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Work and Paths - Thermodynamics - Solved Quiz, Exercises of Thermodynamics

Assam Agricultural UniversityThermodynamics

Some of the topic in thermodynamics are: Property tables and ideal gases, First law for closed and open (steady and unsteady) systems, Entropy and maximum work calculations, Isentropic efficiencies, Cycle calculations (Rankine, refrigeration, air standard) with mass flow rate ratios. This quiz is about: Work and Paths, Expansion to a Volume, Mass, Pressure, Ideal Gas, Constant Pressure

Typology: Exercises

2013/2014

Uploaded on 02/01/2014

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Solution– Work and Paths
1 A mass of 0.1 kg of water fills a piston-cylinder apparatus with an initial volume of 0.01 m3
at a pressure of 1 MPa. The water then undergoes an expansion to a volume of 0.02 m3
following the path equation P = a + b/V where b = –0.02 MPa·m3, and a can be found from
the path equation as a = P1 – b/V1. Find the work and the temperature change of the water.
We can use the path equation to find the work as pathPdV.
1
2
12 ln
2
1V
V
bVVadV
V
b
aPdVW
V
Vpath
We are given the value of b and we can find the value of a = P1 – b/V1 = 1 MPa – (–0.02
MPa·m3)/(0.01 m3) = 3 MPa. We can now find the value for the work.
3
3
333
1
2
12 01.0
02.0
ln02.001.002.03ln m
m
mMPammMPa
V
V
bVVaW
W = 0.016137 MPa·m3 = 0.016137 MJ = 16.137 kJ
We can find the temperature at the initial state P1 = 1 MPa and v1 = V1/m = (0.01 m3)/(0.1 kg) =
0.1 m3/kg. From the saturation table at P1 = 1 MPa we see that this specific volume is between vf
and vg, so we must be in the mixed region. Thus the initial temperature is the saturation
temperature at 1 MPa or T1 = 179.91oC. At the final state, P2 = a + b/V2 = 3 MPa + (–0.02
MPa·m3)/(0.2 m3) = 2 MPa, and v2 = V2/m = (0.02 m3)/(0.1 kg) = 0.2 m3/kg. At this pressure and
specific volume we see that the temperature is very close to 600oC. If we used interpolation we
would find the final temperature as follows:
C
kg
m
kg
m
kg
m
kg
m
CC
CT o
oo
o69.601
19960.02.0
19960.021144.0
600650
600 33
33
The temperature difference, T2 – T1 = 601.69oC – 179.91oC or
T = 421.78oC
2 Repeat problem one if the substance is air instead of water, but all other data are the
same. Treat air as an ideal gas with R = 0.287 kJ/kg·K.
The calculation of the path equation variables and the work did not depend on the property values
for water. So the results from problem 1 for the work and P2 = 2 MPa will not change;
W = 0.016137 MPa·m3 = 0.016137 MJ = 16.137 kJ
Using the ideal gas law gives the temperature difference as follows:
kJ
mMPa
Kkg
kJ
kg
mMPamMPa
mR
VPVP
mR
VP
mR
VP
T
1000
287.0
1.0
01.0102.02
3
33
11221122

T = 1045.3 K = 1045.3oC
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Solution– Work and Paths

1 A mass of 0.1 kg of water fills a piston-cylinder apparatus with an initial volume of 0.01 m 3

at a pressure of 1 MPa. The water then undergoes an expansion to a volume of 0.02 m^3 following the path equation P = a + b/V where b = – 0.02 MPa·m 3 , and a can be found from the path equation as a = P 1 – b/V 1. Find the work and the temperature change of the water.

We can use the path equation to find the work as pathPdV.

1

2 2 1 ln

2

1

V

V

dV aV V b V

b W PdV a

V

path V

We are given the value of b and we can find the value of a = P 1 – b/V 1 = 1 MPa – (–0. MPa·m 3 )/(0.01 m 3 ) = 3 MPa. We can now find the value for the work.

3

3 3 3 3

1

2 2 1

  1. 01

ln 3 0. 02 0. 01 0. 02 ln m

m MPa m m MPa m V

V

W aV V b

W = 0.016137 MPa·m 3 = 0.016137 MJ = 16.137 kJ

We can find the temperature at the initial state P 1 = 1 MPa and v 1 = V 1 /m = (0.01 m 3 )/(0.1 kg) = 0.1 m^3 /kg. From the saturation table at P 1 = 1 MPa we see that this specific volume is between vf and vg, so we must be in the mixed region. Thus the initial temperature is the saturation temperature at 1 MPa or T 1 = 179.91oC. At the final state, P 2 = a + b/V 2 = 3 MPa + (–0. MPa·m 3 )/(0.2 m 3 ) = 2 MPa, and v 2 = V 2 /m = (0.02 m 3 )/(0.1 kg) = 0.2 m 3 /kg. At this pressure and specific volume we see that the temperature is very close to 600oC. If we used interpolation we would find the final temperature as follows:

C

kg

m

kg

m

kg

m

kg

m

C C

T C

o

o o o

  1. 69

3 3

3 3

The temperature difference, T 2 – T 1 = 601.69oC – 179.91oC or  T = 421. o C

2 Repeat problem one if the substance is air instead of water, but all other data are the same. Treat air as an ideal gas with R = 0.287 kJ/kg·K.

The calculation of the path equation variables and the work did not depend on the property values for water. So the results from problem 1 for the work and P 2 = 2 MPa will not change;

W = 0.016137 MPa·m 3 = 0.016137 MJ = 16.137 kJ

Using the ideal gas law gives the temperature difference as follows:

kJ

MPa m

kg K

kJ kg

MPa m MPa m

mR

PV PV

mR

PV

mR

PV

T

3

3 3 2 2 1 1 2 2 11

     T = 1045.3 K = 1045.

o C

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3 How would your

answers to problems 1 and 2 change if the path was a two-step path between the same initial and final states: (a) constant pressure to the final volume then (b) constant volume to the final pressure?

The path for problems one and two and the path for this problem are shown in the plot at the right. Since the area under the two- step path is less than th e area under the P = a + b/V path, we expect that the work will be less for this path.

Since the initial and final states are the same for both paths, regardless of the substance, the temperature temperatures at each state will be the same (although different for air and water) so

the  T values computed in problems 1 and 2 will not change.

For both air and water the work will change with the new path, but it will have the same value for both substances. The only work is done along the constant pressure part of the path for which W

= P 1 (V 2 – V 1 ) = (1 MPa)(0.02 m 3

  • 0.01 m 3 ) or W = 0.01 MPam 3 = 0.01 MJ = 10 kJ.

Quiz Two Paths

0.

0.

0.

0.

1.

1.

1.

1.

2.

2.

0.000 0.005 0.010 0.015 0.020 0.

Volume, V (m3)

Pressure, P (MPa)

P = a + b/V

Two-step path

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