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A rigid tank having a volume of 0.1 m3 initially contains water as a two-phase liquid-vapor...

A rigid tank having a volume of 0.1 m3 initially contains water as a two-phase liquid-vapor
mixture at 1 bar and a quality of 1%. The water is heated in two stages:
Stage 1: Constant-volume heating until the pressure is 20 bar. (treat as closed system)
Stage 2: Continued heating while saturated water vapor is slowly withdrawn from the tank at a
constant pressure of 20 bar. (Note: Use transient analysis with all the water leaving
having a quality of 1.0)
Heating ceases when all of the water remaining in the tank is saturated vapor at 20 bar. For the
water, evaluate the heat transfer, in kJ, for each stage of heating. Ignore KE and PE effects.
(4204, 8725)

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Concepts and reason

Specific internal energy:

Specific internal energy is one of the properties of the material which is referred as the ratio between the internal energy of material and material mass. It is also stated as the amount of kilo-joules obtained by one kilogram of the definite material. It is denoted by и
.

Specific volume:

The volume of a substance per unit mass is known as specific volume. It is also stated as the amount of cubic meters obtained by one kilogram of a definite substance. It is denoted by V
.

First law of thermodynamics

It states that energy cannot be created nor destroyed but can be changed from one form to another. Thus, energy is conserved. For an energy imbalance in a system, the energy change inside the system is difference between energy entering into the system and the energy leaving the system.

Heat transfer:

The phenomenon of transfer of energy from one body to another by virtue of temperature difference with or without the medium is known as heat transfer.

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.

First, obtain the various thermodynamic properties from the property table of “Saturated water-pressure table”. Then, obtain the specific internal energy at state 1 and state 2 to calculate the heat transfer in stage 1. Finally, obtain the specific internal energy and specific enthalpy to calculate the heat transfer in stage 2.

Fundamentals

The expression for specific volume (v)
given as follows:

v=-

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

The expression for specific volume is given as follows:

V=v,+xv,

Here, specific volume is v, specific volume of saturated liquid is and specific volume of saturated liquid vapor mixture is .

The expression for specific internal energy is given as follows:

и
и 3и, + хи

Here, specific internal energy is и
, specific internal energy of saturated liquid is и,
and specific internal energy of saturated liquid vapor mixture is .

The expression for the first law of thermodynamics is given as follows:

Q= AU +W

Here, heat transferred is Q, work done by exiting flow rate is W and the change in internal energy is Δυ
.

In a constant volume process, the work done is zero. Thus, the first law reduces to

= AU
о-m(и, -и,)

Here, the mass of the substance is m, initial internal energy , and final internal energy is .

Draw the Т-v
diagram showing stage 1 and 2 as in Figure (1).

T
20 bar
2
3
1 bar
1
Figure 1

Here, temperature is T
, and specific volume is V
.

From the table “Saturated water-pressure table”, obtain the following properties corresponding to the pressure of 1bar
.

Specific volume of saturated liquid, 0.001043 m/kg|
.

Specific volume of saturated vapor, 1.6941 m/kg
.

Specific internal energy of saturated liquid, = 417.4 kJ/kg
.

Specific internal energy of saturated vapor, 2505.6 kJ/kg
и.
.

Find the specific volume (v.)
at state 1.

V.

Here, dryness fraction is .

Substitute 0.001043 m3/kg
for , 0.01 for and 1.6941 m/kg
n
for .

(0.001043 m/kg)+(0.01)(1.6941-0.001043) m/kg
- 0.018 m/kg

Find the specific internal energy (u,)
at state 1.

n= n
-u

Substitute 417.4 kJ/kg
for ил
, 0.01 for and 2505.6 kJ/kg
for и
.

4(417.4 kJ/kg)+(0.01) (2505.6-417.4)kJ/kg
= 438.282 kJ/kg
=

Find the mass at state 1 (m,)
.

Here, volume of the tank is , and mass of the water at state 2 т,
.

Substitute 0.1 m
3
for V and 0.018 m/kg
for .

0.1 m3
т,
0.018 m/kg
т, 3 т,
m2 = 5.56 kg
т,

From the table of “Saturated water-pressure table”, obtain the following properties corresponding to the pressure of .

Specific volume of saturated liquid, .

Specific volume of saturated vapor, =0.099587 m/kg
.

Specific internal energy of saturated liquid, и, 3906.12 kJ/kg
.

Specific internal energy of saturated vapor, =2599.1 kJ/kg
и,
82
.

Find the quality at state 2 .

2=v
V.
82

Here, dryness fraction at state 2 is .

Substitute 0.018 m/kg
3
for V2
, 0.001177 m2/kg
for and 0.099587 m2/kg
for .

0.018 m/kg 0.001177 m/kg+x (0.099587-0.001177)m/kg
x, = 0.17

Find the specific internal energy at state 2 (и,)
.

и, %3и, +*, (и, —и,
82

Substitute 906.12 kJ/kg
for , 0.17 for х,
and 2599.1 kJ/kg
for u
.

|4, =(906.12 kJ/kg)+(0.17)(2599.1-906.12)kJ/kg
=1193.93 kJ/kg

Find the heat transfer at stage 1 (e)
.

Substitute 5.56 kg
for т
, 1193.93KJ/kg
for и,
and 438.282kJ/kg
for .

From property table of “Saturated water – Pressure table” obtain the following properties corresponding to the pressure of .

=0.099587 m3/kg
=2599.1 kJ/kg

Find the mass at state 3 .

Substitute for V and 0.099587 m/kg
for .

0.1 m3
т,
0.099587 m/kg
=1.004 kg

Find the heat transfer at stage 2 (0,)
.

0, %3 (ти, — т,и, )-h (m, - т,

Substitute 1.004 kg for т,
, 2599.1 kJ/kg
for , for т,
, for and 2798.3kJ/kg
for .

1.004 kgx 2599.1kJ/kg
(-5.56 kgx1193.93KJ/kg /-2798.3kJ/kg(1.004 kg-5.56kg)
2609.5-6638.25+12749.05
-8720.3 kJ/kg

Ans:

The heat transfer at stage 1 (e)
is 4201.4 kJ.

The heat transfer at stage 2 (0,)
is 8720.3 kJ.

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