Final answer:
The kinetic energy of the gas exiting is 16 times greater than the kinetic energy of the gas entering due to the outlet diameter being half of the inlet diameter and the principle of the continuity equation in fluid dynamics.
Step-by-step explanation:
The kinetic energy of the gas exiting a truncated cone, where the outlet diameter is half the diameter of the inlet, can be determined using the principles of fluid dynamics – particularly, the continuity equation and the equation for kinetic energy. According to the continuity equation, since the gas flows at constant density and steady-state, the mass flow rate is conserved. This means that A1V1 = A2V2, where A1 and A2 are the cross-sectional areas of the inlet and outlet, respectively, and V1 and V2 are the velocities. Given that Area is proportional to the square of the diameter, when the outlet diameter is half of the inlet diameter, the outlet area is one-fourth of the inlet area. Subsequently, to conserve mass flow rate, the velocity at the outlet must be four times greater than the velocity at the inlet (V2 = 4V1). Kinetic energy, given by KE = (1/2)mv², will be 16 times greater since velocity is squared in the equation for kinetic energy. Therefore, the kinetic energy of the gas exiting is 16 times greater than the kinetic energy of the gas entering.