# Problem 03 - Bernoulli's Energy Theorem

**Problem 3**

A 300-mm pipe is connected by a reducer to a 100-mm pipe. See Figure 4-02. Points 1 and 2 are at the same elevation, the pressure at 1 is 200 kPa. The discharge Q is 30 liters per second flowing from 1 to 2 and the energy lost from 1 to 2 is equivalent to 20 kPa.

- Compute the pressure at 2 if the liquid is water.
- Compute the pressure at 2 if the liquid is oil (sp gr = 0.80).
- Compute the pressure at 2 if the liquid is molasses (sp gr = 1.5).

**Solution 3**

$Q_1 = Q_2 = Q = 0.03 \, \text{ m}^3\text{/s}$

Head loss

$HL = \dfrac{20}{\gamma} \, \text{ m}$

Velocity heads

$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.3^4)} = 0.0092 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.1^4)} = 0.7436 \, \text{ m}$

Energy equation between 1 and 2

$E_1 - HL = E_2$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2$

$0.0092 + \dfrac{200}{\gamma} + 0 - \dfrac{20}{\gamma} = 0.7436 + \dfrac{p_2}{\gamma} + 0$

$\dfrac{p_2}{\gamma} = \dfrac{180}{\gamma} - 0.7344$

$p_2 = 180 - 0.7344\gamma$

Part a: The liquid is water:

$p_2 = 180 - 0.7344(9.81)$

$p_2 = 172.79 \, \text{ kPa}$ *answer*

Part b: The liquid is oil (sp gr = 0.80):

$p_2 = 180 - 0.7344(0.80 \times 9.81)$

$p_2 = 174.24 \, \text{ kPa}$ *answer*

Part 3: The liquid is molasses (s = 1.5):

$p_2 = 180 - 0.7344(1.5 \times 9.81)$

$p_2 = 169.19 \, \text{ kPa}$ *answer*