Question

Water is floating smoothly through a closed-pipe system. At one point $A$, the speed of the water is $3.0m$ while at another point $B$, $1.0m$ higher, the speed is $4.0m/s$. The pressure at $A$ is $20kPa$ when the flowing $18kPa$ when the water flow stop. Then

• A. the pressure at $B$ when water is flowing is $6.5kPa$
• B. the pressure at $B$ when water is flowing is $8.0kPa$
• C. the pressure at $B$ when water stops flowing is $10kPa$
• D. None of these

Answer 1

Answer: (A), (C)

the pressure at $B$ when water stops flowing is $10kPa$
the pressure at $B$ when water is flowing is $6.5kPa$

Let $p_1,h_1,v_1$ and $p_2,h_2,v_2$ be the pressures, heights and velocities of flow at the two points, respectively.

Then Bernoulli's theorem gives
$p_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=p_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$ ...(I)

Putting $v_1=3.0m/s,v_2=4.0m/s,(h_2-h_1)=1m,p_1=20kPa$,

we get
$p_2=20+[{10}^3\times 10(-1)+\frac{{10}^2}{2}[9-16\left]\right]\times {10}^{-3}$

$=20-10-3.5=6.5kPa$

Also, when the flow stops, $v_1=v_2=0$ and then from Eq. (I)

we have $P_2=20-10=10kPa$