Collection of Solved Problems in Physics

The Venturi Tube

Task number: 1162

The air of velocity 15 m/s and of density 1.3 kg/m³ is entering the Venturi tube (placed in the horizontal position) from the left. The radius of the wide part of the tube is 1.0 cm; the radius of the thin part of the tube is 0.5 cm. The tube of shape U connecting wide and thin part of the main tube (see the picture) is filled with the mercury of the density 13 600 kg/m³. Determine the height difference that stabilizes between the surfaces of the mercury in U-tube.

• Notation

  v1 = 15 m/s	Velocity of the entering air
  r1 = 1.0 cm	Radius of the wide part of the tube
  r2 = 0.5 cm	Radius of the thin part of the tube
  ρ1 = 1.3 kg/m3	Density of air
  ρ2 = 13 600 kg/m3	Density of the mercury
  Δh = ?	Height difference of the surfaces in U-tube

• Hint 1

  What causes different heights of mercury in the arms of the U-tube?

• Solution to Hint 1

  The cause of height differences is the separate air pressure above its surface. In the left arm the pressure air p₁ is higher and is pushing the mercury lower into the tube. On to the contrary the pressure air p₂ is lower in the right arm and the mercury can rise higher.

• Hint 2

  Can you interconnect mathematically the separate air pressure in the arms of the U-tube with the found difference Δh? Examine the size of the pressure in the lower level of the surface h₀.

• Solution to Hint 2

  Arms of the U-tube represent communicating vessels and so the sum of the hydrostatic pressure and the air pressure in both arms must be equal in low surface level h₀. Thus:

  \[p_1\,=\,p_2\,+\,p_h\tag{1}\]

  where p₁, p₂ are air pressures in the arms of the U-tube and p_h the hydrostatic pressure of the mercury column of height Δh. We substitute the known formula for the magnitude of the hydrostatic pressure into the equation (1) express found difference Δh:

  \[p_1\,=\,p_2\,+\,{\Delta}h\varrho_2g\,\Rightarrow\,{\Delta}h\,=\,\frac{p_1\,-\,p_2}{\varrho_2g}\tag{2}\]

  where ρ₂ is the density of the mercury and  g is the gravitational acceleration.

• Hint 3

  Now we will determine the pressure differences in the numerator of expression (2). Express it from Bernoulli’s principle which describes the steady flow in the main tube, and substitute it into relation for Δh.

• Solution to Hint 3

  We can write Bernoulli’s principle for the stabilized flow of the air in the Venturi tube in the form:

  \[p_1\,+\,\frac{1}{2}{\varrho}_1v_1^2\,=\,p_2\,+\,\frac{1}{2}{\varrho}_1v_2^2\tag{3}\]

  Where the left size of the equation describes the place of the radius of tube S₁ and the right size of the radius of tube S₂. At the same time v₁, and v₂ are the velocities of the air flowing in these places, and ρ₁ is the density of the air. (Because the tubes are in the horizontal position, we don’t have to take in consideration the changes of the potential energy of the flowing air in Bernoulli’s principle.) It is given for the found difference of the pressure:

  \[p_1\,-\,p_2\,=\,\frac{1}{2}{\varrho}_1(v_2^2\,-\,v_1^2)\tag{4}\]

  Substituting this relation into equation (2) we get:

  \[{\Delta}h\,=\,\frac{{\varrho}_1(v_2^2\,-\,v_1^2)}{2{\varrho}_2g}\tag{5}\]

• Hint 4

  What quantity from relation (5) do we have to find out yet? Use the equation which describes steady flow fluid and the quantity which we are looking for will be found in the equation. Express this quantity and substitute it into relation (5).

• Solution to Hint 4

  The last unknown quantity in relation (5) is the velocity of air v₂ at the point of radius S₂. We use the continuity equation for steady flow to determine this velocity:

  \[S_1v_1\,=\,S_2v_2\tag{6}\]

  Assume the circular radius of the tube, so you can substitute for the surfaces S₁, S₂ and express the velocity v₂:

  \[{\pi}r_1^2v_1\,=\,{\pi}r_2^2v_2\,\Rightarrow\,v_2\,=\,v_1\frac{r_1^2}{r_2^2}\tag{7}\]

  where r₁, r₂ are assigned radiuses of corresponding parts of the tube. Now substitute the calculated velocity into relation (5) and we get:

  \[{\Delta}h\,=\,\frac{{\varrho}_1}{2{\varrho}_2g}((v_1\frac{r_1^2}{r_2^2})^2\,-\,v_1^2)\,=\,\frac{{\varrho}_1v_1^2}{2{\varrho}_2g}(\frac{r_1^4}{r_2^4}\,-\,1)\tag{8}\]

  Numerically:

  \[{\Delta}h\,=\,\frac{1.3{\cdot}15^2}{2{\cdot}13\,600{\cdot}9.81}(\frac{0.010^4}{0.005^4}\,-\,1)\,\mathrm{m}\,=\,0.016\,\mathrm{m}\,=\,1.6\,\mathrm{cm}\]

  The difference between the levels of height of mercury in arms of U-tube is approximately 1.6 cm.

