2C.6 Rotating cone pump (see Fig. 2C.6). Find the mass rate of flow through this pump as a func- tion of the gravitational acceleration, the impressed pressure difference, the angular velocity of the cone, the fluid viscosity and density, the cone angle, and other geometrical quantities labeled in the figure. Pressure P2 Pressure Pi (poundals ft-2) K 2B Az Direction of flow with mass rate of flow w (lb,,,/s) Fig. 2C.6 A rotating-cone pump. The variable r is the distance from the axis of rotation out to the center of the slit. (a) Begin by analyzing the system without the rotation of the cone. Assume that it is possible to apply the results of Problem 2B.3 locally. That is, adapt the solution for the mass flow rate from that problem by making the following replacements: replace (Po - P₁)/L by -dP/dz replace W by thereby obtaining 2πr 2πz sin ẞ . W = dp B³p 2πz sin B dz μ (2C.6-1) The mass flow rate w is a constant over the range of z. Hence this equation can be integrated to give (P1 - P2) = 3 μω 4π B³p sin ẞ In 는 L₁ (2C.6-2) (b) Next, modify the above result to account for the fact that the cone is rotating with angular velocity . The mean centrifugal force per unit volume acting on the fluid in the slit will have a z-component approximately given by (Fcentrif.) =Kpn²z sin² B (2C.6-3) What is the value of K? Incorporate this as an additional force tending to drive the fluid through the channel. Show that this leads to the following expression for the mass rate of flow: w= 4TB³p sin B (P₁- P₂) + (½KpN² sin² ß)(L{ — L³)] In (L₂/L₁) 3μ Here Pp; +pgL; cos ẞ. (2C.6-4)