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QUESTION 4

4.1 In an inward flow reaction turbine, the supply head is 18 m, and the maximum discharge is [tex]$0.923 \, m^3 / s$[/tex]. The external diameter is twice the internal diameter, and the velocity of flow is constant and equal to [tex]$0.19 \sqrt{2gh}$[/tex]. The runner vanes are radial at the inlet, and the runner rotates at [tex][tex]$340 \, r/min$[/tex][/tex]. The hydraulic efficiency is 86%, and the vanes occupy 13% of the circumference.

Calculate:

4.1.1 The velocity of flow. (2)

4.1.2 The theoretical head. (2)

4.1.3 The inlet tangential velocity of the runner. (2)

4.1.4 The guide vane angle. (2)

4.1.5 The vane angle at the exit for radial discharge.

Sagot :

GinnyAnsweridn
Let's solve each part of the question step-by-step.

### Given Data:
- Supply head [tex]\(H\)[/tex] = 18 meters
- Maximum discharge [tex]\(Q\)[/tex] = 0.923 m³/s
- [tex]\(d_{ext}\)[/tex] = 2 \times [tex]\(d_{int}\)[/tex] (External diameter is twice the internal diameter)
- Rotation speed [tex]\(N\)[/tex] = 340 rotations per minute (rpm)
- Hydraulic efficiency [tex]\(\eta_h\)[/tex] = 0.86 (as a decimal)
- Vane occupy ratio [tex]\(\varepsilon\)[/tex] = 0.13 (as a decimal)
- Gravity acceleration [tex]\(g\)[/tex] = 9.81 m/s²
- Velocity of flow formula: [tex]\(V_f = 0.19 \sqrt{2gh}\)[/tex]

### Part 4.1.1: The velocity of flow
First, we need to calculate the velocity of flow [tex]\(V_f\)[/tex].

Using the formula [tex]\(V_f = 0.19 \sqrt{2gh}\)[/tex]:
[tex]\[ V_f = 0.19 \sqrt{2 \times 9.81 \times 18} \][/tex]
[tex]\[ V_f = 0.19 \sqrt{2 \times 9.81 \times 18} \][/tex]
[tex]\[ V_f \approx 3.570 \, \text{m/s} \][/tex]

### Part 4.1.2: The theoretical head
The theoretical head [tex]\(H_t\)[/tex] can be calculated using the hydraulic efficiency:
[tex]\[ H_t = \eta_h \times H \][/tex]
[tex]\[ H_t = 0.86 \times 18 \][/tex]
[tex]\[ H_t = 15.48 \, \text{meters} \][/tex]

### Part 4.1.3: Inlet tangential velocity of the runner
The inlet tangential velocity of the runner [tex]\(V_r\)[/tex] is calculated using the external diameter and the rotation speed.

First, we need to convert rotations per minute (rpm) to rotations per second (rps):
[tex]\[ N = \frac{340}{60} \approx 5.6667 \, \text{rps} \][/tex]

[tex]\[ V_r = \pi \times d_{ext} \times N \][/tex]

Assuming [tex]\(d_{ext} = 2 \times d_{int}\)[/tex] and given no exact value for [tex]\(d_{int}\)[/tex], the formula itself suffices the calculation. Thus:
[tex]\[ V_r = \pi \times 2 \times N \][/tex]
[tex]\[ V_r = \pi \times 2 \times 5.6667 \][/tex]
[tex]\[ V_r \approx 35.605 \, \text{m/s} \][/tex]

### Part 4.1.4: The guide vane angle
The guide vane angle [tex]\(\alpha\)[/tex] can be calculated using the tangent ratio:
[tex]\[ \tan(\alpha) = \frac{V_r}{V_f} \][/tex]
[tex]\[ \alpha = \tan^{-1}\left(\frac{35.605}{3.570}\right) \][/tex]
[tex]\[ \alpha \approx 84.273^\circ \][/tex]

### Part 4.1.5: The vane angle at the exit for radial discharge
For radial discharge, the vane angle at the exit is zero degrees. This is given directly by the problem.

Therefore, the vane angle at the exit [tex]\( \beta = 0^\circ \)[/tex].
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