Answer :
Sure, let's solve this step-by-step using Bernoulli's equation along with the given loss formula.
### Step 1: Understanding Bernoulli’s Equation
Bernoulli’s equation for two points along a streamline in a steady, incompressible flow is given by:
[tex]\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \][/tex]
Where [tex]\( P \)[/tex] is pressure, [tex]\( \rho \)[/tex] is the density of the liquid, [tex]\( v \)[/tex] is velocity, and [tex]\( h \)[/tex] is the height.
In terms of pressure head (pressure head = [tex]\(\frac{P}{\rho g} + h\)[/tex]), this simplifies to:
[tex]\[ h_{p1} + \frac{v_1^2}{2g} = h_{p2} + \frac{v_2^2}{2g} + h_{\text{loss}} \][/tex]
where:
- [tex]\( h_{p1} \)[/tex] is the pressure head at the smaller end.
- [tex]\( v_1 \)[/tex] is the velocity at the smaller end.
- [tex]\( h_{p2} \)[/tex] is the pressure head at the larger end.
- [tex]\( v_2 \)[/tex] is the velocity at the larger end.
- [tex]\( h_{\text{loss}} \)[/tex] is the loss of head due to friction or other effects.
### Step 2: Given Values
- Length of the tube: 2.0 m (not directly used in Bernoulli’s equation for pressure head but relevant contextually)
- Velocity at smaller end ([tex]\( v_1 \)[/tex]): 5.0 m/s
- Velocity at larger end ([tex]\( v_2 \)[/tex]): 2.0 m/s
- Pressure head at smaller end ([tex]\( h_{p1} \)[/tex]): 2.5 m
- Acceleration due to gravity ([tex]\( g \)[/tex]): 9.81 m/s²
- Loss coefficient: 0.35
- Loss of head formula: [tex]\(\frac{0.35 \left(1 + v_2 \right)^2}{2g} \)[/tex]
### Step 3: Calculate the Loss of Head
The head loss ([tex]\( h_{\text{loss}} \)[/tex]) is given by:
[tex]\[ h_{\text{loss}} = \frac{0.35 \left(1 + v_2 \right)^2}{2g} \][/tex]
Substituting [tex]\( v_2 = 2 \)[/tex]:
[tex]\[ h_{\text{loss}} = \frac{0.35 \left(1 + 2 \right)^2}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{0.35 \times 3^2}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{0.35 \times 9}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{3.15}{19.62} \][/tex]
[tex]\[ h_{\text{loss}} \approx 0.16055 \, \text{m} \][/tex]
### Step 4: Apply Bernoulli's Equation to Find the Pressure Head at the Larger End
Using Bernoulli's equation:
[tex]\[ h_{p1} + \frac{v_1^2}{2g} = h_{p2} + \frac{v_2^2}{2g} + h_{\text{loss}} \][/tex]
Rearranging to solve for [tex]\( h_{p2} \)[/tex]:
[tex]\[ h_{p2} = h_{p1} + \frac{v_1^2}{2g} - \frac{v_2^2}{2g} - h_{\text{loss}} \][/tex]
Substitute the given values:
[tex]\[ h_{p2} = 2.5 \, \text{m} + \frac{5.0^2}{2 \times 9.81} - \frac{2.0^2}{2 \times 9.81} - 0.16055 \][/tex]
Calculate the velocity head differences:
[tex]\[ \frac{5.0^2}{2 \times 9.81} = \frac{25}{19.62} \approx 1.274 \, \text{m} \][/tex]
[tex]\[ \frac{2.0^2}{2 \times 9.81} = \frac{4}{19.62} \approx 0.204 \, \text{m} \][/tex]
[tex]\[ h_{p2} = 2.5 + 1.274 - 0.204 - 0.16055 \][/tex]
Now, sum these values:
[tex]\[ h_{p2} = 2.5 + 1.274 - 0.204 - 0.16055 \][/tex]
[tex]\[ h_{p2} \approx 3.409785 \, \text{m} \][/tex]
### Step 5: Conclusion
The loss of head in the tube ([tex]\( h_{\text{loss}} \)[/tex]) is approximately 0.16055 meters, and the pressure head at the larger end ([tex]\( h_{p2} \)[/tex]) is approximately 3.409785 meters.
