# Problem 02 - Bernoulli's Energy Theorem

**Problem 2**

From Figure 4-01, the following head losses are known: From (1) to (2), 0 m; from (2) to (3), 0.60 m; from (3) to (4), 2.1 m; from (4) to (5), 0.3 m. Make a table showing elevation head, velocity head, pressure head, and total head at each of the five points. How high above the center of the pipe will water stands in the piezometer tubes (3) and (4)?

**Solution 2**

$HL_{1-2} = 0$

$HL_{2-3} = 0.6 \, \text{ m}$

$HL_{3-4} = 2.1 \, \text{ m}$

$HL_{4-5} = 0.3 \, \text{ m}$

Total head loss from (1) to (5)

$HL_{1-5} = HL_{1-2} + HL_{2-3} + HL_{3-4} + HL_{4-5}$

$HL_{1-5} = 0 + 0.6 + 2.1 + 0.3$

$HL_{1-5} = 3 \, \text{ m}$

Note:

$E = \dfrac{v^2}{2g} + \dfrac{p}{\gamma} + z$

Sum up head from (1) to (5)

$E_1 - HL_{1-5} = E_5$

$(0 + 0 + 6) - 3 = \left(\dfrac{{v_5}^2}{2g} + 0 + 0 \right)$

$\dfrac{{v_5}^2}{2g} = 3 \, \text{ m}$

Since the diameter of the pipe is uniform and the opening for the jet is equal to the diameter of the pipe, the velocity heads at any point on the pipe are equal. Thus,

$\dfrac{{v_3}^2}{2g} = \dfrac{{v_4}^2}{2g} = \dfrac{{v_5}^2}{2g} = 3 \, \text{ m}$

Sum up head from (1) to (2)

$E_1 - HL_{1-2} = E_2$

$(0 + 0 + 6) - 0 = \left( 0 + \dfrac{p_2}{\gamma} + 0 \right)$

$\dfrac{p_2}{\gamma} = 6 \, \text{ m}$

Sum up head from (2) to (3)

$E_2 - HL_{2-3} = E_3$

$(0 + 6 + 0) - 0.6 = \left( 3 + \dfrac{p_3}{\gamma} + 0 \right)$

$\dfrac{p_3}{\gamma} = 2.4 \, \text{ m}$

Sum up head from (3) to (4)

$E_3 - HL_{3-4} = E_4$

$(3 + 2.4 + 0) - 2.1 = \left( 3 + \dfrac{p_4}{\gamma} + 0 \right)$

$\dfrac{p_4}{\gamma} = 0.3 \, \text{ m}$

Sum up head from (4) to (5)

$E_4 - HL_{4-5} = E_5$

$(3 + 0.3 + 0) - 0.3 = (3 + 0 + 0)$

$3 = 3 \, \text{(check)}$

Tabulated result

Point | Elevation head (m) | Velocity head (m) | Pressure head (m) | Total head (m) |

1 | 6 | 0 | 0 | 6.0 |

2 | 0 | 0 | 6 | 6.0 |

3 | 0 | 3 | 2.4 | 5.4 |

4 | 0 | 3 | 0.3 | 3.3 |

5 | 0 | 3 | 0 | 3.0 |

Piezometric heights

$h = \dfrac{p}{\gamma} + z$

$h_3 = 0 + 2.4 = 2.4 \, \text{ m}$ *answer*

$h_4 = 0 + 0.3 = 0.3 \, \text{ m}$ *answer*