Give step-by-step solution with explanation and final answer:
A Francis turbine is designed with the following conditions: beta_{1} = 60 deg beta_{2} = 90 deg r_{1} = 5 m, r_{2} = 3 and B = 1m When the discharge is 126 m²³/s and N = 60 rpm, calculate the entrance angle a, that would prevent separation of the streamlines at the entrance of the runner. Determine the maximum power and torque that can be generated under these conditions. Finally, for the same discharge and runner size, can you suggest an improvement to the design of the runner blades.
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Give step-by-step solution with explanation and final answer:
A Francis turbine is designed with the following conditions: beta_{1} = 60 deg beta_{2} = 90 deg r_{1} = 5 m, r_{2} = 3 and B = 1m When the discharge is 126 m²³/s and N = 60 rpm, calculate the entrance angle a, that would prevent separation of the streamlines at the entrance of the runner. Determine the maximum power and torque that can be generated under these conditions. Finally, for the same discharge and runner size, can you suggest an improvement to the design of the runner blades.
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Asked by: varrisanju5@gmail.com
Created at: 2025-10-16 11:04:06
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# Step-by-Step Solution: Francis Turbine Design
## **Given Data**
- Inlet blade angle, \( \beta_1 = 60^\circ \)
- Outlet blade angle, \( \beta_2 = 90^\circ \)
- Inlet radius, \( r_1 = 5\,\text{m} \)
- Outlet radius, \( r_2 = 3\,\text{m} \)
- Width of runner, \( B = 1\,\text{m} \)
- Discharge, \( Q = 126\,\text{m}^3/\text{s} \)
- Rotational speed, \( N = 60\,\text{rpm} \)
## **1. Velocity Diagram at Inlet**
### **a. Angular Velocity**
\[
\omega = \frac{2\pi N}{60} = \frac{2\pi \times 60}{60} = 2\pi\,\text{rad/s}
\]
### **b. Peripheral (Tangential) Velocity at Inlet**
\[
u_1 = \omega r_1 = 2\pi \times 5 = 10\pi \approx 31.416\,\text{m/s}
\]
### **c. Discharge and Flow Velocity**
Area at inlet:
\[
A_1 = 2\pi r_1 B = 2\pi \times 5 \times 1 = 10\pi \approx 31.416\,\text{m}^2
\]
Radial (flow) velocity at inlet:
\[
V_{f1} = \frac{Q}{A_1} = \frac{126}{31.416} \approx 4.012\,\text{m/s}
\]
### **d. Relative Velocity and Velocity Components**
From velocity diagram (see figure below):
- \( V_{f1} \): Flow velocity (axial/radial)
- \( u_1 \): Peripheral velocity
- \( V_{w1} \): Whirl velocity (tangential component)
- \( V_{r1} \): Relative velocity to the blade
\[
\tan \beta_1 = \frac{V_{f1}}{u_1 - V_{w1}}
\]
But at inlet, whirl velocity is generally unknown. However, for no separation, the **absolute velocity at entry should be tangent to the blade**, i.e., entrance angle \( \alpha_1 = \alpha \).
From velocity triangle:
\[
\tan \alpha_1 = \frac{V_{f1}}{V_{w1}}
\]
But,
\[
V_{w1} = u_1 - V_{r1} \cos \beta_1
\]
\[
V_{r1} = \frac{V_{f1}}{\sin \beta_1}
\]
So,
\[
V_{r1} = \frac{4.012}{\sin 60^\circ} = \frac{4.012}{.866} \approx 4.633\,\text{m/s}
\]
Now,
\[
V_{w1} = u_1 - V_{r1} \cos \beta_1 = 31.416 - 4.633 \times .5 = 31.416 - 2.3165 = 29.099\,\text{m/s}
\]
Now, entrance angle \( \alpha \) is:
\[
\tan \alpha = \frac{V_{f1}}{V_{w1}} = \frac{4.012}{29.099} = .1378
\]
\[
\alpha = \arctan(.1378) \approx 7.87^\circ
\]
---
## **2. Maximum Power and Torque**
### **a. Work Done per Second (Power)**
\[
\text{Power} = \rho Q (u_1 V_{w1} - u_2 V_{w2})
\]
Assume water density \( \rho = 100\,\text{kg/m}^3 \).
#### **At Outlet**
- For Francis turbines, \( \beta_2 = 90^\circ \) means flow leaves radially (\( V_{w2} = \)).
\[
u_2 = \omega r_2 = 2\pi \times 3 = 6\pi \approx 18.850\,\text{m/s}
\]
\[
\text{Thus,} \quad \text{Power}_{\text{max}} = 100 \times 126 \times 31.416 \times 29.099
\]
Calculate:
\[
31.416 \times 29.099 = 915.475
\]
\[
\text{Power}_{\text{max}} = 100 \times 126 \times 915.475 = 115,350,000\,\text{W} = 115.35\,\text{MW}
\]
### **b. Torque**
\[
\text{Torque},\ T = \frac{\text{Power}}{\omega}
\]
\[
T = \frac{115,350,000}{2\pi} = \frac{115,350,000}{6.283} \approx 18,368,000\,\text{Nm}
\]
---
## **3. Suggested Improvement**
For the same discharge and runner size:
- The absolute velocity at entrance (\( \alpha \approx 7.87^\circ \)) is quite small, indicating a **very sharp entry**. This may make the turbine sensitive to flow disturbances and could risk separation at off-design conditions.
- **Improvement suggestion:**
- **Increase the inlet blade angle** (\( \beta_1 \)) to allow a higher entrance angle (\( \alpha \)), making the turbine less sensitive to minor flow variations and reducing risk of separation.
- Alternatively, **increase the number of smaller blades** or use **curved blades** optimized for variable load conditions to reduce the risk of separation and improve efficiency.
---
## **Summary Table**
| Parameter | Value |
|--------------------------|------------------|
| Entrance angle \( \alpha \) | \( 7.87^\circ \) |
| Max power | 115.35 MW |
| Max torque | 18,368,000 Nm |
---
## **Velocity Diagram**

*Alt text: Velocity triangle at the inlet of a Francis turbine showing the relationships between u, V_f, V_w, V_r, and entrance angle alpha.*
---
## **Final Answers**
- **Entrance angle \( \alpha \) to prevent separation:**
\( \boxed{7.87^\circ} \)
- **Maximum power generated:**
\( \boxed{115.35\,\text{MW}} \)
- **Maximum torque generated:**
\( \boxed{18,368,000\,\text{Nm}} \)
- **Improvement suggestion:**
Increase blade entry angle or use more, smaller, or curved blades to reduce separation risk and improve efficiency.
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