An unknown liquid flows smoothly through a 6.0-mm-diameter hor...

Question

An unknown liquid flows smoothly through a 6.0-mm-diameter horizontal tube where the pressure gradient is 600 Pa/m. Then the tube diameter gradually shrinks to 3.0 mm. What is the pressure gradient in this narrower portion of the tube?

Accepted Answer

Hi everyone. In this practice problem, we are going to determine a fiscus liquids pressure region. When we have a fiscus liquid flowing steadily through a horizontal duck with a non uniform cross section, the diameter of the horizontal duct decreases from one centimeter at point A to 0.5 or half centimeter to point at point B. The liquid experiences uh 500 pass KS per meter pressure region as it travels through point A and we're being asked to determine the liquid pressure region as it travels through point B. The options given are a 500 pascals per meter. B, one kg pascals per meter, C four kg pascals per meter. And lastly D eight kg pascals per meter. So for this particular practice problem, since the viscosity is not negligible, we will have to solve this problem using the quasi equation. So the quasi equation relates the flow rate to the pressure region, the volume flow rate with the pressure gradient, the viscosity, the radius of the duck and also the length of the tube. So the quasi equation is given by Q equals to pi multiplied by R to the power of four divided by eight U multiplied that by delta B and divided by L. So Q is the politic flow rate. Mu is the fluid fiss R is the radius of the duck. Delta P is the pressure difference that keeps the fiscus fluid fluid flowing. And L is the length of the tube assuming that the liquid is incompressible. The flow is also laminar, we can use the continuity equation to relate the flow rate at points A and point B off the duck. So using the continuity equation, I'm gonna write that down continuity equation, we will have a Q or the folia fluorid at A well equals to Q or the folia fluorid at B Q A and Q B can be replaced using the quasi equation given here. So that we can obtain pi multiplied by R A to the power of four divided by A mu multiplied by delta B divided by L of A. Well equals to pi multiplied by R B to the power of four divided by eight mu multiplied by delta P divided by L of B. So if we notice in this two equation, we are actually solving for delta P divided by L of B which is going to be the pressure gauge as it travels through point B, we can simplify this equation by crossing out the P uh the pi from both sides and a mu also from both sides. So essentially, we will get the equation after the simplification of R A to the power of four, uh multiplied by delta B divided by L of A to be equals to R B to the power of four multiplied by delta B divided by L of B just like. So, so next, we wanna rearrange this equation that we have simplified in order for us to get what we are looking for, which is delta B over or divided by L of B, which will then equals to open parentheses. R A divided by R B. All close parentheses, all to the power of four multiplied by delta B divided by L of A just like. So, so now we can actually substitute all of the known information given to get delta B divided by L. So delta P divided by L of B will then equals to R A is given to be one centimeter because we are just um taking essentially the ratio of R A divided by R B. I'm just going to leave that N centimeter and then R B is given to be 0.5 centimeter and all of that is going to be on to the power of four. It gonna multiply it by delta P divided by L FA which is given to be pastels per meter in the problem statement. So calculating this, this will give us delta P O divided by L of B to B 8000 pascals per meter or in kilo pascals or multiplying it by 10 to the power of negative three kilo pascals divided by one pascals. We will get the final answer for delta B divided by L of B or the pressure region of B or going into um point B to B eight kg pascals per meter. So that will be the final answer which is going to be corresponding to option D in our answer choices. So option D with the liquids pressure gradient as it travels through point B to be eight kg pascals per meter will be the answer to this particular practice problem. So that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

An unknown liquid flows smoothly through a 6.0-mm-diameter horizontal tube where the pressure gradient is 600 Pa/m. Then the tube diameter gradually shrinks to 3.0 mm. What is the pressure gradient in this narrower portion of the tube?

Key Concepts

Continuity Equation

Bernoulli's Principle

Pressure Gradient

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