Question

Need to solve step by step this exercise about fluids..

Answer 1

Given:

The pressure in the lower pipe is,

`P_1=120\text{ kPa}`

The speed of water is,

`v_1=1\text{ m/s}`

The radius of the lower pipe is

`\begin{gathered} r_1=12\text{ cm} \\ =0.12\text{ m} \end{gathered}`

the radius of the upper pipe is

`\begin{gathered} r_2=6\text{ cm} \\ =0.06\text{ m} \end{gathered}`

The height of the upper pipe is,

`h_2=2\text{ m}`

The density of water is,

`\rho=1000kg/m^3`

Using the continuity equation,

`\begin{gathered} \pi r^2_{1^{}}v_1=\pir_2^{}^2_{}v_2 \\ v_2=\frac{r^2_{1^{}}v_1}{r^2_{2^{}}} \\ v_2=(0.12*0.12*1)/(0.06*0.06) \\ v_2=4\text{ m/s} \end{gathered}`

Hence the speed of water is 4 m/s.

Applying Bernoulli's principle we get,

`\begin{gathered} P_1+(1)/(2)\rho(v_1)^2+\rho g*0_{}=P_2+(1)/(2)\rho(v_2)^2+\rhogh_2 \\ 120*10^3+(1)/(2)*1000*1^2+0=P_2+(1)/(2)*1000*4^2+1000*9.8*2 \\ P_2=120500-27600 \\ P_2=92.9\text{ kPa} \end{gathered}`

Hence, the pressure is 92.9 kPa.