Question

Water flows from the pipe shown in FIGURE P14.60 with a speed of $4.0\mathrm{m}/\mathrm{s}$.

a. What is the water pressure as it exits into the air?

b. What is the height $h$of the standing column of water?

Short answer

(a) Atmospheric equal to$1m$

( b) The height of the standing water column is $4.6m$

Step-by-step solution

Introduction (part a)

(a)

Find the water pressure as it exits into the air.

The place where the water exits from the pipe is opened. So, the pressure at the exit point is equal to an atmospheric pressure which is equal to $1m$.

Explanation (part b)

(b)

Calculate the height of the standing column of water. Equation of continuity is,

$A_1{v}_1=A_2{v}_2$

Rearrange this for the speed of the liquid at the entrance,

$v_1=\left(\frac{A_2}{A_1}\right)v_2$

Substitute $5.0{\mathrm{cm}}^2\text{for}{A}_2,10.0{\mathrm{cm}}^2\text{for}{A}_1\text{and}4.0\mathrm{m}/\mathrm{s}\text{for}{v}_2$in above equation

$v_1=\left(\frac{5.0{\mathrm{cm}}^2}{10.0{\mathrm{cm}}^2}\right)(4.0\mathrm{m}/\mathrm{s})$

$=2.0\mathrm{m}/\mathrm{s}$

From Bemoullis equation,(part b)

From Bemoullis equation,

$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2\left(1\right)$

Here $P_1$and $P_2$are the pressures at the two ends, $h_1$and $h_2$are their heights respectively, $\rho$is the density of water and $g$is the acceleration due to gravity.

Pressure at the entrance is equal to the atmospheric pressure plus the pressure due to the water column above it,

$P_1=P_{\mathrm{atm}}+h\rho g\left(2\right)$

Here $p_{atm}$is the atmospheric pressure and$h$is the height of the water column. Also, the pressure at the exit is equal to the atmospheric pressure.

$P_2=P_{\mathrm{atm}}\left(3\right)$

Substitute the equations (2) and (3) in above equation (1).

$P_{\mathrm{atm}}+h\rho g+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_{\mathrm{atm}}+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Simplify and rearrange the above expression for $h$

$hg+\frac{1}{2}{v}_1^2+g{h}_1=\frac{1}{2}{v}_2^2+g{h}_2$

$h=\frac{1}{2g}\left(v_2^2-v_1^2\right)+\left(h_2-h_1\right)$

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2\text{for}g,2.0\mathrm{m}/\mathrm{s}\text{for}{v}_1,4.0\mathrm{m}/\mathrm{s}\text{for}{v}_2,0\mathrm{m}\text{for}{h}_1\text{and}4.0\mathrm{m}\text{for}{h}_2\text{in}$above equation

$h=\frac{1}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\left((4\mathrm{m}/\mathrm{s}{)}^2-(2\mathrm{m}/\mathrm{s}{)}^2\right)+(4\mathrm{m}-0\mathrm{m})$

$=4.6\mathrm{m}$