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 (
or dV = |
|
) and pressure axis intercept of C as shown in figure.
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%
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