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Dmitry_Shevchenko [17]
3 years ago
10

Air can be humidified by passing through a long tube whose inside diameter is lined with a wick saturated with liquid water. The

device is shown below. In the process, air containing 1.0 mol % water vapor at 40°C and 1 atm total pressure enters a 3.0 cm inside diameter tube at a velocity of 3 m/s. Water evaporates from the wick as the air passes over it. Water adsorbent liner with 1.0 mm thick 0.6 g water/cubic cm liner. Gas stream velocity of 3 m/s, 1 atm and 40C D=3.0 cm L=8m There is no diffusion resistance of water vapor through the wick itself. The entire process is maintained at 40°C where the saturated vapor pressure of water is 55.4 mmHg. The viscosity of air under these conditions is 1.91 x 104_8_ and the density of air is 1.13 x 10-38. The cm-S diffusivity is 0.28
(i) Using a shell balance, develop a differential equation to predict the concentration of water vapor exiting the tube.
(ii) Integrate the equation you developed in (a) with the appropriate boundary conditions (or limits) to obtain the final form of the equation to predict the concentration of water vapor exiting the tube.
(iii) If the thickness of the wick lining the inner surface of the tube is 1.0 mm and it initially contains 0.6 g of water per cm'. If the tube is 8.0 m long and humidification occurs at 40°C and 1.0 atm pressure, determine the concentration of water vapor in the outlet air.
(iv) Determine how long the process can operate before the liquid water is depleted.
Physics
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

IV because the process of water is equal to 5,8 to 783253.23 to the hendroxagram of 4.

Explanation:

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A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
2 years ago
Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets
schepotkina [342]
I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is:  There's a <em>pair</em> of forces, <em>in both directions</em>, between
every two masses.

-- The strength of the force depends on the <em>product</em> of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal. 
It's the product that counts.  Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers.  More distance => weaker force.  Actually, (more distance)² ==> weaker force.

-- The forces are <em>equal in both directions</em>.  Your weight on Earth is exactly equal to
the Earth's weight on you.  You can prove that.  Turn your bathroom scale face down
and stand on it.  Now it's measuring the force that attracts the Earth toward you. 
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth. 
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal.  But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

--  This works exactly the same for every pair of masses in the universe.  Gravity
is everywhere.  You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity".  It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon 
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons.  (That's
about 2.205 pounds).  The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram.  In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.
7 0
3 years ago
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

5 0
3 years ago
Read 2 more answers
Which of the following statements is true?
Alborosie
My guess is A. I'm not 100% positive but i'm pretty sure.
8 0
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