Let's break down problem step by according to the. We’ll on **Question2, Part3** (ial ducted fan): ### ** Data** -Pressure coefficient (\( \psi \** = .28 **Mass flow rate (\( \{m} \))** =2 kg/s- **Rot speed (\( \))** 650 rpm - **Hub (\( Dhub} \))** = .15 m - **Tip diameter (\( D_{tip} \))** = .65 m - **Ambient pressure** = 1. bar = 100,000 Pa - **Temperature (\( T \))** = 28°C = 301 K - **Efficiency (\( \eta \))** = 85% = .85 Let's solve each sub-question step by step. --- ## **a) Air Angles at Entry and Exit** We need to find the **blade angles** at the entry and exit, but for that, we need velocities at the hub and tip. #### **Step 1: Find the Mean Diameter and Blade Speed** \[ D_{mean} = \frac{D_{hub} + D_{tip}}{2} = \frac{.15 + .65}{2} = .40\,m \] \[ r_{mean} = \frac{D_{mean}}{2} = .20\,m \] \[ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 650}{60} \approx 68.07\, \text{rad/s} \] \[ U = \omega \times r_{mean} = 68.07 \times .20 = 13.61\, m/s \] #### **Step 2: Find the Axial Velocity (\( V_a \))** The area at the mean diameter: \[ A = \frac{\pi}{4}(D_{tip}^2 - D_{hub}^2) = \frac{\pi}{4}(.65^2 - .15^2) = \frac{\pi}{4}(.4225 - .0225) = \frac{\pi}{4}(.4) \approx .314\,m^2 \] Assume air is ideal gas, so at 1 bar, 28°C: \[ \rho = \frac{P}{RT} = \frac{100,000}{287 \times 301} \approx 1.15\,kg/m^3 \] \[ V_a = \frac{\dot{m}}{\rho A} = \frac{2}{1.15 \times .314} \approx \frac{2}{.3611} \approx 5.54\, m/s \] #### **Step 3: Find Blade Angles** The **blade angle at entry (\( \beta_1 \))** and **exit (\( \beta_2 \))** (assuming velocity triangles): \[ \tan \beta = \frac{V_a}{U} \] \[ \beta_{mean} = \arctan \left( \frac{V_a}{U} \right) = \arctan \left( \frac{5.54}{13.61} \right) \approx \arctan(.407) \approx 22.1^\circ \] Thus, **air angles at entry and exit at mean radius ≈ 22.1°** (assuming symmetric design for simplicity). --- ## **b) Pressure Developed in mm G** Given pressure coefficient (\( \psi \)): \[ \psi = \frac{\Delta P}{.5 \rho U^2} \] \[ \Delta P = \psi \cdot .5 \rho U^2 = .28 \times .5 \times 1.15 \times (13.61)^2 \] \[ = .28 \times .5 \times 1.15 \times 185.27 \] \[ = .28 \times .5 \times 213.06 \] \[ = .28 \times 106.53 \approx 29.83\, Pa \] Convert to mm of water gauge (\(1\,mm\,H_2O = 9.81\,Pa\)): \[ \Delta P_{mm\,H_2O} = \frac{29.83}{9.81} \approx 3.04\, mm\, H_2O \] --- ## **c) Fan Efficiency** Fan efficiency (\( \eta \)) is given as **85% (.85)**. --- ## **d) Power Required to Drive the Fan** \[ \text{Power}_{\text{fluid}} = \Delta P \cdot Q \] First, find the volume flow rate: \[ Q = \frac{\dot{m}}{\rho} = \frac{2}{1.15} \approx 1.74\, m^3/s \] \[ \text{Power}_{\text{fluid}} = 29.83\,Pa \times 1.74\,m^3/s = 51.9\,W \] The actual **shaft power required** (input to the fan) considering efficiency: \[ \text{Shaft Power} = \frac{\text{Power}_{\text{fluid}}}{\eta} \] \[ = \frac{51.9}{.85} \approx 61.1\, W \] --- # **Summary Table** | Parameter | Value | |-----------------------------|------------------------| | Blade Angle (mean) | 22.1° | | Pressure Developed | 3.04 mm H₂O | | Fan Efficiency | 85% | | Power Required (to shaft) | 61.1 W | --- ### **Final Answers** - **a)** Air Angles at entry and exit (at mean radius): **22.1°** - **b)** Pressure developed: **3.04 mm H₂O** - **c)** Fan Efficiency: **85%** - **d)** Power required to drive the fan: **61.1 W** --- Let me know if you need detailed derivations or diagrams for the other sub-questions!

