Section II – First Law Applied to Closed Systems Page 14
SECTION II – FIRST LAW APPLIED TO CLOSED SYSTEMS
0.2 kg of fluid, initially at a temperature of 165C, expands reversibly at a constant pressure of 7 bars until the volume is doubled. Find the final temperature, work done and heat transferred if the fluid is steam with an initial quality (dryness fraction) of 70%.
T2 = ?
W12 = ?
Q12 = ?
· Initial state is in the liquid-vapour saturation region since x1 = 0.7
· Final state: P2 = 7 bars v2m v2 = 2v1 = 0,3727 m3/kg
steam is a superheated vapour at the final state. Interpolation is necessary!
T = 300V, P = 0.6 MPa ® v = 0.4344 m3/kg, u = 2801.0 kJ/kg
T = 300C, P = 0.8 MPa ® v = 0.3241 m3/kg, u = 2797.2 kJ/kg
T = 350C, P = 0.6 MPa ® v = 0.4742 m3/kg, u = 2881.2 kJ/kg
T = 350C, P = 0.8 MPa ® v = 0.3544 m3/kg, u = 2878.2 kJ/kg
· Process in P-v and T-v diagrams
· Work done: W12 = m
· First Law: Q12 – W12 = U2 – U1
Q12 = W12 + (U2-U1) = W12 + m(u2-u1)
= mP(v2-v1) + m(u2–u1)
Q12 = 26.8 + 0.2(2807.3-2009.7) = 186.3 kJ
· Alternatively, Q12 = mP(v2-v1) + m(u2-u1)
= m[(u2 + P2v2)-(u1+P1v1)]
Q12 = m(h2-h1)
T = 300C, P = 0.6 MPa, h = 3061.6 kJ/kg
T = 300C, P = 0.8 MPa, h = 3056.5 kJ/kg
(at P = 0.7 MPa, T = 300C)
T = 350C, P = 0.6 MPa, h = 3165.7 kJ/kg
T = 350C, P = 0.8 MPa, h = 3161.7 kJ/kg
(at P = 0.7 MPa, T = 350C)
Q12 = m(h2-h1) = 0.2 (3069.7 – 2143.6) ® Q12 = 185.2 kJ
The cylinder shown below is fitted with a piston that is restrained by a spring so arranged that for zero volume in the cylinder the spring is fully extended. The spring force is proportional to the spring displacement and the weight of the piston is negligible. The enclosed volume in the cylinder is 120 litres when the piston encounters the stops. The cylinder contains 4 kg of water initially at 350KPa, 1% quality, and the water is then heated until it exists as a saturated vapour. Show this process on a P-diagram and determine:
(a) The final pressure in the cylinder
(b) The work done by the water during the process
(c) Heat added during the process
(d) Change in internal energy
P1 = 350 kPa
x1= 0.01
m = 4 kg
T1 = Tsat = 138.88C
v1 = *1-0.01)(0.001079) + 0.0l(0.5243)
= 0.00631121 m3/kg
· Process in the P-diagram.
· Consider a free body diagram of the piston
For any equilibrium position of the piston,
PA – PatmA – Fs = 0 ® P = Patm + Fs/A
Fs = ksy ® P = Patm + ks y/A
= Ay ® P = Patm + ks /A2
· Process 1 ® 2: P1 = Patm + ®
ks = 9960 A2 KN/m2
P2 = Patm + ks(2-0)/A2 = 101 + 9960A2
P2 = 1296.2 KPa
(a) Process 2 ® 3: v = const ® v3 = vg = 0.12/4 = 0.03 m3/kg
Interpolate for P3 with v3 = 0.03 m3/kg between P = 6 MPa, vg = 0.03244 m3/kg and P = 7 MPa, vg = 0.02737 m3/kg in the saturated water Pressure table.
(b)
(since
W13 = 78.2 kJ
(c) Heat added during process
First Law: Q13 – W13 = U3 –U1
Q13 = W13 + (U3-U1) = 78.2 + 7926.8 = 8005.0 kJ
(d) Change in the internal energy
U = U3 – U1 = m (U3 – U1)
U3 = Ug @ P = 6.481 MPa; interpolation in the saturated water Pressure table.
