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Andru [333]
2 years ago
14

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.09 m3/s and at a velocity of 3 m/s, and it leaves

in the normal direction along the pump casing. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction. Take the density of water to be 1000 kg/m3.
Physics
1 answer:
Angelina_Jolie [31]2 years ago
8 0

To develop this problem it is necessary to apply the concepts related to Mass Flow, as a unit dependent on density and volumetric flow rate, as well as apply the sum of Forces on the flow. We will start by determining the mass flow:

\dot{m} = \rho \dot{V}

Where,

\rho = Density of water

\dot{V} = Volumetric flow rate

Replacing our values we have

\dot{m} =1000*0.09

\dot{m} = 90Kg/s

Calculate the force acting on the shaft by using the moment equation for steady one-dimensional flow,

\sum \vec{F} = \sum \limit_{out} \beta \dot{m} \vec{V} - \sum_{in} \beta \dot{m} \vec{V}

(-F_R)_x = -\beta \dot{m}V

Where

\beta=momentum flux Correction factor

V = Velocity

Replacing we have then

\beta = 1

\dot{m} = 90kg/s

V = 3m/s

-(F_R)_x = (1\times 90\times 3)

(F_R)_x = 270N

Therefore the force acting on the shaft is 270N

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Dmitriy789 [7]

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance C=8.5nF=8.5\times 10^{-9}f

Potential difference across capacitor V = 12 volt

Current through capacitor i=3.33mA=3.33\times 10^{-3}A

Capacitive reactance will be equal to X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm

Capacitive reactance is equal to X_c=\frac{1}{\omega C}

3603.60=\frac{1}{\omega\times  8.5\times 10^{-9}}

\omega =32647.091rad/sec

2\pi f=32647.091

f=5198.98Hz

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

3 0
2 years ago
The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in
Kisachek [45]

Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

w(t)=25-t^2

We have to find the rate of change of water in the tank at t = 3 minute

For rate of change we have to differentiate both side

So \frac{dw}{dt}=0-2t

At t = 3 minute

\frac{dw}{dt}=0-2\times 3=-6gallon/minute

8 0
2 years ago
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xxTIMURxx [149]

Answer:

The correct answer would be B. 18 to 26%.

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It is estimated that human muscles have an efficiency of about 18% to 26%.

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2 years ago
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MAVERICK [17]
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

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2 years ago
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2 years ago
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