LESSON-8 NUMERICAL PROBLEMS ON FIRST LAW OF THERMODYNAMICS FOR CLOSED SYSTEM

Problem 8.1: Calculate the final temperature, pressure, work done and heat transfer if the fluid is compressed reversibly from volume 6m3 to 1m3 in a cylinder. Comment on results when the initial temperature and pressure of fluid as 20°C and 1 bar. The index of compression may be assumed as 1, 1.3 and 1.4 respectively. Take Cp = 1.005 and Cv = 0.718 kJ/kgK and R = 0.287 kJ/kg K.

Solution:

Given: V1 = 6 m3,   V2 = 1 m3 ,  T1 = 20+273 = 293K,    p1 = 1 bar = 1 x 105 N/m2

              = 1.4

 (a) For n = 1 (i.e. isothermal process):

(i)  To determine final temperature

Formula: p1V1 = p2V2

             1 x 105 x 6 = P2 x 1

         Answer:  p2 = 6 x 105  N/m2 = 6 bar

(ii)   To determine final pressure,

              Formula: T2 = T1

       Answer: T2 = 20+273 = 293K

 

(iii)   To determine work done,

              Formula: 1W2= P1V1ln

        Answer:      1W2  = 1 x 105 x 6 x ln = -1075 kJ

                            Here negative sign means work done on the system.

(iv) To determine heat transfer,

               Formula: Heat transfer, 1Q2 =    1W2 

      Answer:   1Q2 = -1075 kJ

(b)  For n = 1.3 (i.e. Polytrophic process)

(i)    To determine final temperature

  Formula: p1V11.3 = p2V2 1.3 = Constant

               1 x 105 x 61.3 = P2 x 11.3

           Answer: p2 = 61.3 x 105   = 10.27x 105   N/m2 = 10.27 bar

(ii)    To determine final pressure,

        Formula: Also,

                      or    

         Answer: = 501.54 K

(iii)    To determine work done,

                    Formula: 1W2 =

       Answer:   1W2  =    = -1423.33 kJ

Here negative sign means work done on the system.

(iv)      To determine heat transfer,

                     Formula: Heat transfer, 1Q2 =

             Answer:         1Q2   =  = - 355.83 kJ

(c) For n = 1.4 (i.e. reversible adiabatic process)

(i)    To determine final temperature

      Formula:    P1V11.4 = P2V2 1.4 = Constant

                              1 x 105 x 61.4 = P2 x 11.4

     Answer:    P2 = 61.4 x 105   = 12.286x 105   N/m2 = 12.286 bar

(ii) To determine final pressure,

     Formula: Also, 

                         or   

       Answer:   =   599.97K

(iii)  To determine work done,

Formula:    1W2 =

      Answer:   1W2   =  = -1571.5 kJ

(iv) To determine heat transfer,

      Answer: Heat transfer, 1Q2 = 0 kJ

Problem 8.2. A system containing 0.2 m3 of air at a pressure of 4 bar and 160°C expands reversible adiabatically to pressure of 1.06 bar and after this the gas is heated at the constant pressure till the enthalpy increases by 65 kJ. Calculate the work done.

Now, imagine that these processes are replaced by a single reversible polytropic process producing the same work between initial and final state; find the index of expansion in this case. Cp of air = 1.005 kJ/k and R = 0.287 kJ/kg K.

Solution:

Given data:  V1 = 0.2 m3,   p1 = 4 bar = 4 x 105 N/m2,     T1 = 160+273 = 433K,   

p2 = 1.06 bar = 1.06 x 105 N/m2,    

p3 = p2 = 1.06 bar = 1.06 x 105 N/m2,     h3-h2 = 65 kJ

Cp= 1.005    R= 0.287 kJ/kg K

(a) For isentropic process 1-2,

      To determine work done, 1W2

             Formula:  1W2 =

                       Finding unknown V2 :

                         V2 =  = 0.5164 m3

      Answer: Workdone, 1W2 = =  = 63.154  kJ                                          

(b) For process 2-3 (constant pressure)

     To determine work done, 2W3

             Formula:  2W3=p (V3 – V2)

                     Finding unknownV3: 

                V3= V2

                                Finding unknown T2:   

    For process 1-2, We have

    T2  =  =  296.27 K

Finding unknown T3: 

     Heat addition, 2Q3 = m (h3 – h2) = m Cp (T3-T2)

                 Where,   m =   = 0.6437 kg

or   65 = 0.6437 x 1.005 (T3 – 296.27)

T3 = 396.74 K

Therefore, V3= V2 = = 0.6915 m3

Answer:  2W3 =   p(V3 – V2) = 1.06 x 105 (0.6915 - 0.5164) =  18.561  kJ

(c)  For process 1-3 (Reversible polytropic process)

 To determine index of expansion

      Work done for polytropic process, 1W3 = =

According to the question,    1W2 + 2W3 = 1W3

Put 1W2,2W3 and 1W3 in above equation,   

63.154 + 18.561 =

        Answer:    n =1.082

Example 8.3:  A cylinder of 7.5 cm inside diameter has a spring loaded piston as shown in figure. The stiffness of the spring is 14 N/mm of compression. The initial pressure, volume and temperature of air in the cylinder are 2.8 bar, 0.00005 m3 and 25 °C respectively. Determine the amount of heat added to the system so that the piston moves by 40 mm. Assume Cv = 0.71 kJ/kg °K and R = 0.28 kJ per kg °K.

