--- title: "Volumetric efficiency from orifice air measurement" subject: "Internal Combustion Engines" chapter: 8 key: "semester_6/internal_combustion_engines:volumetric-efficiency-orifice-air-consumption" source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/volumetric-efficiency-orifice-air-consumption site: BCT Engineering --- # Volumetric efficiency from orifice air measurement Internal Combustion Engines · Chapter 8: Engine performance and testing of engines - Asked in: year not given, 6 marks - Video: [Internal Combustion Engines: The volumetric Efficiency | Dr. Samer Ali](https://www.youtube.com/watch?v=9kyO8lSRlb0) — Professor Samer Ali - Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/volumetric-efficiency-orifice-air-consumption - Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-8 ## Question A four-cylinder, four-stroke engine of bore 90 mm and stroke 110 mm runs at 2500 rpm. Air consumption is measured by a sharp-edged orifice of diameter 48 mm and coefficient of discharge 0.62, fitted to an air box; the water manometer reads 100 mm. Ambient air is at 101.3 kPa and 300 K. The fuel consumption is 9.5 kg/h. Find the volumetric efficiency and the air-fuel ratio. Take $R = 287$ J/kg K. ## Short answer Air density 1.177 kg/m3, orifice head 85.0 m air, flow 0.0458 m3/s. Swept volume rate 0.0583 m3/s gives ηv =78.6%. Air mass flow 194.1 kg/h, fuel 9.5 kg/h, A/F =20.4. ## Answer #### Air density and orifice head $$ \begin{aligned} \rho_a &= \frac{p}{RT} = \frac{101\,325}{287 \times 300} = 1.177\ \text{kg/m}^3 \\ H_a &= h_w \frac{\rho_w}{\rho_a} = 0.100 \times \frac{1000}{1.177} = 84.97\ \text{m of air} \end{aligned} $$ #### Air flow through the orifice $$ \begin{aligned} A_o &= \frac{\pi}{4}(0.048)^2 = 1.810 \times 10^{-3}\ \text{m}^2 \\ \dot{V}_a &= C_d A_o \sqrt{2 g H_a} = 0.62 \times 1.810 \times 10^{-3} \times \sqrt{2 \times 9.81 \times 84.97} \\ &= 0.62 \times 1.810 \times 10^{-3} \times 40.83 = 0.04581\ \text{m}^3/\text{s} \end{aligned} $$ Air mass flow $= 0.04581 \times 1.177 = 0.05391$ kg/s $= 194.1$ kg/h. #### Swept volume rate $$ \begin{aligned} V_s &= \frac{\pi}{4}(0.09)^2(0.11) = 6.998 \times 10^{-4}\ \text{m}^3 \\ \dot{V}_s &= 4 \times 6.998 \times 10^{-4} \times \frac{2500}{2 \times 60} = 0.05832\ \text{m}^3/\text{s} \end{aligned} $$ #### Volumetric efficiency and A/F $$ \begin{aligned} \eta_v &= \frac{0.04581}{0.05832} = 0.7855 \\ A/F &= \frac{194.1}{9.5} = 20.4 \end{aligned} $$ An air box is used between the engine and the orifice to damp the pulsation of the intake, so that the orifice reading is steady. **Answer:** $\eta_v = 78.6\%$ and $A/F = 20.4$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the manometer reading converted to head of air? A. H_a = h_w x rho_w / rho_a = 84.97 m B. H_a = h_w x rho_a / rho_w C. H_a = h_w = 0.1 m D. H_a = h_w x g Answer: A. 0.100 x 1000/1.177 = 84.97 m of air. (Key point: Orifice head in air) 2. What expression gives volume flow through the orifice? A. Cd A sqrt(2 g H_a) = 0.0458 m^3/s B. A sqrt(H_a) with no Cd C. Cd A x H_a D. Cd A / sqrt(2 g H_a) Answer: A. V = 0.62 x 1.810e-3 x sqrt(2 x 9.81 x 84.97) = 0.04581 m^3/s. (Key point: Air flow through orifice) 3. How is volumetric efficiency calculated? A. Actual air volume flow / swept volume rate (with N/2 for four-stroke) = 78.6% B. Fuel flow / air flow C. Swept volume / actual air D. Air mass / fuel mass Answer: A. eta_v = 0.04581/0.05832 = 0.786, the swept rate using N/2 cycles per second times 4 cylinders. (Key point: Volumetric efficiency) 4. Why is an air box used between the engine and the orifice? A. To heat the air B. To damp the intake pulsations so the orifice reading is steady C. To mix fuel with air D. To measure the fuel Answer: B. The air box smooths the pulsating intake flow so the manometer reading is steady. 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