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joja [24]
3 years ago
5

A steady state filtration process is used to separate silicon dioxide (sand) from water. The stream to be treated has a flow rat

e of 50 kg/min and contains 0.22% sand by mass. The filter has a cross-sectional area of 9 square meters and successfully filters out 90% of the input sand by mass. As the filter is used, a cake forms, which we will assume is pure sand (SG of sand = 2.25). The filter needs to be replaced once this cake has a thickness of 0.25 meters.
Required:
How long can a new filter be used before it needs to be replaced, in units of days?
Engineering
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

3.19 days

Explanation:

<em>Given data :</em>

stream flow rate = 50 kg/min

stream contains ; 0.22% sand by mass

Cross sectional area of filter = 9 m^2

Filter successfully filters out 90% of the input sand by mass

SG = 2.25

thickness of cake formed = 0.25 meters

<u>Determine how long a new filter can be used before replacement </u>

Given that for every 1 minute  5.43*10^-5 m thickness of sand layer(cake) forms on the filter and the replacement of filter is done once the cake thickness = 0.25 meters

To determine the number of days ( X ) before replacing filter we apply the relationship below

5.43 * 10^-5 = 1 min

0.25 m = X

hence ; X = 0.25 / (5.43 * 10^-5 )  = 4604.051 minutes ≈ 3.19 days

<u />

Attached below is the beginning part of the detailed solution

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The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet
Crank

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, I_s = 2 A

power supplied, P_s = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A

The number of turns on the input side is calculated as;

\frac{N_p}{N_s} = \frac{I_s}{I_p}  \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns

4 0
2 years ago
Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False
Valentin [98]

Answer:

I would say false but I am not for sure

8 0
2 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
vodomira [7]

Answer:

I = 0.625 A

Explanation:

Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

5 0
2 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate
RUDIKE [14]

Answer:

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Explanation:

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T_1 = 32^0 C,  m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0

if we take  

The mass flow rate of the second stream = m_2(kg/min)

The mass flow rate of mixed exit stream = m_3 (kg/min)

Now from mass conservation

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Therefore the mass flow rate of second stream will be 4.5 kg/min.

7 0
3 years ago
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