Chapter 10, Problem 46P

To determine

The flow velocity and the gauge pressure in the given pipe on the top floor are to be determined.

Answer to Problem 46P

The flow velocity and the gauge pressure in the given pipe on the top floor are $2.22m/s$ and $2.03atm$ respectively.

Explanation of Solution

Given info:

Pressure at stream level, $P=3.8atm$

Pipe’s diameter, $d_1=5cm=0.05m$

Water’s speed, $v_{water}=0.60m/s$

Second diameter, $d_2=2.6cm=0.026m$

Height, $h_1=18m$

The density of water, $\rho =1000Kg/m^3$

Formula used:

• The equation of continuity is,

  $A_1{v}_1=A_2{v}_2$

• The Bernoulli’s equation is,

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

Calculation:

Form the equation of continuity,

  $\begin{array}{l}{A}_1{v}_1=A_2{v}_2\\ \pi r_1{}^2{v}_1=\pi r_2{}^2{v}_2\\ r_1{}^2{v}_1=r_2{}^2{v}_2\end{array}$

Then the flow velocity at the top floor is,

  $v_2=\frac{r_1{}^2{v}_1}{r_2{}^2}=\frac{{\left(0.025\right)}^2\times 0.60m/s}{{\left(0.013\right)}^2}=\frac{3.75\times {10}^{-4}}{1.69\times {10}^{-4}}=2.21893m/s$

Now, taking the Bernoulli’s principle,

  $\begin{array}{l}{P}_1+g{h}_1\rho +\frac{1}{2}\rho v_1{}^2=P_2+g{h}_2\rho +\frac{1}{2}\rho v_2{}^2\\ P_1+\frac{1}{2}\rho v_1{}^2=P_2+gh\rho +\frac{1}{2}\rho v_2{}^2\left(\because h_1=0and{h}_2=h\right)\\ P_2=P_1+\frac{1}{2}\rho v_1{}^2-\frac{1}{2}\rho v_2{}^2\end{array}$

Now convert the gauge pressure into Pa,

  $P_1=3.8atm\times \left(1.01\times {10}^5\frac{Pa}{atm}\right)=3.838\times {10}^{5}Pa$

Now use this value in above equation

  $\begin{array}{l}{P}_2=3.838\times {10}^{5}Pa+\frac{1}{2}\rho \left(v_1{}^2-v_1{}^2\right)\\ P_2=3.838\times {10}^{5}Pa+\frac{1}{2}\times \left(1000kg/m^3\right)\left({\left(0.60\right)}^2-{\left(2.22\right)}^2\right)\\ =205115.8Pa\end{array}$

Convert the gauge pressure into atmospheric pressure,

  $P_2=\frac{205115.8Pa}{\left(1.01\times {10}^5\frac{Pa}{atm}\right)}=2.03atm$

Conclusion:

The gauge pressure is $2.03atm$ .