Question

Water of density \( \rho = 1000 \, {kg/m}^3 \) flows with a velocity \( V = 50 \, {m/s} \) through a 180° curved tube of uniform cross-section as shown in the figure. If the flow rate is \( 0.06 \, {m}^3/{s} \), the magnitude of the reaction force \( F_x \) required to keep it stationary is ________ kN (rounded off to one decimal place).

Hint:

In curved flow problems, the force required to keep the flow stationary is related to the change in momentum due to the change in velocity direction.

Answer 1

Step 1: Understanding the problem. We need to find the reaction force \( F_x \) required to keep the curved pipe stationary. We can use the principle of conservation of momentum and apply it to the control volume that contains the curved section of the tube. 
Step 2: Calculate the mass flow rate. The mass flow rate \( \dot{m} \) is given by: \[ \dot{m} = \rho \times Q, \] where \( \rho = 1000 \, {kg/m}^3 \) is the density and \( Q = 0.06 \, {m}^3/{s} \) is the flow rate. Thus: \[ \dot{m} = 1000 \times 0.06 = 60 \, {kg/s}. \] Step 3: Apply the principle of momentum. The change in momentum for the 180° curved section is calculated by: \[ F_x = \dot{m} \times (V_1 - V_2), \] where \( V_1 = V_2 = 50 \, {m/s} \) (velocity is the same at both points). In a 180° turn, the velocity direction changes, so we use the velocity vector change rather than the magnitude. The change in velocity in the direction of the reaction force is \( \Delta V = 2V \), as the direction of the velocity vector flips. Thus, the reaction force \( F_x \) is: \[ F_x = \dot{m} \times 2V = 60 \times 2 \times 50 = 6000 \, {N}. \] Step 4: Convert to kN and round off. \[ F_x = \frac{6000}{1000} = 6.0 \, {kN}. \] Final Answer: The magnitude of the reaction force required to keep the pipe stationary is \( \boxed{6.0} \, {kN} \).