Chapter 7, Problem 147P

A continuous column flash system is separating 100 kmol/h of a saturated liquid feed that is 45 mol% methanol and 55 mol% water at 1.0 atm. Operate with L/V = 1.5 and the outlet bottoms at xN = 0.28.  Find the values of FL, FV, y1, and the number of equilibrium stages required. Find the value of Q used to vaporize FV. For a normal flash with the same feed and the same V/F, find the values of x and y.

A beer still is being used to separate ethanol from water at 1.0 atm. The saturated liquid feed flow rate is F = 840.0 kmol/h. The feed is 44.0 mol% ethanol. The saturated vapor steam is pure water with ratio of steam flow rate S to feed rate, S/F = 2/3. We desire a bottoms product that is 4.0 mol% ethanol. CMO is valid.  Find the mole fraction of ethanol in the distillate vapor, yD,E. Find the number of equilibrium stages required. If the feed is unchanged and the S/F ratio is unchanged, but the number of stages is increased to a very large number, what is the lowest bottoms mole fraction of ethanol that can be obtained?

3.1 Convert the following base-2 numbers to base-10: (a) 1011001, (b) 110.0101, and (c) 0.01011.

Consider the forces acting on the handle of the wrench in (Figure 1). a)  Determine the moment of force F1={−F1={−2i+i+ 4 jj −−8k}lbk}lb about the zz axis. Express your answer in pound-inches to three significant figures.   b)  Determine the moment of force F2={F2={3i+i+ 7 jj −−6k}lbk}lb about the zz axis. Express your answer in pound-inches to three significant figures.

I need you to explain each and every step (Use paper)

Calculate the Moment About the Point A -20"- 5 lb 40 N D 1.5 m 40 N 4.5 m A 15 lb. 150 mm 52 N 5 12 100 mm 15 lb. 26 lb. 12 5 34 lb. 13 8 15 77777 36 lb.

Calculate the Moment About the Point A -20"- 5 lb 40 N D 1.5 m 40 N 4.5 m A 15 lb. 150 mm 52 N 5 12 100 mm 15 lb. 26 lb. 12 5 34 lb. 13 8 15 77777 36 lb.

Formala for Hunzontal component= + cos & Vertical Component: Fsin t Find the vertical and horizontal components for the figure bellow: 30° 200 N 77 200 cos 30 = 173 N // 200 sin 30 = 100 N YA a₂+b₂ b₂ (b₁,b₂) a+b 20haits (a+b₁,a+b) Magnitude a and b a = lbl = 2o unite rugle of vector a wt Horisontal Axis = 30 11 vector & wt Honzontal Axis - 60° b b a= |a| Cas 30 a2 (a1, a2) ag = 10 bx = /b/ cos a 1 20 cos 80 = 17.32 Sia 30 = 20 sin 30. 60 = 10 = 20 Cos 60 = It by = 161 sin 60 = 20 sia 60 = 17.32 b₁ Rx ax +bx = 17.32 +10=2732 a₁ a₁+b₁ X By = ou + by= + + by = 10 + 17.32 =27.32 Magnitude = 38.637 Find the Vector a +b the Resultans The angle of the vector with the horizontal axle is 30 degrees while the angle of the vector b is 60 degrees. The magnitude of both vectors is 20 (units) angle of the Resultant vector = tam- " (14) 45

The net force exerted on the piston by the exploding fuel-air mixture and friction is 5 kN to the left. A clockwise couple M = 200 N-m acts on the crank AB. The moment of inertia of the crank about A is 0.0003 kg-m2 . The mass of the connecting rod BC is 0.36 kg, and its center of mass is 40 mm from B on the line from B to C. The connecting rod’s moment of inertia about its center of mass is 0.0004 kg-m2 . The mass of the piston is 4.6 kg. The crank AB has a counterclockwise angular velocity of 2000 rpm at the instant shown. Neglect the gravitational forces on the crank, connecting rod, and piston – they still have mass, just don’t include weight on the FBDs. What is the piston’s acceleration?