A venturi meter is often used to measure the flow rates of liquids passing throu

Question

A venturi meter is often used to measure the flow rates of liquids passing through pipes. To accomplish this kind of measurement, one needs to vary the pressure within the liquid by causing the speed of the liquid to vary as it flows. This is done using a pipe which tapers in its radius from large to small. The diameter of the larger pipe shown in the diagram below is 42 cm while the smaller pipe has a diameter of 24 cm. You measure the ‘head’ (aka the difference between the heights h1 and h2 in the measuring tubes) to be 6 cm.

A.) Assuming the liquid is water, find an expression for the speed of the water as it flows through pipe 2 as a function of the speed with which it flows through pipe 1. Hint: This is a liquid which is effectively incompressible, so ‘flow rate in’ should equal ‘flow rate out’.

B.) Use Pascal’s 2nd Law to get an expression for the pressure at the center of pipe 1 in terms of the column height h1. Do the same for the pressure at the center of pipe 2 as a function of the column height h2. Note that both the measuring tubes are open to their environment. Lastly, find the difference between these pressures, P1 – P2, as a function of the fluid density and the heights h1 and h2. Hint: You don’t need to know h1 nor h2 individually…you can get away with knowing the difference between them instead!

C.) Coming at this pressure difference from another route, apply Bernoulli’s Equation to find the difference P1 – P2 as a function of fluid density and the speeds v1 and v2. Note that the y-coordinates describe the position of the bit of water flowing, NOT the heights of columns in the measurement tubes. The water flowing through the center of the pipe along its axis is at the same y-coordinate at both point 1 and point 2 (define that y-coordinate to be zero if so desired).

D.) Knit the two approaches together to find an equation that includes ONLY v1 and v2 as variables. Remember, you know the difference in the column heights within the measuring tubes to be 6 cm. Does it matter if we used water as our flowing liquid or would any liquid produce the same results?

E.) The equation you derived in part ‘d’ above has two unknowns in it (the speeds at each point). Fortunately we have another equation with these same two variables in it courtesy of part ‘a’. Use these two equations to solve for the two speeds.

F.) Finally, use the speeds from the previous step to double check that the flow rates are the same through points 1 and 2. Calculate these flow rates explicitly, complete with proper units. Are they the same (within rounding errors from your calculators)?

Explanation / Answer

Here ,

as water is incompressible

using equation of continuity

A1 * v1 = A2 * v2

v2 = A1 * v1/A2

the expression of speed in pipe 2 is A1 * v1/A2

B)

using pascal's second law ,

P1 = p * g * h1

p is the density of water

for the pressure 2 ,

P2 = p * g * h2

for the difference in pressure

P1 - P2 = p * g * h1 - p * g * h2

P1 - P2 = p * g * (h1 - h2)

C)

Using bernoulli's equation

P1 - P2 = 0.5 * p * (v2^2 - v1^2)

the pressure difference is P1 - P2 is 0.5 * p * (v2^2 - v1^2)

d)

for the equations in c and b

p * g * (h1 - h2) = 0.5 * p * (v2^2 - v1^2)

g * (h1 - h2) = 0.5 * (v2^2 - v1^2)

(v2^2 - v1^2) = 9.8 * 0.06 * 2

(v2^2 - v1^2) = 1.176

this is the equation ,

No , the liquid does not matter , as long as it is incompressable

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