Does the V Discharge, Flow Rate, or P2 get affected if we remove the taper between V Inlet and V Discharge? For example, what would be the effects if we use a bushing going from a 2in/50A to a 1in/25A instead of a reducer like in the diagram used above?
Referencing Bernoulli: Then the decrease in the rho*g*h term (that cancelled out in the beginning) is going to tend to counteract the decrease in Static pressure
So I'm an absolute beginner trying to get a handle on fluid dynamics. One thing that confuses me is I thought a thinner pipe would normally give you a higher pressure if all other factors remained the same, but in this case there is less pressure in the smaller section of pipe. Why is that? I just can't get the two things to align in my head.
I am pretty sure that is due to the velocity. A1v1=A2V2, if A1 is greater than A2, V2 will be greater than V1. the smaller pipe has a greater velocity. To prove it just set the values, and you will see that the pressure in the smaller diameter pipe is smaller.
You are absolutely correct. The issue people stumble on is not what the continuity and bernoulli equations yield, but rather having a conceptual understanding mastery of the pressure drop independent of the equations.
Yes you can. You draw a controlled volume at the tip of the discharge where it's diameter is the same as the previous one. Essentially, you just have sqrt of 2P / density. Notice that this is the same as converting total static pressure into dynamic pressure (velocity). In other words, you are just solving bernoulli equation with the right side equals to zero..... At state two, it doesn't have kinetic energy or pressure or potential energy.
and what about situation when input and output diameters are equal and still we read from gauges a certain pressure drop? thanks for answer in advance, great video, and whole channel is great also! greetings from Poland
In a frictionless environment then there will not be a change in pressure. Once we add friction into the picture, then we find there are MANY more considerations that need to be taken into account.
@INTEGRALPHYSICS do you mean velocity and pressure or are you talking about the positions of cross section area in the formulas? I can't seem to reconfigure bernuollis law to get a calculatable answer. Change in velocity must be negative, change in pressure must be negative.
So, put simply, flow rate and pressure are inversely related. As one rises, the other decreases. At least, this is what I am understanding here. This relationship reminds of of basic electrical theory, where Amps and Volts are inversely related; as Volts decrease in a transformer, Amps increase.
If Q=A1V1=A2V2 isn't Q suppose to be 0.51x0.05^2 or 3.20x0.02^2 which is 0.00128 for the flow rate. How did you get 4.03? Okay, got it now, using pie and then conversion to L made it 4.003
Best ever explanation of Bernoulli's equation. Thanks!
Glad it was helpful!
So the equation only works if there is difference in A1 & A2? otherwise that bottom term becomes 0.
Does the V Discharge, Flow Rate, or P2 get affected if we remove the taper between V Inlet and V Discharge? For example, what would be the effects if we use a bushing going from a 2in/50A to a 1in/25A instead of a reducer like in the diagram used above?
Good work, and easy to understand.
I am confuse why the industries pumps discharge always smaller than suctions?
This is a gross generalization but typically a large inlet helps prevent cavitation at the impeller.
What if the water in pipes is moving from up to down instead of from left to right in the pipe ?
Referencing Bernoulli: Then the decrease in the rho*g*h term (that cancelled out in the beginning) is going to tend to counteract the decrease in Static pressure
@@INTEGRALPHYSICSin
So I'm an absolute beginner trying to get a handle on fluid dynamics. One thing that confuses me is I thought a thinner pipe would normally give you a higher pressure if all other factors remained the same, but in this case there is less pressure in the smaller section of pipe. Why is that? I just can't get the two things to align in my head.
Thats a great quesiton that comes up alog. The outcome is incredibly non intuitive. I'll write out an explanation and post on the channel here soon.
@@INTEGRALPHYSICS Plz don't forget about me 🙏 This is still driving me nuts lol.
Havent forgot, I was mulling this one over today.
I am pretty sure that is due to the velocity. A1v1=A2V2, if A1 is greater than A2, V2 will be greater than V1. the smaller pipe has a greater velocity. To prove it just set the values, and you will see that the pressure in the smaller diameter pipe is smaller.
You are absolutely correct. The issue people stumble on is not what the continuity and bernoulli equations yield, but rather having a conceptual understanding mastery of the pressure drop independent of the equations.
Hi can I use this method to calculate the flow rate with one end open to the atmosphere (Pa=0)? if not, what could be the reason?
Yes you can. You draw a controlled volume at the tip of the discharge where it's diameter is the same as the previous one. Essentially, you just have sqrt of 2P / density. Notice that this is the same as converting total static pressure into dynamic pressure (velocity).
In other words, you are just solving bernoulli equation with the right side equals to zero..... At state two, it doesn't have kinetic energy or pressure or potential energy.
Great job, man.
and what about situation when input and output diameters are equal and still we read from gauges a certain pressure drop? thanks for answer in advance, great video, and whole channel is great also! greetings from Poland
In a frictionless environment then there will not be a change in pressure. Once we add friction into the picture, then we find there are MANY more considerations that need to be taken into account.
Do you have a example with reversed direction ?
If you mean going from a small ID to a large ID. Just flip the values, the eqn still holds up.
what do you mean by flip the values?
@@Razan220he meant the velocity and the pressure.
@INTEGRALPHYSICS do you mean velocity and pressure or are you talking about the positions of cross section area in the formulas?
I can't seem to reconfigure bernuollis law to get a calculatable answer.
Change in velocity must be negative, change in pressure must be negative.
Thank you. You solved my problem. ❤🔥❤🔥❤🔥❤🔥❤🔥❤🔥
I love this problem. Im glad you found it helpful.
Does this work for differential pressure between inlet and outlet of a pump?
Careful, there is an influx of energy at the pump impeller, you can't apply Bernoulli's across the pump like that.
So, put simply, flow rate and pressure are inversely related. As one rises, the other decreases. At least, this is what I am understanding here.
This relationship reminds of of basic electrical theory, where Amps and Volts are inversely related; as Volts decrease in a transformer, Amps increase.
True, Untill the flow rate is less than sonic velocity
I wonder how v2 is 3.20m/s? It is not 3.45m/s?
I think you forgot to square your area when solving for velocity.
I mean you did square it for the answer, so your answer is right but not write it out
If Q=A1V1=A2V2 isn't Q suppose to be 0.51x0.05^2 or 3.20x0.02^2 which is 0.00128 for the flow rate. How did you get 4.03?
Okay, got it now, using pie and then conversion to L made it 4.003
I skipped a bunch of algebra steps to cram the problem onto one page... admittedly not my slickest work.
Q=A1V1=A2V2, How a throttling valve controls the flow, Please someone help me to understand
what if A1=A2?
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