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2 years ago

7

Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un

til the tank contains only saturated vapor. Determine the final mass of vapor in the tank, in kg, and the final pressure, in bar.

1 answer:

elena55 [62]2 years ago

7 0

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

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$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

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Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

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initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

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$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

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$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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