Answered

A Pelton wheel is rotated at 400 rev/min, generating 67.5 W under a head of 60 m of water. The inlet pipe's diameter at the base of the single nozzle is 15200 mm. The operating conditions are as follows:

- Coefficient of velocity ([tex]\( C_0 \)[/tex]) = 0.97
- Ratio of wheel velocity to jet velocity ([tex]\( \phi \)[/tex]) = 0.40
- Machine efficiency ([tex]\( \eta \)[/tex]) = 0.83

Determine:

i) The volume flow rate
ii) The diameter of the jet
iii) The wheel diameter
iv) The pressure in the inlet pipe at the nozzle base

(3 marks)
(4 marks)
(4 marks)
(4 marks)



Answer :

GinnyAnswer
Absolutely! Let's walk through the solution step-by-step for the given conditions.

### Given Data:
- Revolutions per minute (RPM) = 400
- Power output = 67.5 W
- Head (H) = 60 meters of water
- Inlet diameter at the base of the nozzle = 15200 mm
- Coefficient of velocity (C_v) = 0.97
- Ratio of wheel velocity to jet velocity = 0.40
- Efficiency = 0.83

First, let's convert some of the units to the appropriate measurements (e.g., diameter from mm to meters).

Conversions:
- [tex]\( \text{Inlet diameter} = 15200 \text{ mm} = 15.2 \text{ m} \)[/tex]

### 1. Volume Flow Rate (Q):
The volume flow rate can be calculated using the velocity of the jet and the area of the inlet pipe.

#### a) Calculate Jet Velocity (V_j):
Jet velocity [tex]\( V_j \)[/tex] can be determined using the formula:
[tex]\[ V_j = C_v \sqrt{2 g H} \][/tex]
Where:
- [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex] (acceleration due to gravity)
- [tex]\( H = 60 \, \text{m} \)[/tex]

With the given coefficient of velocity [tex]\( C_v = 0.97 \)[/tex]:
[tex]\[ V_j = 0.97 \times \sqrt{2 \times 9.81 \times 60} = 0.97 \times \sqrt{1177.2} \][/tex]

#### b) Area of the Inlet Pipe:
[tex]\[ \text{Area of inlet (A)} = \pi \left(\frac{\text{inlet diameter}}{2}\right)^2 = \pi \left(\frac{15.2}{2}\right)^2 \][/tex]

#### c) Volume Flow Rate:
[tex]\[ Q = A \times V_j \][/tex]

Using these formulas, the volume flow rate [tex]\( Q \)[/tex] is obtained as:
[tex]\[ Q = 6039.12 \, \text{m}^3/\text{s} \][/tex]

### 2. Diameter of the Jet (D_j):
The diameter of the jet can be calculated using the flow rate and jet velocity.

#### a) Area of the Jet:
[tex]\[ \text{Area}_{\text{jet}} = \frac{Q}{V_j} \][/tex]

#### b) Diameter of the Jet:
[tex]\[ D_j = \sqrt{\frac{4 \times \text{Area}_{\text{jet}}}{\pi}} \][/tex]

Based on this, the diameter of the jet [tex]\( D_j \)[/tex] is:
[tex]\[ D_j = 15.2 \, \text{m} \][/tex]

### 3. Wheel Diameter (D_w):
The wheel diameter can be calculated using the relationship between wheel velocity and jet velocity, and then determining the wheel circumference.

#### a) Wheel Velocity (V_w):
[tex]\[ V_w = \text{ratio} \times V_j \][/tex]
[tex]\[ V_w = 0.40 \times V_j \][/tex]

#### b) Wheel's Circumference:
[tex]\[ \text{Circumference of the wheel} = \frac{V_w}{RPM / 60} \][/tex]

#### c) Diameter of the Wheel:
[tex]\[ D_w = \frac{\text{Circumference of the wheel}}{\pi} \][/tex]

From these, the diameter of the wheel [tex]\( D_w \)[/tex] is:
[tex]\[ D_w = 0.63562 \, \text{m} \][/tex]

### 4. Pressure in the Inlet Pipe at the Nozzle Base:
Using Bernoulli's equation, assuming that the velocities change from zero to the jet velocity and using the density of water.

#### a) Inlet Pressure:
[tex]\[ \text{Pressure}_{\text{inlet}} = \frac{1}{2} \times \rho \times V_j^2 \][/tex]

Where [tex]\( \rho = 1000 \, \text{kg/m}^3 \)[/tex].

With the given values, the pressure at the inlet pipe is:
[tex]\[ \text{Pressure}_{\text{inlet}} = 553813.74 \, \text{Pa} \][/tex]

### Summary:
1. Volume Flow Rate (Q): [tex]\( 6039.12 \, \text{m}^3/\text{s} \)[/tex]
2. Diameter of the Jet (D_j): [tex]\( 15.2 \, \text{m} \)[/tex]
3. Wheel Diameter (D_w): [tex]\( 0.63562 \, \text{m} \)[/tex]
4. Pressure in the Inlet Pipe at the Nozzle Base: [tex]\( 553813.74 \, \text{Pa} \)[/tex]

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