Water flows from the tank shown through a very short pipe. Assume the flow is quasi-steady. Esti

step 1 Here we will first write the head loss equation. By Bernoulli equation we can write the head los

Water flows from the tank shown through a very short pipe....

Key Concepts

System Design for Flow Improvement

Improving the flow system to achieve a larger flow rate involves design modifications such as increasing the effective cross-sectional area of the exit pathway, enhancing the pressure head (for example through a higher fluid level or additional pressure sources), and reducing friction or energy losses in the system by streamlining transitions and minimizing sharp bends or constrictions. These changes can significantly enhance the fluid's discharge rate by reducing resistance and maximizing the utilization of available potential energy.

Bernoulli's Principle

Bernoulli's Principle is a fundamental concept in fluid mechanics that relates pressure, velocity, and elevation in a flowing fluid. It implies that an increase in the speed of the fluid occurs simultaneously with a decrease in either the pressure or the gravitational potential energy of the fluid, assuming incompressible, non-viscous, and steady flow. This principle is often employed to analyze flow from tanks and through orifices.

Torricelli's Law

Torricelli's Law is derived from Bernoulli's Principle and provides a simple method to estimate the exit velocity of a fluid leaving an orifice under the force of gravity. It states that the speed of efflux is equivalent to the speed that a free-falling object would acquire when falling from the same height, expressed as the square root of twice the product of gravitational acceleration and the height of the fluid column.

Flow Rate Estimation

Flow rate estimation in fluid mechanics involves calculating the volumetric flow rate by multiplying the cross-sectional area of the outlet (or pipe) by the fluid's velocity. This method is widely used in practical applications to determine how much fluid is being transported per unit time, assuming conditions such as quasi-steady flow and negligible losses.

Transcript

00:01 Here we will first write the head loss equation.

00:05 By bernoulli equation we can write the head loss equation.

00:07 So it will be p1 divided by row plus alpha 1 multiplied by v1 squared divided by 2 plus gz1 minus p2 divided by row plus v2 plus alpha 2 multiplied by v2 plus gz2 plus gz2 is equal to head loss due to friction which can be written as f l v f l v square f l v2 square divided by 2d plus head loss due to head loss due to sudden contraction which can be written as kv square kv2 squared divided by 2 here since the 5 is horizontal so we can write we can write z1 is equal to z2 also let us assume the kinetic kinetic kinetic kinetic energy correction factor is equal to 1 so alpha 1 will be equal to alpha 2 alpha 1 will be alpha 2 so we can write we can write from above equation v1 squared divided by 2 plus g z 1 gz 1 minus v2 squared divided by 2 is equal to kv2 square divided by 2 let this is equation number 2 and this is equation number 1 this is equation number 1 now from the now since the flow is re -interrant type, hence from the table of minor loss coefficient for the pipe entrance, we can write k is equal to 0 .78.

01:44 Also, also, from continuity equation, we can write a1 v1 is equal to a2 v2, hence the velocity v1 will be equals to v2 multiplied by a2 divided.

02:00 By a 1 a 2 divided by a 1 now from equation 2 from equation 2 we can write we can write v2 square v2 divided by 2 multiplied by a 2 divided by a 1 whole square whole square plus g z 1 is equal to minus minus v2 squared divided by 2 is equal to kv2 squared divided by 2 so from here we can write v2 square is equal to 2 g z1 divided by 1 plus k minus a 2 by a 1 whole square so we will put all the values and we will get 2 multiplied by 9 .81 1 multiplied by 0m .8 is 1 meter divided by 1 plus the value of k is 0 .78 minus a 2 is 350 and a 1 is 3 ,500 this is square so from here we get the value v2 is equal to 3 .3 meter per second hence discharge q can be written as a2 multiplied by v2 a2 is given as 350 multiplied by 10 to the power of minus 6 multiplied by 3 .33 3 .33 so it will come out to be 1 .1 1 .17 multiplied by 10 to the power of minus 3 meter cube per second minus 3 meter cube per second so so so if if a large flow rate is required if q to be large to be large then then the outlet area then the outlet area or the outlet pipe diameter a2, a2 must be increased.

04:10 A2 has to be increased.

04:15 Has to be increased...

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