For reference, the Bernoulli equation is:

$latex E_t = E_p + E_v + E_z\newline

\indent =\frac{p}{\rho}+\frac{v^{2}}{2}+zg\hspace{20px}^{(SI)}\newline

\indent =\frac{p}{\rho}+\frac{v^{2}}{2g_c}+\frac{zg}{g_c}\hspace{20px}^{(US)}&s=2$

## Example 1

A reservoir contains an outlet that consists of a pipe that discharges water 50 ft below the top of the reservoir.

What is the velocity of the water exiting the pipe (at point B)?

### Solution

If you define elevation zero at point B:

- Point A has only potential energy
- Point B has only kinetic energy

The Bernoulli equation states that total energy must be the same, therefore:

U.S. Units | Metric |
---|---|

$latex E_{tA} = E_{tB}\newline\newline 0 + 0 + \frac{z_A g}{g_c} = 0 + \frac{v_B^{2}}{2g_c} + 0\newline\newline\frac{(50 ft)(32.2 ft/s^{2})}{32.2 \frac{lbm\cdot ft}{lbf\cdot s^{2}}} = \frac{v^{2}}{2(32.2 \frac{lbm\cdot ft}{lbf\cdot s^{2}})}\newline\newline\newline v = 56.7 ft/s&s=2$ | $latex E_{tA} = E_{tB}\newline\newline 0 + 0 + z_A g = 0 + \frac{v_B^{2}}{2} + 0\newline\newline (15.2 m)(9.81 m/s^{2}) = \frac{v^{2}}{2}\newline\newline\newline v = 17.3 m/s&s=2$ |

## Example 2

A pump is pumping water up a hillside into a reservoir. The pump rate is 4 ft/s (1.22 m/s) with a pressure of 100 psi (689 kPa). What is the maximum elevation of the reservoir that the pump can handle? (Water = 62.4 lbm/ft^{3} or 1000 kg/m^{3}).

### Solution

- The pump has both pressure and kinetic energy
- The reservoir has only potential energy

According to Bernoulli’s principle, the sum of the energy must be the same at both locations.

U.S. Units | Metric |
---|---|

$latex E_{t\cdot pump} = E_{t\cdot reservoir}\newline\newline \frac{p}{\rho} + \frac{v^{2}}{2g_c} + 0 = 0 + 0 + zg_c\newline\newline \frac{100 psi}{62.4 lbm/ft^{3}} + \frac{(4 ft/s)^{2}}{2(32.2 \frac{lbm\cdot ft}{lbf\cdot s^{2}})}= \newline\newline \hspace{50px}= \frac{z\cdot (32.2 ft/s^{2})}{32.2 \frac{lbm \cdot ft}{lbf\cdot s^{2}}}\newline\newline\newline z = 231.0 ft&s=2$ | $latex E_{t\cdot pump} = E_{t\cdot reservoir}\newline\newline \frac{p}{\rho} + \frac{v^{2}}{2} + 0 = 0 + 0 + zg\newline\newline \frac{689 kPa}{1000kg/m^{3}} + \frac{(1.22 m/s)^{2}}{2}= \newline\newline\hspace{50px}= z\cdot (9.81 m/s^{2})\newline\newline\newline z = 70.3 m&s=2$ |

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