U = 4(2585.3 – 603.6) = 7926.8 kJ
Air at 1 atm and 20C occupies on initial volume of 1000 cm3 in a cylinder. The air is confined by a piston which has a constant restraining force so that the gas pressure always remain constant. Heat is added to the air until its temperature reaches 260C. Calculate the heat added, the work the air does on the piston, and the change in internal energy of the air.
· System
1 atm = 101.3 KPa
20C = 293 K
260C = 533 K
1000 cm3 = 10-3 m3
Q12 = ?
W12 = ?
U = ?
· Can air be treated as an ideal gas under the given conditions?
® Air can be treated as an ideal gas
· Mass in the system
· Volume at the final state
· Work done by the air on the piston
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· Internal energy at the initial and final states
U = m(U2-U1) = 1.205 x 10-3 (384.09-209.06) = 0.215 kJ
· Heat transfer
First Law: Q12 – W12 = U ® Q12=W12 + U
Q12 = (0.083 + 0.215) kJ = 0.298 kJ = 298 J
· Process in the P-diagram
Air in a closed vessel of fixed volume, 0.14 m3, exerts a pressure of 10 bars at 250C. If the vessel is cooled so that the pressure falls to 3.5 bars, determine the final temperature and heat transferred.
· System
Assumptions
· System is stationary
· Potential energy datum is at the system
level
· Can air be treated as an ideal gas under the given conditions?
® Air can be treated as an ideal gas under the given conditions.
· Mass of air in the system
· Final temperature
· Heat transferred
First Law: Q12 – W12 = U2 – U1
0.9 kg of air, initially at a pressure of 15 bars and a temperature of 250C expands reversibly and polytropically to 1.5 bars. Find the final temperature, work done and heat transferred if the index of expansion is 1.25.
· System
Assumptions
• System is stationary
• Potential energy datum is at the level of the system.
· Can air be treated as an ideal gas under the given conditions?
Z1 = 0.975 » 1; since is so small, Z2 » 1 for all Tr values ® air can be treated as an ideal gas under the given conditions.
· Properties at the initial state
· Properties at the final state
Polytropic Process:
Ideal gas:
· Work done by the air
· Heat transferred
First Law: Q12 – W12 = U2 – U1
Q12 = W12 + m(U2 – U1) = 199.3 + 0.9(235.47 – 376.60)
Q12 = 72.3 kJ
Alternatively, we can write
NOTE:
Q12 = 199.3 + 0.9(-141.04) = 72.4 kJ
· Process in the P-diagram
Polytropic processes involving ideal gases
· Process that obeys Pvn = const is a polytropic process.
- n: polytropic index of expansion or compression
- polytropic processes not limited to ideal gases
· If n = 0, Pvn = const becomes P = const ® ISOBARIC PROCESS
· If n = , v = becomes v = const ® ISOCHORIC PROCESS
· If n = 1, Pvn = const becomes Pv = const and for an ideal gas, Pv = RT ® T = const ® ISOTHERMAL PROCESS
· If n = k, where k = cp/cv ® ADIABATIC PROCESS (NO HEAT TRANSFER)
(for ideal gas)
· When n > 0 (i.e. positive), P and v cannot increase or decrease simultaneously i.e. P corresponds to v and vice versa. Compression and expansion processes fall into quadrants 2 and 4 respectively with n positive and an initial state at A.
· It is possible, physically, for processes to fall into quadrants 1 and 3. n will then be negative and P and v will increase or decrease simultaneously. However, such processes do not often occur in practical systems.
· One example of a polytropic process with n < 0 is the expansion of a gas behind a piston that is attached to a linear spring.
Free-body diagram of piston.
At any intermediate equilibrium state (recall process is reversible and therefore quasistatic), Fx=0 ® Fp – Fs - = 0 ® Fp = + Fs
Fp = P.A, = PatmA, Fs = kX
P = Patm + k
x ® P and x ® v
- P and v increase simultaneously ® n < 0
- Process will fall in quadrant 1 starting at A
· Relationships between P, v and T for polytropic processes involving ideal gases
- Pvn = const ® ®
- Pv = RT ® P1v1 = RT1, P2v2 = RT2 ®
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