# Fluid Machinery Assignment Solution ## Problem Overview This assignment focuses on analyzing an axial ducted fan's performance parameters, including air angles, pressure developed, fan efficiency, and power required to drive the fan. ### Given Data - **Pressure Coefficient (\( \psi \))**: 0.28 - **Mass Flow Rate (\( \dot{m} \))**: 2 kg/s - **Rotational Speed (\( N \))**: 650 rpm - **Hub Diameter (\( D_{hub} \))**: 0.15 m - **Tip Diameter (\( D_{tip} \))**: 0.65 m - **Ambient Pressure**: 1 bar = 100,000 Pa - **Temperature (\( T \))**: 28°C = 301 K - **Overall Efficiency (\( \eta \))**: 85% = 0.85 ## Step-by-Step Solution ### a) Air Angles at Entry and Exit #### Step 1: Calculate Mean Diameter and Blade Speed \[ D_{mean} = \frac{D_{hub} + D_{tip}}{2} = \frac{0.15 + 0.65}{2} = 0.40\,m \] \[ r_{mean} = \frac{D_{mean}}{2} = 0.20\,m \] \[ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 650}{60} \approx 68.07\, \text{rad/s} \] \[ U = \omega \times r_{mean} = 68.07 \times 0.20 \approx 13.61\, m/s \] #### Step 2: Calculate Axial Velocity (\( V_a \)) Area at mean diameter: \[ A = \frac{\pi}{4}(D_{tip}^2 - D_{hub}^2) \approx 0.314\,m^2 \] Density of air (\( \rho \)): \[ \rho = \frac{P}{RT} = \frac{100,000}{287 \times 301} \approx 1.15\,kg/m^3 \] Axial velocity: \[ V_a = \frac{\dot{m}}{\rho A} \approx 5.54\, m/s \] #### Step 3: Calculate Blade Angles Using the velocity triangle: \[ \tan \beta = \frac{V_a}{U} \] \[ \beta_{mean} = \arctan \left( \frac{V_a}{U} \right) \approx 22.1^\circ \] ### b) Pressure Developed in mm H₂O Using the pressure coefficient: \[ \Delta P = \psi \cdot 0.5 \rho U^2 \] \[ \Delta P \approx 29.83\,Pa \] Convert to mm H₂O: \[ \Delta P_{mm\,H_2O} = \frac{29.83}{9.81} \approx 3.04\, mm\, H_2O \] ### c) Fan Efficiency Given as \( \eta = 85\% \). ### d) Power Required to Drive the Fan Calculate volume flow rate: \[ Q = \frac{\dot{m}}{\rho} \approx 1.74\, m^3/s \] Calculate fluid power: \[ \text{Power}_{\text{fluid}} = \Delta P \cdot Q \approx 51.9\,W \] Calculate shaft power: \[ \text{Shaft Power} = \frac{\text{Power}_{\text{fluid}}}{\eta} \approx 61.1\, W \] ## Summary of Results | Parameter | Value | |-----------------------------|------------------------| | Blade Angle (mean) | 22.1° | | Pressure Developed | 3.04 mm H₂O | | Fan Efficiency | 85% | | Power Required (to shaft) | 61.1 W | ## Final Answers - **a)** Air Angles at entry and exit: **22.1°** - **b)** Pressure developed: **3.04 mm H₂O** - **c)** Fan Efficiency: **85%** - **d)** Power required to drive the fan: **61.1 W** Feel free to ask for further clarifications or additional questions!