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grandymaker [24]

2 years ago

5

22.5-4. Minimum Liquid Flow in a Packed Tower. The gas stream from a chemical reactor contains 25 mol % ammonia and the rest are

inert gases. The total flow is 181.4 kg mol/h to an absorption tower at 303 K and 1.013 × 105 Pa pressure, where water containing 0.005 mol frac ammonia is the scrubbing liquid. The outlet gas concentration is to be 2.0 mol % ammonia. What is the minimum flow L min ⁡ ′ Using 1.5 times the minimum, plot the equilibrium and operating lines. Ans. L min ⁡ ′ = 262.6kg mol/h

1 answer:

Firdavs [7]2 years ago

5 0

Answer:

check attachments for answers

Explanation:

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Murrr4er [49]

Answer: Instrumentality;low

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2 years ago

True or False: Drag and tailwind are examples of a contact force.<br> tyy guyss

zloy xaker [14]

Answer:

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Explanation:

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2 years ago

The force exerted on a bridge pier in a river is to be tested in a 1:10 scale model using water as the working fluid. In the pro

Len [333]

Answer:

Force on the prototype is 5000 N

Solution:

As per the question:

Depth of water, x = 2.0 m

Flow velocity, v' = 1.5 m/s

Width of the river, w = 20 m

Force on the bridge pier model, F' = 5 N

Pressure, Ratio = Ratio of scale length

Scale = 1:10

Now,

$\frac{P'}{P} = \frac{x}{w} = \frac{2.0}{20}$

where

P' = pressure on model

P = pressure on prototype

$\frac{\frac{F'}{A'}}{\frac{F}{A}} = \frac{1}{10}$

where

F' = Force on model

F = Force on prototype

A' = Area of model

A = Area of prototype

Now:

$\frac{F'}{F}.\frac{A}{A'} = \frac{1}{10}$

$\frac{5}{F}.\frac{1}{\frac{1}{10}}.\frac{1}{\frac{1}{10}} = \frac{1}{10}$

F = 5000 N

3 0

2 years ago

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

monitta

The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

3 0

2 years ago

A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs

Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)= 50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)= 50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg = $13.6\times 10^3\ kg/m^3$

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6 x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0

2 years ago