A distillation column with a partial reboiler and a total condenser is being used to separate a Xmixture of Propane (1), Propylene (2) , and Butylene (3). The feed is 30.0% mole Propane, 30.0% mole Propylene and 40.0% mole Butylene. The feed is input as a saturated vapor. We desire 94.00% recovery of the Propylene in the bottoms and 98.50% recovery of the Propane in the distillate. The reflux is returned as a saturated liquid, and CMO can be assumed. Equilibrium can be represented as constant relative volatilities. Choosing Propylene (2) as the reference component, a12 = 1.48 and a32 = 0.65. Use a basis of 100 moles of feed/h. Use a %3D basis of 100 moles of feed/h. Assuming this system has a class II pinch possible roots (O) have been determined as 0.9321, 1.2962 and 1.5424. Find the minimum reflux ratio, Rmin:

Determine the mass, in kg, of 0.1 m3 of Refrigerant 134a at 4 bar, 100⁰C.

Cylindrical Coordinates (r, 0, z): Continuity equation ap 10(prv,) 10(pve), a(pv₂) + + + at r ər ae дz Navier-Stokes Equation for constant p and u 10² vr (avr avr να δν, avr vo + p² მ02 r-dir: at ər r 20 дz r 1 ap (ave ave ve ave aveva) = + a (1a(rve) 1 0² ve + arr Ər 2 მ02 + + 1 - 0-dir: P at ər r 20 дz r rae avz ap əvz ve dv₂ д 1 0² vzd²v₂) r² 20² дz2 {+ - ( 1² / (+²) + r Əvz at + z-dir: Р + Vr + дz ər r 20 дz rər ər Stokes viscosity relation 1 dur dve, Tez-Tze=μ[+1 duz T₁z=z=μ(dur + dr Tre Ter= dz r de r de A Newtonian fluid flows inside an annular (¹) cylinder with height H as shown in Figure p3. The radius of the inner cylinder is kR; while the outer cylinder with radius R. The fluid is filled between the inner and outer cylinders. The inner cylinder rotates with a constant angular velocity Qi (Voi = Ni kR) and the outer cylinder rotates with a constant angular velocity No (V₂0= No R). Assume V₁ #0, dp/d 0 = 0, d V₁/dz = 0, and g₁ = 0. (a) Analyze the continuity equation. (b)…

2. Two incompressible viscous fluids of the same density p flow, one on top of the other, down an inclined plane making an angle a with the horizontal. Their viscosities are µi and µ2, the lower fluid is of depth hị and the upper fluid is of depth h2. Show that: u,(y) = (hj +h2. 2 (t)- )y - 1 2 gsin a V1 so that the velocity of the lower fluid u¡(y) is dependent on the depth h2, but not the viscosity, of the upper fluid. Why is this?