# Forces on Jet Deflectors (Thrust Reversers) Let's break down the physics of the two jet deflectors shown: one **round (circular arc)** and one with **straight panels**. Both are designed to redirect the jet exhaust at 45° angles relative to the engine axis. ## **Given Data** - **Velocity of gases, \( V \) = 500 m/s** - **Nozzle area, \( A \) = 0.2 m²** - **Density, \( \rho \) = 0.5 kg/m³** - **Angle of deflection, \( \theta \) = 45°** - **Flow is incompressible** --- ## **Step 1: Find the Mass Flow Rate (\( \dot{m} \))** \[ \dot{m} = \rho A V = 0.5 \times 0.2 \times 500 = \boxed{50~\text{kg/s}} \] --- ## **Step 2: Momentum Change (Force) for Each Deflector** ### **General Principle** - The **force** exerted by the jet on the deflector is equal to the **rate of change of momentum** of the gases as they are redirected. - We analyze the \( x \) (horizontal) and \( y \) (vertical) components. --- ## **A. Round Deflector (Circular Arc)** - **Exhaust redirected symmetrically 45° up and 45° down** - **No net vertical force** (y-components cancel) - **Both streams contribute to reversing the x-momentum** ### **Velocity Components After Deflection** Each stream (half the total mass flow) is redirected at 45°: \[ V_x = V \cos(45^\circ) = 500 \times \frac{1}{\sqrt{2}} \approx 353.6~\text{m/s} \] \[ V_y = V \sin(45^\circ) = 500 \times \frac{1}{\sqrt{2}} \approx 353.6~\text{m/s} \] ### **Force Calculation** - **Initial x-momentum:** \( \dot{m} V \) (all in -x) - **Final x-momentum:** Each half goes at \( +353.6~\text{m/s} \) and \( -353.6~\text{m/s} \) (but both at 45° to the axis). - **But both jets are reversed, so the net x-momentum is \( \dot{m} V_x \) in +x for each, but since both are symmetric, add both:** \[ \text{Total force in x:} \] \[ F_x = \dot{m} \left[0 - (-V)\right] = \dot{m} V \] But after deflection, the x-component for each jet is: \[ \text{Each jet:}~~ V \cos(45^\circ) = 353.6~\text{m/s} \] Total x-momentum out: \[ \text{Total x-momentum out:} = 2 \times \frac{1}{2} \dot{m} V \cos(45^\circ) = \dot{m} V \cos(45^\circ) \] **So, the net force on the deflector:** \[ F_x = \dot{m} \left[ V \cos(0) - V \cos(45^\circ) \right] \] \[ = 50 \left[500 - 353.6\right] = 50 \times 146.4 = \boxed{7320~\text{N}} \] This is the **force in the direction of motion** (slowing the plane). ### **No Sideways Force (Y-Direction)** The vertical (y) components cancel due to symmetry. --- ## **B. Straight Panel Deflector** The calculation is the **same** as above because the **exit direction is the same** (45° up and 45° down). The geometry doesn't affect the net change in momentum (assuming a perfect deflection without losses). \[ F_x = 50 \left[500 - 353.6\right] = \boxed{7320~\text{N}} \] --- ## **Summary Table** | Deflector Type | Force Opposing Motion (\( F_x \)) | Vertical Force (\( F_y \)) | |------------------|------------------------------------|-----------------------------| | Round (Arc) | 7320 N | 0 N | | Straight Panels | 7320 N | 0 N | --- ## **Physical Interpretation** - **Both deflectors reverse the exhaust at 45° angles**, so the **change in x-momentum (opposing the airplane motion) is the same** for both. - **No net vertical force** due to symmetry (equal upward and downward jets). - The **force calculated (7320 N)** is the **reverse thrust** generated by the deflectors to slow down the airplane. --- ### **Let me know if you want the vector breakdown or a diagram explanation!**