Air flow rate in a test of an internal combustion engine is...

Air flow rate in a test of an internal combustion engine is to be measured using a flow nozzle installed in a plenum. The engine displacement is 1.6 liters, and its max. imum operating speed is 6000 rpm. To avoid loading the engine, the maximum pressure drop across the nozzle should not exceed $0.25 \mathrm{m}$ of water. The manometer can be read to $\pm 0.5 \mathrm{mm}$ of water. Determine the flow nozzle diameter that should be specified. Find the minimum rate of air flow that can be metered to ±2 percent using this setup.

Key Concepts

Measurement Resolution and Uncertainty

This concept deals with the precision and accuracy of instruments used to measure physical quantities. In the context of differential pressure flow measurement, the smallest readable pressure drop (or the instrument’s resolution) determines the minimum detectable change in flow rate. The instrument’s uncertainty has to be low enough compared to the measurement range to ensure that the flow rate can be determined within the desired accuracy level, such as within a specified percentage error.

Flow Nozzle Sizing

This concept involves designing the geometry of a nozzle so that it causes an appropriate pressure drop in the flow while still allowing sufficient mass flow. The sizing is determined by balancing the need for a measurable differential pressure (using principles like Bernoulli’s equation) with restrictions imposed by the system, such as preventing excessive back pressure or flow disruption. It typically requires relating the diameter of the nozzle to the flow velocity and the allowable pressure drop.

Differential Pressure Flow Measurement

This principle is based on the fact that a change in fluid velocity through a constriction leads to a drop in pressure, as described by Bernoulli’s equation. By measuring the pressure difference across the nozzle, the flow rate can be determined using appropriate discharge coefficients and correction factors. This method is widely applied in fluid dynamics for accurately gauging mass or volume flow rates in various systems.

Engine Air Flow Requirements

Understanding the air flow requirements of an engine is crucial when designing measurement systems for internal combustion engines. It involves calculating the expected flow rate based on engine displacement and operating speed, ensuring that the measurement device (like a nozzle) can accurately capture the engine’s full range of airflow without imposing undue restrictions or loads on the engine.

Transcript

00:01 The displacement of the engine q is given as 1 .61 liter 1 .61 liter which can be written as 1 .61 multiplied by 10 to minus 3 meter cube.

00:17 Now the operating speed is given as omega is equals to 6000 rpm rpm so it will be equals to 6 ,000 divided by 60 revolution per second it will be equal to 100 revolution per second now the pressure drop pressure drop is given at stalled h is equal to 0 .25 meter of water is equal to 250 mm of water in manometer it will have a variation of is equal to plus minus 0 .5 mm now now let's calculate the mass flow rate it will be row multiplied by q let us take this equation as equation 1 so let's put the value of row of air is equal to 1 .23 multiply the hue is given as 1 .61 multiplied by 10 to the power of minus 3 divided by 2 multiplied by 100 here n is equal to 2 mass flow rate will be halved for n is equal to 2 mass flow rate way will be out so m dot will be 0 .098 4 kg per second now the mass flow rate equation for a flow meter is given by m actual is equal to cat cat divided by square root of 1 minus beta 2d power of 4 multiplied by root over 2 row del p.

02:15 Let this equation be question to assume high reynolds number re is high higher than 2 multiplied by 10 to the power of 5 so so the coefficient of discharge will be 0 .99 now assume assume beta is equal to 0 del p is equal to row g h so mass flow rate m -dess will be c 80 c80 square root of 2 row multiplied by row water g del h so it will come out to be so from here we get 80 is equal to m dot divided by c root over 2 row 2 row h2 o g del edge so this will be equal to m dot is given as 0 .0984 divided by c the value of c is 0 .99 0 .99 multiplied by square root of square root of 2 multiplied the density is 1 .23 multiplied by row of water is 10 to part 3 5 9 .8 1 multiplied by 0 .85 so so after calculating we get the 80 is equal to 1 .28 multiplied by 10 20 power of minus 3 1 .28 multiplied by 10 to the power of minus 3 meter square so this is the throat area so it can be written as pi divided 4 d t square is equal to 1 .28 multiplied by 10 to the power of minus 3 hence from here we get d t is equal to 40 .3 40 .3 m m 40 .3 m 40.

04:38 40 .3 m this is the throat diameter.

04:42 This is throat diameter.

04:44 Throat diameter.

04:46 Now, now, given manometer can read plus minus, plus minus of water.

04:58 If there is plus minus 2 % error, so if error is plus minus 0 .02, then, then, then, relative uncertainty, relative unsurtenity, relative unsurrected...

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