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lianna [129]
2 years ago
10

The distillation column in Figure 3 is set up for so-called boil-up (V) control. It has

Engineering
1 answer:
11111nata11111 [884]2 years ago
3 0

A distillation column is an essential item used in the distillation of liquid mixtures to separate the mixture into its component parts, or fractions, based on the differences in volatilities. Fractionating columns are used in small scale laboratory distillations as well as large scale industrial distillations.

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Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
The Chinese had a toy helicopter some 1000 years ago. True Or False ?
Vadim26 [7]

Answer:

f

Explanation:

for the lazy people

5 0
3 years ago
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your
olchik [2.2K]

Solution:

The given formula,

x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}

y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}

\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}

From the table,

1) \(\quad x=2 \cdot 87, \quad y=0.96\)\\\(\frac{x}{y}=\frac{2187}{0.96}\)22895\\\\2) \(x=3 \cdot 466 \cdot y=6 \cdot 998\)\\\(\frac{x}{y}=\frac{3 \cdot 466}{0.898}\)\(=3 \cdot 4729\)

3\(x=3 \cdot 229, y=1 \cdot 08\)\\\(\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}\)\\=2.9898\)\\\\4) \(x=6.525, y=1.094\)\\\(\frac{x}{y}=\frac{5.625}{1.094}\)\\=5.0502

8 0
3 years ago
technician a says symbols are used to represent components in electrical wiring diagrams. technician b says components often loo
MArishka [77]
Technician a. Components should always look the same so they are known universally
3 0
3 years ago
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