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Volgvan
3 years ago
6

Steam at 0.6 mpa, 200 oc, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 mpa and a veloci

ty of 600 m/s. Determine the final temperature if the steam is superheated in the final state, and the quality if it is saturate.
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and

temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m(h1 + \frac{v1^2}{2} +gz1) + Q = \dot m(h2 + \frac{v2^2}{2} +gz1) + W

\dot m(h1 + \frac{v1^2}{2} = \dot m(h2 + \frac{v2^2}{2}

(h1 + \frac{v1^2}{2} = (h2 + \frac{v2^2}{2}[tex]2850.6 + \frac{50^2}{2} *\frac{1 kJ/kg}{1000 m^2/s^2} = h2 + \frac{600^2}{2} *\frac{1 kJ/kg}{1000 m^2/s^2}

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
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Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

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