Fluid Mechanics seventh edition by Frank M. White · Problem P3.65
Integral Relations for a Control Volume — Frank M. White
Physics — Fluid Mechanics · linear-momentum-equation
Problem
The box in Fig. P3.65 has three 0.5-in holes on the right side. The volume flows of 20°C water shown are steady: 0.1 ft/s out the top hole, 0.2 ft/s out the middle hole, and 0.1 ft/s out the bottom hole. The details of the interior are not known. Compute the force, if any, that this water flow causes on the box.Intuitive explanation
This problem tests the momentum equation for a control volume—specifically, how flowing fluid exerts force on a container. The key insight is that the net force on the box equals the rate of momentum flowing in minus the rate flowing out. Since water enters through one port and exits through three holes at different heights, you'll need to account for the momentum of each exit stream (mass flow rate times velocity) and apply Newton's second law in integral form to find the total force the box must exert to change the flow's direction and speed.
Intuitive explanation
This problem tests the momentum equation for a control volume—specifically, how flowing fluid exerts force on a container. The key insight is that the net force on the box equals the rate of momentum flowing in minus the rate flowing out. Since water enters through one port and exits through three holes at different heights, you'll need to account for the momentum of each exit stream (mass flow rate times velocity) and apply Newton's second law in integral form to find the total force the box must exert to change the flow's direction and speed.