are theoretically consumed? Answer: Fuel consumption per cubic minute —— 3×1.2=3.6mg/m3; 40 cubic meters of fuel consumption per minute ——40×3.6/1000=0.144g; Fuel consumption for 1000 hours of operation ——1000×60×0.144=8640g=8.64kg; Convert to volume. That is 8.64/0.8=10.8L (The density of lubricating oil is about 0.8) Of course, the above are only theoretical oil consumption. In practice, it is larger than this value. Because the oil filter element filtration ability continues to decline. Assume to caculate it according to 4000 hours. A 40m³/min air compressor runs at least 40 liters of oil. Usually 40m³/min air compressors are refueled with about 10~12 barrels (18 liters/barrel) per maintenance. Oil consumption is about 20%.

3. On the plateau, air compressor air consumption calculation

Calculate the air flow of the air compressor from the plain to the plateau: Reference formula: V₁/V₂=R₂/R₁; V₁ = air flow in the plain area; V₂ = air flow in the plateau area; R₁ = compression ratio of the plain; R₂ = compression ratio of the plateau; Example: There is an 110kw air compressor. Exhaust pressure is 8bar. And air flow 20m³/min. What is the discharge air flow of this machine at an altitude of 2000 meters? (Assume the atmospheric pressure at 2000 meters is 0.85bar.) Solution: According to the formula V₁/V₂=R₂/R₁; (label 1 is plain, 2 is plateau) V₂=V₁R₁/R₂; R₁=9/1=9; R₂=(8+0.85)/0.85=10.4; V₂=20×9/10.4=17.3m³/min; Therefore, the dicharge volume of this air compressor at an altitude of 2000 meters is 17.3m³/min. It means that if this air compressor is used in the plateau area, the air flow will be significantly attenuated.

4. Calculation of Air Consumption of Pneumatic Tools

The calculation method of the air consumption of the air source system when each pneumatic device works intermittently: Q=0.5ψ×K1×K2×K3×Qmax m³/min; Among them: Qmax – the actual maximum air consumption required. ψ – utilization factor. It takes into account factors that all pneumatic equipment will not use at the same time. The empirical value of the coefficient is 0.95 to 0.65. In general, the more pneumatic devices there are, the less they are used at the same time. Then its value is small. On the contrary, its value is large. For 2 devices, take 0.95. For 4 devices, take 0.9. If 6 devices, take 0.85. For 8 devices, take 0.8. For more than 10 devices, take 0.65. K1 – leakage coefficient. The value is selected in the range of 1.2 to 1.5. K2 – Spare factor. The value is selected in the range of 1.2 to 1.6. K3 – Uneven coefficient. In addition, it is considered that there are uneven factors in the calculation of average air consumption. In order to ensure its maximum usage, and its value is selected within the range of 1.2 to 1.4.

5. When the air flow is insufficient, calculate the air consumption difference

Sometimes, the increase in the pneumatic equipment will lead to insufficient air supply. To maintain the rated working pressure, how much air compressor needs to be added. Formula: ΔQ=Q actual – Q original=Q original*(P original – P actual)/P actual + 0.1013 Among them: Q actual – the air compressor flow required by the system in the actual state; Q original – the volume flow of the original air compressor; P actual – the pressure MPa that can be achieved under actual conditions; P original – the working pressure MPa that can be achieved by the original use; ΔQ – air flow to be increased (m3/min). Example: The original air compressor is 10m³/min and 8 bar. User added pneumatic equipment. The current air compressor pressure can only hit 5 bar. So what size of air compressor you need to add to meet the air demand of 8 bar. According to the formula, △Q=10*(0.8-0.5)/(0.5+0.1013)≈4.99m³/min. Therefore, it at least needs an air compressor with 4.99m³/min and 8bar. In fact, the principle of this formula is as follows: Calculate the difference from the target pressure as a proportion of the current pressure. Apply this ratio to the flow rate of the air compressor currently in use. That is, the difference from the target flow is obtained.

6. Calculate the specific power of the screw air compressor

The only criterion for determining whether the screw air compressor is energy-saving is the “specific power”. Suppose there is a screw air compressor machine. Its parameters are as follows: air flow = 24m/min, working pressure = 7Bar; motor rated power (rated output power or rated shaft power): P = 132kw; η = 94.7% power factor: COSφ=0.892, service factor S.F=1.15 (other manufacturers also choose the service factor S.F=1.2). Based on the above parameters, we can know: The nominal rated input power of the motor of this machine (without considering the service factor and at full load): P1 = (motor rated output power P ÷ motor efficiency η) = 132kw ÷ 94.7% = 139.39kw The nominal rated power input of this machine (considering the service factor and at full load): P2 = (main motor rated output power P ÷ main motor efficiency η) x (service factor S.F-0.05)= (132kw ÷ 94.7%) x (1.15 – 0.05)= 153.33kw (Note: In theory, we should consider a 5% margin when calculating the service factor. And it cannot be calculated in full) The nominal specific power of the screw air compressor (at 7bar, considering the service factor and full load): PB1 = P2 ÷ 24 m3/min= 6.39kw/( m/min) In the case of an air-cooled compressor, it also needs to consider the input power of the fan. If the screw air compressor is air-cooled, the rated power of the fan motor is 4.5kw, and the efficiency is 85%, the input power consumption of the fan motor is: PF = 4.5kw ÷ 85% = 5.29kw Therefore, the nominal total input power of the screw air compressor machine (considering the fan power consumption and the service factor and at full load): PZ = P2 + PF = 153.33 + 5.29 = 158.62 kw The nominal specific power of the screw air compressor air-cooled machine (at 7bar, considering the service factor and full load): PB2 = PZ ÷ 24 m3/min = 158.62 ÷ 24 = 6.61kw/(m/min).