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QUESTION 24

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Concepts and reason

Spring force:

The force exerted by a spring due to compression or stretching when it is attached to an object is called spring force.

Properties of water:

The various phases of the pure water are as shown in figure (1).

4
3.
Т
V
Figure 1
P const

State 1: Compressed liquid

The state of the water which comprises of only liquid phase at a time is known as compressed liquid.

State 2: Saturated liquid

The state of the water which comprises of liquid phase and it will start to vaporize on further heating is known as saturated liquid

State 3: Saturated liquid vapor mixture

The state of the water which comprises of both liquid and gaseous phases simultaneously at a time is known as saturated liquid vapor mixture.

State 4: Saturated vapor

The state of the water which comprises of gaseous phase and it will start to condense into liquid on further cooling is known as saturated vapor.

State 5: Superheated vapor

The state of the water which comprises of only gaseous phase at a time is known as superheated vapor.

Isobaric Process:

Isobaric process is a thermodynamic process in which, the pressure is kept constant during heat input given to the system or heat rejection from the system takes place. The work done in an isobaric process is the product of pressure and change in volume of the system.

Property tables:

Property tables are the thermodynamic tables which provide the data about temperature, pressure, specific enthalpy, specific entropy and specific volume at certain saturation temperature, saturation pressure or at combined temperature and pressure in superheat condition.

Fundamentals

Pressure:

Pressure is the force acting perpendicular to the area and it is given by,

F
P =
A

Here, the pressure is Р
, force is F and the area is A.

Spring force:

According to Hooke’s law, the spring force exerted by a spring is given by,

Here, the spring force is , spring constant is k and deformation in spring length is x.

Specific volume

The volume per unit mass of the substance is known as specific volume.

v=-

Here, specific volume is V
, volume of the pure substance is , and mass of the pure substance is т
.

Isobaric process:

The work done(w)
in an isobaric process 1-2
is given by,

W P(V-)

Here, the final volume is and the initial volume is .

(a)

Calculate the final pressure of the water.

P P
2

Here, the final pressure is 3
, the pressure until expansion takes place without spring force is Р.
and the displacement in piston position is x.

Substitute 250 kPa
for Р.
, 100 kN/m
for k
, 20 сm
for x, and 0.1 m2
for A
.

1 m
(100 kN/m) 20 cmx 100 cm,
1 kPa
250 kPa
P
(0.1 m2)
1 kN/m2
P 450 kPa

From the “Steam tables of water”, note the initial specific volume of water at initial pressure of 250 kPa
and temperature of 25°C
.

Initial specific volume, 4= s@sc =0.001003 m3/kgj
.

Calculate the initial volume(V)
.

И, 3 ту,
=

Here, the mass of water is m.

Substitute 50 kg for m and 0.001003 m3/kg
for .

=50 kg (0.001003 m/kg)
V 0.05015 m

Calculate the final volume in the piston cylinder (V,)
.

Here, volume when piston reaches linear spring is .

Substitute for , for x and for .

Calculate the final specific volume .

3
т

Substitute for 3
and 50 kg for m.

From the “Saturated water-Pressure tables”, note the specific volumes at ,

On observation, the final state of final specific volume is,

Thus, note the saturated temperature at as the final temperature (7)
from the “Saturated water-Pressure” table.

T =Tsat@450 kPa
=147.9°C

(b)

Draw the P-V diagram from the values of pressure and volume as in figure (1).

The work done during the process is nothing but the area under the lines 1-2 and 2-3.

Thus, calculate the work done (W
b,out
as,

Substitute for , for , for , for , for , and for 3
.

Ans: Part a

The final pressure is .

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