• Complete solution

  The cause of height differences is the separate air pressure above its surface. In the left arm the pressure air p₁ is higher and is pushing the mercury lower into the tube. On to the contrary the pressure air p₂ is lower in the right arm and the mercury can rise higher. Because the arms of the U-tube represent communicating vessels, sum the hydrostatic pressure and the pressure of air in both arms must be equal in low level of surface h₀. Thus:

  \[p_1\,=\,p_2\,+\,p_h\tag{1}\]

  where p₁, p₂ are air pressures in the arms of the U-tube and p_h ph the hydrostatic pressure of the mercury column of height Δh. We substitute the known formula for the magnitude of the hydrostatic pressure into the equation (1) express found difference Δh:

  \[p_1\,=\,p_2\,+\,{\Delta}h\varrho_2g\,\Rightarrow\,{\Delta}h\,=\,\frac{p_1\,-\,p_2}{\varrho_2g}\tag{2}\]

  where ρ₂ is the density of the mercury and g is the gravitational acceleration.

  Now your task is to express the difference of the pressure in the numerator in relation (2). Use Bernoulli’s principle which for steady flow in the Venturi tube has this form:

  \[p_1\,+\,\frac{1}{2}{\varrho}_1v_1^2\,=\,p_2\,+\,\frac{1}{2}{\varrho}_1v_2^2\tag{3}\]

  Where the left size of the equation describes the place of the radius of tube S₁ and the right size of the radius of tube S₂. At the same time v₁, v₂ are the velocities of the air flowing in these places, and ρ₁ is the density of the air. (Because the tubes are in the horizontal position, we don’t have to take in consideration the changes of the potential energy of the flowing air in Bernoulli’s principle.) It is given for the found difference of the pressure:

  \[p_1\,-\,p_2\,=\,\frac{1}{2}{\varrho}_1(v_2^2\,-\,v_1^2)\tag{4}\]

  Substituting this relation into equation (2) we get:

  \[{\Delta}h\,=\,\frac{{\varrho}_1(v_2^2\,-\,v_1^2)}{2{\varrho}_2g}\tag{5}\]

  The last unknown quantity in relation (5) is the velocity of air v₂ at the point of radius S₂. We use the continuity equation for steady flow to determine this velocity:

  \[S_1v_1\,=\,S_2v_2\tag{6}\]

  Assume the circular radius of the tube, so you can substitute for the surfaces S₁, S₂ and express the velocity v₂:

  \[{\pi}r_1^2v_1\,=\,{\pi}r_2^2v_2\,\Rightarrow\,v_2\,=\,v_1\frac{r_1^2}{r_2^2}\tag{7}\]

  where r₁, r₂ are assigned radiuses of corresponding parts of the tube. Now substitute the calculated velocity into relation (5) and we get:

  \[{\Delta}h\,=\,\frac{{\varrho}_1}{2{\varrho}_2g}((v_1\frac{r_1^2}{r_2^2})^2\,-\,v_1^2)\,=\,\frac{{\varrho}_1v_1^2}{2{\varrho}_2g}(\frac{r_1^4}{r_2^4}\,-\,1)\tag{8}\]

  Numerically:

  \[{\Delta}h\,=\,\frac{1.3{\cdot}15^2}{2{\cdot}13\,600{\cdot}9.81}(\frac{0.010^4}{0.005^4}\,-\,1)\,\mathrm{m}\,=\,0.016\,\mathrm{m}\,=\,1.6\,\mathrm{cm}\]

  The difference between the levels of height of mercury in arms of U-tube is approximately 1.6 cm.

• Answer

  \[{\Delta}h\,=\,\frac{{\varrho}_1v_1^2}{2{\varrho}_2g}(\frac{r_1^4}{r_2^4}\,-\,1)\]

  Difference between the heights of the surfaces of mercury in the arms of U-tube is approximately 1.6 cm.

• Using the Venturi tube

  In practise the Venturi tube is used in planes to measure the flight velocity. It provides more exact measurement than for example the common Pitot tube but it is intended rather for smaller flight velocities – the speed of sound is reached in the thinner part at small velocities.