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Liono4ka [1.6K]
3 years ago
6

Air at 38oC and 97% relative humidity is to be cooled to 14oC and fed into a plant area at a rate of 510 m3/min. Calculate the r

ate (kg/min) at which water condenses. (Round the answer to the nearest whole number)
Physics
1 answer:
bekas [8.4K]3 years ago
8 0

Answer:

mass flow rate at  water condenses is 36.72 kg/min

Explanation:

given data

temperature t1 = 38°C

temperature t2 = 14°C

humidity ∅= 97 % = 0.97

rate v = 510 m³/min

to find out

mass flow rate at  water condenses

solution

by gas equation we find here mass flow rate  that is

pv = mRT

put here value and p is 0.066626 bar at 38°C and find m

m = 0.06626 × 10^{5} × 510  / 287×311

m = 37.85 kg/min

so at water condenses mass flow rate is express as

∅ = M / m

Mass flow rate M = ∅ × m

M = 0.97 × 37.85

mass flow rate = 36.72 kg/min

so mass flow rate at  water condenses is 36.72 kg/min

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You wrote. E = m g h.
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Now divide each side by (g h), and you'll have the formula for mass:

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5 0
3 years ago
a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

#SPJ4

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1 year ago
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Andru [333]
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The correct answer is B

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Nataly [62]

Answer:

The correct answer is the second option

Explanation:

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