Question

According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something funny or surprising. In spit takes, a gauge pressure is applied in the mouth, p1, so that liquid flows through pursed lips forming a column of liquid with radius r2 = 4 mm. If the liquid travels at v2 = 3.1 m/s outside the body, and if the column's area is 10x larger inside the mouth, what is p1 in Pa?

Answer 1

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

     The gauge pressure at the mouth is  $p_1$

     The radius of the column is  $r_2 = 4 \ mm = 0.004 \ m$

    The speed of the liquid outside the body is  $v_2 = 3.1 \ m/s$

      The area of the column is  $A_2$

       The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

       $v_1 A_1 = v_2 A_2$

=>       $v_ 1 = v_2 * \frac{A_2}{A_1}$

=>      $v_ 1 = 3.1 * \frac{1}{10}$

=>        $v_ 1 = 0.31 \ m/s$

So

      $A_1 = 10A_2$

=>   $\pi * r_1^2 = 10(\pi * r_2^2)$

=>   $r_1 = 10 * r_2$

substituting values

        $r_1 = 10 * 0.004$

        $r_1 =0.04 \ m$

Now the height of inside the mouth is  $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is  $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's  equation

        $p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now  $\rho = 1000 \ kg m^{-3}$ which is the density of water

        $p_2$ is the gauge pressure of the atmosphere which is  zero

 So

       $p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

                                                  $[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$                          

       $p_1 = 4051.4 \ Pa$

Answer 2

The initial pressure of the gauge pressure at the mouth is 4757 pascals.

The principle of continuity equation asserts that in a steady flow state, the quantity of fluid flowing at the inlet is equivalent to the quantity of fluid at the outlet given that there is a constant mass flow rate.

It can be expressed by using the formula:

$\mathbf{A_1v_1=A_2v_2}$

where;

• The speed of the liquid inside the body is $\mathbf{v_1}$ = ???
• The speed of the liquid outside the body is $\mathbf{v_2}$ = 3.1 m/s
• Area outside the column is = A₂
• Area inside the column is $\mathbf{A_1}$ = 10A₂

∴

Making v₁ the subject of the formula:

$\mathbf{v_1 = \dfrac{A_2}{A_1}\times v_2}$

$\mathbf{v_1 = \dfrac{1}{10}\times3.1}$

$\mathbf{v_1 =0.31 \ m/s}$

Since the area are equivalent to each other

$\mathbf{A_1 = 10A_2}$

$\mathbf{\pi r^2_1 = 10\times \pi r^2_2}$

$\mathbf{ r^2_1 = 10\times r^2_2}$

where;

• r₂ = radius of the column. = 4mm = 0.004 m

∴

$\mathbf{ r_1 = 10\times 0.004}$

$\mathbf{ r^2_1 = 0.04 \ m}$

However, in fluid dynamics, Bernoulli's equation can be applied for the estimation of the initial gauge pressure by using the expression:

$\mathbf{\dfrac{1}{2} \rho v_1^2 + h_1 \rho g + p_1 = \dfrac{1}{2}\rhov_2^2 + h_2 \rho g +p_2}$

Since the gauge pressure of the atmosphere $\mathbf{p_2}$ = 0

∴

The initial gauge pressure applied at the mouth can be determined as:

$\mathbf{p_1 = \dfrac{1}{2} \rho ( v_2^2-v_1^2)}$

here;

$\mathbf{\rho = 1000 \ kg/m^3}$

$\mathbf{p_1 = \dfrac{1}{2} \times 1000 \ kg/m^3 ( 3.1^2-0.31^2)}$

$\mathbf{p_1 =500 \ kg/m^3 (9.5139)}$

$\mathbf{p_1 =4756.95 \ Pa}$

$\mathbf{p_1 \simeq4757 \ Pa}$

Learn more about the principle of continuity equation here:

https://brainly.com/question/14619396