Chapter 11, Problem 11.1PP

Water at $10°C$ flows from a large reservoir at the rate of $1.5\times {10}^2{m}^3/s$ through the system shown in Fig. 11.13 . Calculate the pressure at B.

To determine

To Calculate:

The pressure at B.

Answer to Problem 11.1PP

Pressure at B:

   $p_B=85.1kPa.$

Explanation of Solution

Point A at tank surface:

   $\frac{p_A}{\gamma }+z_A+\frac{v_A^2}{2g}-h_L=\frac{p_B}{\gamma }+z_B+\frac{v_B^2}{2g}$

   $p_A=0,v_A=0$

   $p_B={\gamma }_w\left[z_A-z_B\right]$

   $p_B={\gamma }_w\left[\frac{v_B^2}{2g}-h_L\right]$

   $p_B={\gamma }_w\left[12m-h_{v_B}-1.0h_{v_B}-3f_T\left(30\right)h_{v_B}-f\frac{L}{D}{h}_{v_B}\right]$

Where,

   $h_{v_B}$, velocity head in pipe:

   $h_{v_B}=\frac{v_B^2}{2g}=\frac{1}{2g}{\left(\frac{Q}{A}\right)}^2=\frac{{\left(1.99m/s\right)}^2}{2\left(9.81m/s^2\right)}=0.202m$

   $N_R=\raisebox{1ex}{$vD$}\!\left/ \!\raisebox{-1ex}{$\text{v}$}\right.=\raisebox{1ex}{$\left(1.99\right)\left(0.098\right)$}\!\left/ \!\raisebox{-1ex}{$\left(1.30\times {10}^{-6}\right)$}\right.$

   $N_R=1.50$

   $\raisebox{1ex}{$D$}\!\left/ \!\raisebox{-1ex}{$\epsilon $}\right.=\raisebox{1ex}{$0.098$}\!\left/ \!\raisebox{-1ex}{$1.5\times {10}^{-6}$}\right.=65300\to f=0.0165$

   $f_T\approx 0.010$ in zone of complete turbulence.

   $p_B=\frac{9.81\text{kN}}{m^3}\left[12m-0.202-0.202-\left(3\right)\left(0.010\right)\left(30\right)\left(0.202\right)-\left(0.0165\right)\frac{80.5}{0.098}\left(0.202\right)\right]$

   $p_B=85.1\text{kN}/m^2$

   $p_B=85.1kPa$