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DiKsa [7]
2 years ago
7

Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet

pipe is round with a diameter of 5.2 cm. Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3). For steady-state operation, determine the inlet velocity, in m/s, and the mass flow rate at each exit, in kg/s.
Physics
1 answer:
yaroslaw [1]2 years ago
7 0

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2)  and the remainder leaves at (3)

The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

m_{3}=m_{1}-m_{2}

m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

We need to calculate the inlet velocity

Using formula of velocity

v=\dfrac{m}{\rho A}

Put the value into the formula

v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

v=21.9\ m/s

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

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LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

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4 0
2 years ago
An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration
ollegr [7]

Answer:

<h3>The answer is 4.29 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken acting on it can be found by using the formula

acceleration =  \frac{velocity}{time}  \\

From the question

velocity = 30 m/s

time = 7 s

We have

acceleration =  \frac{30}{7}  \\  = 4.285714...

We have the final answer as

<h3>4.29 m/s²</h3>

Hope this helps you

7 0
2 years ago
When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

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Let's analyze the situation of the globe

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We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

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Ugo [173]

Answer:

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Explanation:

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a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

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P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

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d.

P rad  = I / us

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ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0
2 years ago
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V125BC [204]

Answer:

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Given: R1 = R2 = 8 Ω

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Substitute into equation 1

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7 0
2 years ago
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