Solution:    

Given:   Diameter of cylinder, d = 7.5 cm = 0.075 m;

             P1 = 2.8 bar;

V1 = 0.00005 m3      

T1 = 25 °C+273 = 298K;

Spring stiffness, K = 14 N/mm = 14000 N/m;

Piston movement (X2-X1) = 40 mm = 0.04 m; 

Cv = 0.71 kJ/kg °K and

R = 0.28 kJ per kg °K

Assumptions:

Let us assume an arbitrary datum O-O from which the position of the lower face of the piston is to be measured.

Also assume that the piston weight is zero.

Given:   When, X = Xo, the spring length is its free length. At this position there is vacuum in side the cylinder and other side of the cylinder around the spring.

When  X = X1,    the pressure of air in side the cylinder is 2.8 bar.

Determine the amount of heat added to the system (cylinder) when X = X2 ,  1Q2  ;

        Formula: 1Q2  =(U2 – U1)+ 1W2

                     Finding unknown, 1W2:

Force balance for the piston when piston is at any X position between X1 and X2 as shown in figure:

         A  p = K ( X – Xo )                                                  ……………………….(8.1)

             where, K is the stiffness of the spring.

With the heat transfer to the gas, let the pressure inside the cylinder increase by dP forcing the piston to move upwards by distance dX. Now the force balance for the piston is:

         A (p + dp) = K (X + dX – Xo)                                             …………………(8.2)

From equation (8.1) and (8.2), we get

         A .dp = K dX  = K                                               [ because,   dV = A . dX  ]

   dp = dV                                                            …………………….      (8.3)

     

The P-V relation for the process is a straight line having a slope of (  ) and pressure axis intercept of C as shown in figure.

 

  or   dV =  dp

                       =       =       =         ...............................…(8.4)

                      Finding unknown, p2;

                       From  eqn. (8.3)         p2 –p1 =  

                        p2  = p1 +   (V2 – V1)

                       Finding unknown, V2;

Final volume (V2) = Initial volume (V1) + Change in volume after adding heat in air for X = X1 to X = X2 change.

V2 = V1 + . A . (X2 – X1)

V2= 0.00005  +   × (0.075)2 x 0.04  =  0.000226 m3

p2  = 2.8 x 105 +   (0.000226 – 0.00005) =  4.073 x 105 N/m2

              Therefore from eqn. (8.4),   

              1W2 =   (p22 - p12)

                            =  ( 4.073 x 105  - 2.8 x 105) = 60.142 J

                     Finding unknown, U2 – U1:          

                 U2 – U1 = m Cv (T2 – T1)

                                       Finding unknown,T2 and m:

   

    T2 =     = 1940 K

    m =  =  0.0001637 Kg

                      U2 – U1 = m Cv (T2 – T1) = 0.0001637 x 0.71 x (1940 – 298)

                                                            = 0.1908 KJ = 190.8 Nm

Answer:  1Q2  = (U2 – U1)+ 1W2  = 190.8 + 60.142 = 250.942 Nm or J

Example 8.4: The initial volume inside the cylinder is 0.2 m3, and at this state the pressure inside is 100 kPa (1 bar) which just balance the atmospheric pressure outside and weight of the piston with spring exerts no force on piston. To change state of gas, the gas is now heated until volume is one and half times the initial. The final pressure of gas is 300 kpa (3 bar) and during the process the spring force is proportional to displacement of piston from the initial position. Determine the total work done by the gas and the percentage of work done against the spring.

Solution:    

Given:   V1 = 0.2 m3; 

             p1 = (atmospheric pressure + pressure due to piston weight) = 100 kPa or kN/m2 (1 bar);  

V2 = 1.5 x 0.2 = 0.3 m3;      

 p2 = 300 kpa or kN/m2  (3 bar);  

(a) Determine work done by gas

     Formula: Work done by gas = work done against atmosphere and piston weight + work done against spring

                                                = W atm, w  + Wspring

                           Finding unknown, Watm, w;

                                 W atm, w = pdV = p1 (V2 – V1) = 100 x (0.3 – 0.2) = 10 KJ

Finding unknown,Wspring;

Wspring = K .X .dX  = K .  =   . K . (Xf - Xo) 

                                     Finding unknown, Δ X and K;

                   The pressure exerted by the spring at the end of compression, ps

                          = Gas pressure  – (atmospheric pressure + pressure due to piston weight)

                          = p2 – p1=  300 – 100 = 200 Kpa

                   Thus spring force at the end of compression, Fs = ps  .  A   =  200 A   kN.

                                                                                             where, A is the area of piston

                   This force gives spring displacement (compression)

                  Δ X = Xf – Xo =     = 0.1/A

                   Fs = K(Xf – Xo)

                   K =   = 2000A2kN/m

                        Wspring =  . K . (Δ X)2  =   . 2000 A2  .    = 10 KJ

     Answer: Thus, Work done by gas = W atm, w + Wspring =  10 + 10   =   20 KJ

(b) Determine the percentage of work done against the spring

     Answer:  The percentage work done against spring 50% 

Last modified: Saturday, 7 September 2013, 5:35 AM