Net pressure force

The pressure distribution given by Eq. (3-9) is an odd function of 0, so that the particle experiences a net pressure force or form drag. Integration of the pressure over the surface of the particle leads to a drag component given by... 

This result may be contrasted with potential flow past a sphere, where the streamlines again have fore-and-aft symmetry but p is an even function of 9 so that there is no net pressure force (see Chapter 1). Additional drag components arise from the deviatoric normal stress ... 

These are the Navier-Stokes equations for steady constant fluid property flow. They are sometimes termed the x-, y-, and z-momentum equations, respectively. Basically these equations state that die net rate of change of momentum per unit mass in any direction (the left-hand side) is the sum of the net pressure force and the net viscous force in that direction. 

We are interested only in the momentum in the x direction because the forces considered in the analysis are those in the x direction. These forces are those due to viscous shear and the pressure forces on the element. The pressure force on the left face is p dy, and that on the right is - [p + (dp/dx) dx] dy, so that the net pressure force in the direction of motion is... 

Figure 3-4. A pictorial representation of the force balance on the fluid within an arbitrarily chosen section of a circular tube in steady Poiseuille flow. Pressure forces act on the two cross-sectional surfaces at z = 0 and z = L, while a viscous stress acts at the cylindrical boundary and exactly balances the net pressure force.

However, the quantity on the right-hand side is just the net pressure force in the z direction acting on the fluid between z = 0 and z = L the force Pz=0jtR2 acting on the upstream surface at z = 0 minus the opposite-directed force Pz=lttR2 acting on the downstream surface at z = L. Thus we have demonstrated that the net pressure and viscous forces acting on any macroscopic body of fluid in steady, unidirectional Poiseuille flow must be in exact balance. The same conclusion is true for any steady, unidirectional flow. 

Example 2.7. A layer of rainwater 4 in deep collects on the roof of the oil storage tank of Example 2.6. What net pressure force does it exert on the roof of the tank ... 

This result is most welcome, because it allows us to neglect the acceleration of die fluid as it passes around and between the RBCs, and to establish a continuous balance between the local net pressure force acting on an element of fluid and the viscous stresses acting on the same fluid element. The equation to be solved is therefore... 

There is no net pressure force acting on this section as the projected area is zero. 

The negative sign indicates that the vertical component of Ng is opposite to that assumed in Fig. 13.10c and is in compression rather than tension. This is caused by the column of liquid above the cone whose weight is greater than the net pressure force at section c-c. 

The height of the gasket is usually set so that the net pressure force does not exceed the seating force. Thus,... 

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