A piping system transporting water from reservoir 1 to reservoir 2. The diameter of the stainlesssteel pipe is 25 mm and the length is 60 m. The system is composed of a sharp entrance, two 45- degree elbows, and a sharp exit. a) Sketch the piping system connecting reservoir 1 to reservoir 2. b) Taking the free surface of the lower tank as the reference level (reservoir 2), estimate the elevation h1 of the higher reservoir if the flow rate at is 83.2 L/min. (Note: Ɛ = 0.002, f=0.018, Kelbow= 0.4 Kentrance=0.5, Kexit=1)



Answer :

Answer:

To estimate the elevation \( h_1 \) of reservoir 1, we can proceed with the following steps:

### Step-by-Step Solution:

1. **Understand the components and losses in the piping system:**

- Pipe diameter \( D = 25 \) mm = \( 0.025 \) m

- Pipe length \( L = 60 \) m

- Flow rate \( Q = 83.2 \) L/min = \( 0.0832 \) m³/min = \( 0.001387 \) m³/s

2. **Calculate the Reynolds number (Re):**

\[

\text{Re} = \frac{\rho V D}{\mu}

\]

Assuming water properties (\( \rho = 1000 \) kg/m³, \( \mu = 0.001 \) Pa·s):

\[

V = \frac{Q}{A} = \frac{Q}{\pi \left(\frac{D}{2}\right)^2} = \frac{0.001387}{\pi \left(\frac{0.025}{2}\right)^2} \approx 2.231 \text{ m/s}

\]

\[

\text{Re} = \frac{1000 \times 2.231 \times 0.025}{0.001} \approx 112310

\]

3. **Calculate the friction factor (f):**

Given \( f = 0.018 \) (assuming it's the Darcy-Weisbach friction factor).

4. **Calculate the head loss due to pipe friction:**

\[

h_f = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g}

\]

Where \( g = 9.81 \) m/s²:

\[

h_f = 0.018 \cdot \frac{60}{0.025} \cdot \frac{(2.231)^2}{2 \times 9.81} \approx 4.85 \text{ m}

\]

5. **Calculate the head loss coefficients for components:**

- Entrance loss coefficient \( K_{\text{entrance}} = 0.5 \)

- Exit loss coefficient \( K_{\text{exit}} = 1 \)

- Elbow loss coefficient (each elbow, assuming 45° elbows): \( K_{\text{elbow}} = 0.4 \times 2 = 0.8 \)

Total minor loss coefficient \( \Sigma K = K_{\text{entrance}} + K_{\text{exit}} + K_{\text{elbow}} = 0.5 + 1 + 0.8 = 2.3 \)

6. **Calculate the total head loss (h_L):**

\[

h_L = h_f + \frac{\Sigma K \cdot V^2}{2g}

\]

\[

h_L = 4.85 + \frac{2.3 \cdot (2.231)^2}{2 \cdot 9.81} \approx 5.90 \text{ m}

\]

7. **Calculate the difference in reservoir elevations:**

Since \( h_L = h_1 - h_2 \), where \( h_2 = 0 \) (reference at reservoir 2):

\[

h_1 = h_L + h_2 = 5.90 + 0 = 5.90 \text{ m}

\]

### Conclusion:

The elevation \( h_1 \) of reservoir 1, relative to reservoir 2, is approximately \( 5.90 \) meters. This calculation takes into account the major and minor losses in the piping system between the two reservoirs.

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