Chemistry, 23.08.2019 05:10 caprisun1491

Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. the outlet volumetric flowrate (m3/sec) from the tank is 0.01 times of the liquid height (in meter) in the tank. if the cross section area of is 0.01 m2 how long does it take to fill the tank up to 0.6 meter from the bottom? how long does it take to fill the tank up to 2 meter from the bottom?

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Answer from: venny23

Explanation:

Apply the mass of balance as follows.

Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

$\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

$\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

$\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

$A_{c} = 0.01 m^{2}$

$\frac{dh}{dt}$ + h = 1

$\frac{dh}{dt}$ = 1 - h

$\frac{dh}{1 - h}$ = dt

$\frac{ln(1 - h)}{-1}$ = t + C

Given at t = 0 and V = 0

$A \times h$ = 0

or, h = 0

-ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

C = 0

So, -ln(1 - h) = t

or, t = $ln (\frac{1}{1 - h})$ ........... (1)

(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

t = $ln (\frac{1}{1 - h})$

t = $ln (\frac{1}{1 - 0.6})$

= $ln (\frac{1}{0.4})$

= 0.916 seconds

(b) As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$ .

1 - h = exp (-t)

1 - h = 0

h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

Answer from: Quest

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

Answer from: Quest

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

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Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank...

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