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VIPSolutions · Accepted Answer

**Subject:** Fluid Mechanics  
**Topic:** Pressure Calculation in a Pipe Using Manometer Heights

---

### Step 1: Given Data & Conceptual Introduction

**Given Data:**
- Height $ h_2 = 3 $ in above exit
- Height $ h_3 = 4 $ in (assumed)
- Pipe diameter = 0.75 in
- Pipe length = 21 in
- Need to determine the pressure at point (1) ($ p_1 $) to produce the flow

**Conceptual Introduction:**  
This problem involves determining the required pressure at the inlet of a pipe (point 1) to generate a specific fluid flow, given height differences in a manometer setup. The pressure difference is typically found using the hydrostatic relation and the properties of the manometric fluid.

**Explanation Block:**  
We are given the heights in the manometer and pipe dimensions to find the pressure at point 1, using the hydrostatic equation. This setup translates height differences into pressure differences needed to move the water.

---

### Step 2: Step-by-Step Calculation (Hydrostatic Pressure Difference)

**Step 2.1:**  
Find the height difference between points 2 and 3:
$$
h_{23} = h_3 - h_2 = 4\,\text{in} - 3\,\text{in} = 1\,\text{in}
$$

**Step 2.2:**  
Convert inches to feet (if needed for standard units):
$$
1\,\text{in} = \frac{1}{12}\,\text{ft}
$$

**Step 2.3:**  
Pressure difference (water column):
$$
\Delta P = \gamma_{\text{water}} \cdot h_{23}
$$
Where $ \gamma_{\text{water}} = 62.4\,\text{lb/ft}^3 $.

$$
\Delta P = 62.4\,\text{lb/ft}^3 \times \frac{1}{12}\,\text{ft} = 5.2\,\text{lb/ft}^2
$$

**Step 2.4:**  
Convert $ \text{lb/ft}^2 $ to psi ($ 1\,\text{psi} = 144\,\text{lb/ft}^2 $):
$$
\Delta P = \frac{5.2}{144}\,\text{psi} \approx 0.036\,\text{psi}
$$

**Explanation Block:**  
We calculated the height difference as 1 inch and converted it to the corresponding pressure using the specific weight of water. This gives us the pressure difference needed at point 1 over atmospheric pressure.

---

### Step 3: Total Pressure at Point (1)

**Step 3.1:**  
If the system is open to the atmosphere at the exit, the required gauge pressure at point 1 is just the calculated difference:
$$
p_1 = 0.036\,\text{psi}
$$

**Explanation Block:**  
Since the exit is open to the atmosphere, the gauge pressure at point 1 must be just enough to overcome the hydrostatic head represented by the manometer height difference. Any additional pressure would be due to flow losses, which are assumed negligible here.

---

### Final Answer:

**The pressure needed at point (1) to produce the flow is approximately $ \boxed{0.036 \ \text{psi}} $.**

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