Act like a helpful tutor and exlain me :(¢) Water lows between two tanks through a 100 mm diameter, 30 m long pipe with K = 0.15 mm.
The pipe junction with both reservoirs is sharp edged, there is a

Question

Act like a helpful tutor and exlain me :(¢) Water lows between two tanks through a 100 mm diameter, 30 m long pipe with K = 0.15 mm.
The pipe junction with both reservoirs is sharp edged, there is a 902 elbow 5 m from the inet and
a half open gate valve at 10 m from the nlet.

f the discharge is 15 €/s, calculate the head losses at each fitting and in the pipe to determine the
total head loss between the two tanks. Use the pipe chart in Appendix 8 and the table
below. (5 marks)

Fitting K
‘Sharp entrance 05
90" elbow 06
Half open gate valve 40
‘Sudden enlargement 10

VIPSolutions · Accepted Answer

# Step-by-Step Solution

Let's break down the problem and solve for the **total head loss** between the two tanks.

---

## **Given Data**

- **Pipe diameter** ($D$): 100 mm = 0.1 m
- **Pipe length** ($L$): 30 m
- **Pipe roughness** ($K$): 0.15 mm = 0.00015 m
- **Discharge** ($Q$): 15 L/s = 0.015 m³/s
- **Fittings:**
  - **Sharp entrance:** K = 0.5
  - **90° elbow (5 m from inlet):** K = 0.6
  - **Half open gate valve (10 m from inlet):** K = 4.0

---

## **Step 1: Calculate Velocity in the Pipe**

$$
A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.1)^2 = 0.00785\ \text{m}^2
$$

$$
V = \frac{Q}{A} = \frac{0.015}{0.00785} = 1.91\ \text{m/s}
$$

---

## **Step 2: Calculate Head Losses at Fittings**

Head loss for each fitting:
$$
h_{L,fitting} = K \frac{V^2}{2g}
$$
Where $g = 9.81\ \text{m/s}^2$

### **Sharp Entrance**
$$
h_{L,entrance} = 0.5 \times \frac{(1.91)^2}{2 \times 9.81} = 0.093\ \text{m}
$$

### **90° Elbow**
$$
h_{L,elbow} = 0.6 \times \frac{(1.91)^2}{2 \times 9.81} = 0.112\ \text{m}
$$

### **Half Open Gate Valve**
$$
h_{L,valve} = 4.0 \times \frac{(1.91)^2}{2 \times 9.81} = 0.749\ \text{m}
$$

---

## **Step 3: Calculate Head Loss in the Pipe (Major Loss)**

Use Darcy-Weisbach equation:
$$
h_{L,pipe} = f \frac{L}{D} \frac{V^2}{2g}
$$

### **Find Reynolds Number ($Re$):**

Kinematic viscosity of water ($\nu$) at 20°C ≈ $1 \times 10^{-6}$ m²/s

$$
Re = \frac{V D}{\nu} = \frac{1.91 \times 0.1}{1 \times 10^{-6}} = 1.91 \times 10^5
$$

### **Relative Roughness:**
$$
\frac{K}{D} = \frac{0.00015}{0.1} = 0.0015
$$

### **Estimate Friction Factor ($f$):**

For turbulent flow, use the **Colebrook-White equation** or **Moody Chart**. For $\frac{K}{D} = 0.0015$ and $Re = 1.91 \times 10^5$, from charts, $f \approx 0.021$.

### **Calculate Pipe Loss:**
$$
h_{L,pipe} = 0.021 \times \frac{30}{0.1} \times \frac{(1.91)^2}{2 \times 9.81}
$$
$$
= 0.021 \times 300 \times \frac{3.6481}{19.62}
$$
$$
= 6.3 \times 0.186
$$
$$
= 1.172\ \text{m}
$$

---

## **Step 4: Total Head Loss**

$$
h_{L,total} = h_{L,entrance} + h_{L,elbow} + h_{L,valve} + h_{L,pipe}
$$
$$
= 0.093 + 0.112 + 0.749 + 1.172 = 2.126\ \text{m}
$$

---

## **Summary Table**

| Component             | K Value | Head Loss (m) |
|-----------------------|---------|--------------|
| Sharp Entrance        | 0.5     | 0.093        |
| 90° Elbow             | 0.6     | 0.112        |
| Half Open Gate Valve  | 4.0     | 0.749        |
| Pipe (major loss)     | —       | 1.172        |
| **Total**             | —       | **2.126**    |

---

## **Final Answer**

**The total head loss between the two tanks is approximately $2.13$ meters.**

Let me know if you want to see the calculation steps for the friction factor or any part in more detail!

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