To solve for the total energy head $H$ above the given datum for each fluid, we will use the Bernoulli equation in terms of head. The energy head above a datum is given by the sum of the pressure head, velocity head, and elevation head:

$H = \frac{P}{\gamma} + \frac{V^2}{2g} + z$

where:

• $H$ = total energy head
• $P$ = pressure of the fluid (4 lb/in²)
• $\gamma$ = specific weight of the fluid
• $V$ = velocity of the fluid (8 ft/s)
• $g$ = acceleration due to gravity (32.2 ft/s²)
• $z$ = elevation of the pipe center above the datum (10 ft)

We will calculate $H$ for each fluid type: water, oil, and gas.

Step 1: Convert Pressure $P$ to lb/ft²

Given $P = 4 \, \text{lb/in}^2$, convert this to lb/ft²: $P = 4 \, \text{lb/in}^2 \times (144 \, \text{in}^2/\text{ft}^2) = 576 \, \text{lb/ft}^2$

Step 2: Calculate the Velocity Head

Using the velocity head formula: $\frac{V^2}{2g} = \frac{8^2}{2 \times 32.2} = \frac{64}{64.4} \approx 0.994 \, \text{ft}$

Step 3: Calculate for Each Fluid

Now we will plug in the specific weights of each fluid to find $H$.

(a) Water

For water, the specific weight $\gamma_{\text{water}} = 62.4 \, \text{lb/ft}^3$.

1. Pressure Head: $\frac{P}{\gamma_{\text{water}}} = \frac{576}{62.4} \approx 9.23 \, \text{ft}$

2. Total Energy Head: $H_{\text{water}} = \frac{P}{\gamma_{\text{water}}} + \frac{V^2}{2g} + z = 9.23 + 0.994 + 10 = 20.22 \, \text{ft}$

(b) Oil (Specific Gravity = 0.82)

For oil, we calculate the specific weight by using the specific gravity: $\gamma_{\text{oil}} = 0.82 \times 62.4 = 51.168 \, \text{lb/ft}^3$

1. Pressure Head: $\frac{P}{\gamma_{\text{oil}}} = \frac{576}{51.168} \approx 11.26 \, \text{ft}$

2. Total Energy Head: $H_{\text{oil}} = \frac{P}{\gamma_{\text{oil}}} + \frac{V^2}{2g} + z = 11.26 + 0.994 + 10 = 22.25 \, \text{ft}$

(c) Gas (Specific Weight = 0.042 lb/ft³)

1. Pressure Head: $\frac{P}{\gamma_{\text{gas}}} = \frac{576}{0.042} = 13714.29 \, \text{ft}$

2. Total Energy Head: $H_{\text{gas}} = \frac{P}{\gamma_{\text{gas}}} + \frac{V^2}{2g} + z = 13714.29 + 0.994 + 10 \approx 13725.28 \, \text{ft}$

Summary of Results

1. Water: $H_{\text{water}} = 20.22 \, \text{ft}$
2. Oil: $H_{\text{oil}} = 22.25 \, \text{ft}$
3. Gas: $H_{\text{gas}} = 13725.28 \, \text{ft}$

Would you like further clarification on any part of this process?

---

Related Questions:

1. How does changing pipe diameter affect the velocity head component?
2. Why does a gas with low specific weight produce such a high total energy head?
3. How would increasing fluid velocity impact the total head?
4. Can specific gravity values be used to estimate other fluid properties?
5. What factors might affect pressure head in a real-world fluid system?

Tip: Specific gravity is a quick way to find specific weight by referencing water's properties, which can simplify many fluid calculations.