### Step 1: Understanding Bernoulli’s Equation
Bernoulli’s equation for two points along a streamline in a steady, incompressible flow is given by:
[tex]\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \][/tex]
Where [tex]\( P \)[/tex] is pressure, [tex]\( \rho \)[/tex] is the density of the liquid, [tex]\( v \)[/tex] is velocity, and [tex]\( h \)[/tex] is the height.
In terms of pressure head (pressure head = [tex]\(\frac{P}{\rho g} + h\)[/tex]), this simplifies to:
[tex]\[ h_{p1} + \frac{v_1^2}{2g} = h_{p2} + \frac{v_2^2}{2g} + h_{\text{loss}} \][/tex]
where:
- [tex]\( h_{p1} \)[/tex] is the pressure head at the smaller end.
- [tex]\( v_1 \)[/tex] is the velocity at the smaller end.
- [tex]\( h_{p2} \)[/tex] is the pressure head at the larger end.
- [tex]\( v_2 \)[/tex] is the velocity at the larger end.
- [tex]\( h_{\text{loss}} \)[/tex] is the loss of head due to friction or other effects.
### Step 2: Given Values
- Length of the tube: 2.0 m (not directly used in Bernoulli’s equation for pressure head but relevant contextually)
- Velocity at smaller end ([tex]\( v_1 \)[/tex]): 5.0 m/s
- Velocity at larger end ([tex]\( v_2 \)[/tex]): 2.0 m/s
- Pressure head at smaller end ([tex]\( h_{p1} \)[/tex]): 2.5 m
- Acceleration due to gravity ([tex]\( g \)[/tex]): 9.81 m/s²
- Loss coefficient: 0.35
- Loss of head formula: [tex]\(\frac{0.35 \left(1 + v_2 \right)^2}{2g} \)[/tex]
### Step 3: Calculate the Loss of Head
The head loss ([tex]\( h_{\text{loss}} \)[/tex]) is given by:
[tex]\[ h_{\text{loss}} = \frac{0.35 \left(1 + v_2 \right)^2}{2g} \][/tex]
Substituting [tex]\( v_2 = 2 \)[/tex]:
[tex]\[ h_{\text{loss}} = \frac{0.35 \left(1 + 2 \right)^2}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{0.35 \times 3^2}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{0.35 \times 9}{2 \times 9.81} \][/tex]
[tex]\[ h_{\text{loss}} = \frac{3.15}{19.62} \][/tex]
[tex]\[ h_{\text{loss}} \approx 0.16055 \, \text{m} \][/tex]
### Step 4: Apply Bernoulli's Equation to Find the Pressure Head at the Larger End
Using Bernoulli's equation:
[tex]\[ h_{p1} + \frac{v_1^2}{2g} = h_{p2} + \frac{v_2^2}{2g} + h_{\text{loss}} \][/tex]
Rearranging to solve for [tex]\( h_{p2} \)[/tex]:
[tex]\[ h_{p2} = h_{p1} + \frac{v_1^2}{2g} - \frac{v_2^2}{2g} - h_{\text{loss}} \][/tex]
Substitute the given values:
[tex]\[ h_{p2} = 2.5 \, \text{m} + \frac{5.0^2}{2 \times 9.81} - \frac{2.0^2}{2 \times 9.81} - 0.16055 \][/tex]
Calculate the velocity head differences:
[tex]\[ \frac{5.0^2}{2 \times 9.81} = \frac{25}{19.62} \approx 1.274 \, \text{m} \][/tex]
[tex]\[ \frac{2.0^2}{2 \times 9.81} = \frac{4}{19.62} \approx 0.204 \, \text{m} \][/tex]
[tex]\[ h_{p2} = 2.5 + 1.274 - 0.204 - 0.16055 \][/tex]
Now, sum these values:
[tex]\[ h_{p2} = 2.5 + 1.274 - 0.204 - 0.16055 \][/tex]
[tex]\[ h_{p2} \approx 3.409785 \, \text{m} \][/tex]
### Step 5: Conclusion
The loss of head in the tube ([tex]\( h_{\text{loss}} \)[/tex]) is approximately 0.16055 meters, and the pressure head at the larger end ([tex]\( h_{p2} \)[/tex]) is approximately 3.409785 meters.
