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					Thermodynamics         Chapter Four   Lecture: Noor M. Jasim




                 THERMODYNAMICS




                   Chapter Four
                   Work and Energy


                     Noor M. Jasim




                          2009
Thermodynamics                                             Chapter Four                   Lecture: Noor M. Jasim

                                                    Chapter Four
                                                   Work and Energy

Example (3-1)

   Unit of mass of fluid at a pressure of 3 bar, and specific volume of 0.18 m3/kg expands
reversibly in the piston cylinder assembly according to the processes (a) constant pressure process
until the volume is doubled, (b) according to the relation PV=c until the final volume will be 0.36
kg/m3, and (c) according to the relation pv2=c until the final volume reaches to 0.36 m3/kg.
Determine the work done in each case.

Solution

   According to Fig. (3.4)
a) for constant pressure process               P=c , then

                                                        kN        
                                                  W  P 2
                                                                  V 2  V1 ( m 3 )
                                                                   
                                                  1 2 m          
                                  m3              m3
 3  1.013  10 2 ( 1kg  0.36       1kg  0.18    ) = 54.702           kJ
                                  kg              kg




                          p
                                         (a)     p=c
                                  1                           2

                                         (b)
                                               T=C

                                         (c)                  2\

                                                 pvn = C
                                                                   1
                                                             2\\


                                                                           v
                                  0.18                     0.36


                                           Fig. (3. 4)




b) for constant temperature process T = c , then
                           V2           V                               kN        m3   
            W  mRT n         mp1V1 n 2  1( kg )  3  10 2  1.013 2   01.8
                                                                                    kg
                                                                                          n 2  37.9( kJ )
                                                                                          
            1 2          V1           V1                             m               

c) For pvn = constant where n=2
Thermodynamics                                            Chapter Four              Lecture: Noor M. Jasim
                                                                 p 2V 2  p 1V1
                                                      W m
                                                      1 2         1n

                      n

since p 2   V1  and then p2  p1  1   75.97(kN / m 2 ) , Hence
                                                  2

                                   
      p1  V2 
                                  2
        75 .97  0.36  3  101 .3  0.18 
W  1kg                                     13 .7kJ
1 2
                       2                  



Example (3.2)
Piston cylinder assembly contains 1 kg of gas at initial pressure of 40 bar. The gas expands
reversibly according to a law (PV1.3 = c) until the volume is doubled. The gas is then cooled
reversibly at constant pressure until the piston firmly locked in position until the pressure rises to
the original value of 40 bar. Calculate the net work done by the gas if the initial volume equals 0.02
m3
Solution

   According to figure (3-5)




                               p
                                     1
                p1 = 40x102 kN/m2
                                                        Pv1.3 = 0



                                                               Net work = 17.5 kJ




              p2 = 16.25x102 kN/m2
                                              P                        2
                                     3


                                     0.02                           0.04      v


                                            Fig. (3.5)




Since, p 1V 11.3  p 2V 21.3 , therefore
                1.3          1.3
          V          1
p 2  p 1  1   40   16.25bar , hence
          V 
           2         2
         P V  P1V1          2  16.25  0.04  40  0.02 
W  m 2 2             ( 10 )                              50kJ and
1 2        1 n                           0.3          
W  p 2 V3  V2  , since V3 = V1, hence
2 3
W  16.25  10 2 0.02  0.04  32.5 kJ , and W  0
23                                                     3 1
the net work done = W  W  W  W = 50 – 32.5 + 0 = 17.5 kJ
                         1 2 2 3 34
Thermodynamics                                        Chapter Four                              Lecture: Noor M. Jasim
Example: (3.3)

    Consider, the main shaft of the gear box transmitted power as a system. If the torque applied is 1kJ and the shaft
rotates at speed of 2400 rpm. Determine the power transmitted by the shaft.

Solution
                                                            2400
                                W shaft  2n ( kW )  2        1( kJ / s )  251.2kW
                                                              60
Example (3.4)

    A volume of 0.02m3 of air is contained in piston-cylinder assembly with pressure of 100kPa. At
this position a linear spring that has K=100kN/m is touching the piston but exerting no force on it.
If a heat is transferred to the gas, the gas is then expanded and the piston rises and compresses the
spring until the volume inside the cylinder is doubled. If the cross section area of the piston is
0.2m2. Calculate (a) the final pressure inside the cylinder, (b) the total work done by the mass, and
(c) the fraction of this work done against the spring to compress it.

Solution

Given: piston cylinder device contains gas, spring connected between the piston and the cylinder
       externally.

Find: p2 , w, Wspring

Schematic and Given data: schematic diagram and p-V diagrams for given data is shown in Figure
     (3.8).
                k = 100 kN/m

                                         p
                                                                           2
                               p2 = 150 kPa


                               pm = 125 kPa                                        Wspring
                                                               I

      A = 0.2 m2                                                               Wwith no
                               p1 = 100 kPa 1           P
     P1 = 100 kPa                                                               spring action
                                                               II
     V1 = 0.02 m3
                                                                                       v
                                         0.02                       0.04



                                     Fig. (3.8)



Assumptions: The gas is a closed system.

Analysis:
  The enclosed volume at the final state is

                                                V2  2V1  2  0.02  0.04m 3
hence, the displacement of the piston (and the spring) becomes
                                             ( V 2  V1 ) 0.04  0.02m 3
                                        x                                 0.1m
                                                   A           0.2 m 2
Thermodynamics                                           Chapter Four                                Lecture: Noor M. Jasim


   The force applied by the linear spring at the final state is determined using equation (3.12), then
                                          F  kx  100kN / m  0.1m  10kN

   The additional pressure added by the spring on the gas at this state is
                                                        F 10kN
                                                P               50kN / m 2
                                                        A 0.2m 2

    Then the final pressure of gas inside the cylinder rises by the additional pressure that created from the effect of the
spring, then the final pressure is:

                            Pf = Pi + Pdue to spring action ( kPa ) = 100 + 50 = 150 kPa
         2
b) W   pdV
   1 2  1

   There is no exact relation between P and V given directly from the given data, but we know that the relation
between the force exerted on the spring and its elongation or compression is linear within the elastic limits then the
mean pressure is
                                             pi  p f    100  150
                                       pm                         125kPa
                                                 2           2

then the W can be determined as follows
             1 2
                                                              
                                     W  pm V f  Vi  1250.04  0.02  2.5 kJ
                                     1 2


c) The spring work can be determined as:
                                W spring 
                                             1
                                             2
                                                  2
                                                           1
                                               k x 2  x 1   100 ( 0.1 ) 2  0  0.5 kJ
                                                         2
                                                            2
                                                                                              or

The spring work equal to the triangle area on p-V diagram Figure (3.8)
                                    W spring 
                                                    1
                                                      125  100kPa0.04  0.02  05kJ
                                                    2




Example (3.6)

   A car has a mass of 1ton need to be accelerated from a velocity of 40 km/h to 90km/h is 5 minutes on a level road.
Determine the power required for acceleration.
Solution
                                                            1
   The acceleration work = kinetic energy =                   m(V22  V12 )
                                                            2
                                                                 
                                                                     2
                                                                        40  10 3  
                                                                                    2
                                              90  10 3
                            KE    1000
                                    1                                              250.8  10 3 kJ
                                    2       3600                     3600  
                                                                                


The power required is determined as:
                                                                        250.8  10 3
                                   Power = ( k .E ) / time                          836kW
                                                                          5  60
Thermodynamics   Chapter Four   Lecture: Noor M. Jasim
Thermodynamics   Chapter Four   Lecture: Noor M. Jasim

				
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