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Dahasolnce [82]
3 years ago
7

A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema

ins. The initial volume is 0.114 ft3 . During the process, 16.2 Btu of heat are added. Find the work done. Also draw this process in the P-v space indicating the initial and final states with respect to the liquid-vapor dome.

Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

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Steam enters an insulated turbine at 100 bar, 400oC. At the exit, the pressure and quality are 200 kPa and 0.47, respectively. D
OverLord2011 [107]

Answer:

power produced = 3098.52 kW

Explanation:

given data

insulated turbine = 100 bar

temperature = 400°C

pressure = 200 kPa

mass flow rate = 1.99 kg/s

solution

we use here steam table for At 100 bar and 400°C  

h1 = 3096.5 KJ/Kg

and  

at P2 = 200 Kpa

h2 = hf + 0.47 hg

h2 = 504.7 + 0.47 ×  2201.6  

h2 = 1539.452 KJ/Kg

so here

power produced  is express as

power produced = m × (h1 - h2)    .................1

power produced = 1.99 × ( 3096.5 - 1539.452 )

power produced = 3098.52 kW

8 0
3 years ago
Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This
Art [367]

Answer:

vA = -2.55 m/s

vB = 0.947 m/s

Explanation:

Given:-

- The initial angle of rope, α = 30°

- The angle of rope just before impact or wedge angle, θ = 20°

- The weight of sphere, Ws = 1-lb

- The initial position velocity, vi = 4 ft/s

- The coefficient of restitution, e = 0.7

- The weight of the wedge, Ww = 2-lb

- The spring constant, k = 1.8 lb/in

- The length of rope, L = 2.6 ft

Find:-

 Determine the velocities of A and B immediately after the impact.

Solution:-

- We can first consider the ball ( acting as a pendulum ) to be isolated for study.

- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.

- According to the principle of conservation for the initial point and the point just before impact.

Let,

              vA : The speed of sphere ball before impact

               

                  Change in kinetic energy = Change in potential energy

                  ΔK.E = ΔE.P

                  0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )

                  uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2

                  uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822

                  uA = 5.316 ft/s

- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:

                 e^2 = ( K.E_after impact / K.E_before impact )

- The respective Kinetic energies are:

               

                K.E_after impact = K.E_sphere + K.E_block

                                             = 0.5*ms*vA^2 + 0.5*mb*vB^2

                K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )

                                                         = 1*2.6*( cos ( 20 ) - cos ( 30 ) )

                                                         = 0.1915 J

                32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2  

                6.00544 = vA^2 + 2*vB^2  ... Eq1

- From conservation of linear momentum we have:

                vB = e*( uA - uB )*cos ( 20 ) + vA

                vB = 0.7*( 5.316 - 0 )*cos ( 20)   + vA

                vB = 3.49678 + vA  .... Eq 2

- Solve two equation simultaneously:

               

               6.00544 = vA^2 + 2*(3.49678 + vA)^2

               6.00544 = 3vA^2 + 13.98*vA + 24.455

               3vA^2 + 14.8848*vA + 18.4495 = 0

               vA = -2.55 m/s

               vB = 0.947 m/s

                                 

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Uhm is there a multiple choice?
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Why do automotive technicians need to have good computer skills?​
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