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Triss [41]
3 years ago
10

For binary flash distillation, we discussed in class that there are 8 variables (F, ZA, V, ya, L, XA, P and T) and 4 equations d

erived from VLE and mass balances. Thus, we typically require 4 of these variables to be given so that we can obtain a unique solution to the problem. Let's say, your manager tells you that he has a feed mixture with 2 components (given F, za) and he requires you to come up with a flash column that can produce a certain desired amount of Vapor product (thus V, ya are specified). Identity of both components is known and all VLE data has been provided to you. Has the manager given you enough data? If yes, give a step-by-step description of how would you go about designing the flash column (basically find P and T)? If no, why?
Engineering
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

yes

Explanation:

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A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu
Temka [501]

Answer:

(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN

Explanation:

Solution

Given

A HSS152.4 × 101.6 × 6.4 structural steel is used as a column

Actual length of the column , L= 6 m

The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties  of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

(a)  If the end condition is pinned - pinned

The effective  length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is  fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

=63.33kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 63.33/2

31.665 kN

Therefore the maximum safe for the column is 31.665 kN

(c) If the end condition is fixed- pinned

The effective  length factor K =0.7

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²

=516979.2 8N

=516.98 kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 516.98 kN/2

=258.49 kN

Therefore the maximum safe for the column is 258.49 kN

(d) If the end condition is fixed -fixed

The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

=1013279.4 N

=1013.28 kN

The maximum safe load,  P allow = Pcr/F.S

P allow = 1013.28 / 2

= 506.64 kN

P allow = 506.64 kN

Therefore the maximum safe for the column is 506.64 kN

8 0
4 years ago
Represent each of the following units as a combination of primitive
zimovet [89]

Answer:

a. unit of length: [L]

b. unit of volume: [L^3]

c. unit of pressure:P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]} [ML^{-1}T^{-2}]

d. unit of power: N.m.s^{-1}\equiv [ML^2T^{-3}]

e. unit of force: [kg.m/s^2]\equiv [MLT^{-2}]

f. unit of power: N.m.s^{-1}\equiv [ML^2T^{-3}]

Force: F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t

Power: P=\frac{W}{t}=\frac{F.x}{t}

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

3 0
3 years ago
The transfer function of a typical tape-drive system is given by
maw [93]

Answer:

the range of K can be said to be :  -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}

calculating the characteristics equation; we have:

1 + KG(s) = 0

1+   \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0

{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as  follows:

S^5             1                     5.1                          2+ K

S^4            1.9                   6.2                           4K

S^3             1.83            \dfrac{1.9 (2+K)-4K}{1.9}          0

S^2        \dfrac{11.34-1.9(X)}{1.83}       4K                         0

S          \dfrac{XY-7.32 \ K}{Y}        0                            0

\dfrac{1.9 (2+K)-4K}{1.9} = X

 

\dfrac{11.34-1.9(X)}{1.83}= Y

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34  - 1.9(X) > 0

11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0

11.34  - (3.8 - 2.1K)>0

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

(\dfrac{3.8+1.9K-4K}{1.9})[11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be :  -3.59 < K< 0.35

4 0
3 years ago
How to draw a location/site plan​
leva [86]

Answer: dont know

Explanation:

6 0
4 years ago
Read 2 more answers
A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z                        ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³     .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × 10^{6} / 3

so d = 0.0133 m

so diameter is 14 mm

3 0
4 years ago
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