1. BITS Pil i
Pilani
Pilani Campus
Lecture 9 – Fi t Law for Control Mass
L t First L f C t l M

2. First Law for Control Mass
As seen above, for a closed system
∆U = U2 – U1 = - 1W2 (adiabatic, Q = 0).
∆U = U2 – U1 = 1Q2 (W = 0)
What if both work and heat terms present? Since U is a
state function
∆U = U2 – U1 = 1Q2 - 1W2
More generally since there may also changes in kinetic
energy of the system, and its potential energy in the
gravitational field, we use the total energy E = U + KE + PE,
and write
∆E = E2 – E1 = 1Q2 - 1W2, the conservation of energy
principle extended to include thermodynamic variables and
p
processes
Strictly, we must indicate the path also for W and Q
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3. First Law for Control Mass
• What if a control mass undergoes a cyclic process? Then
∆U = 0, and so Q = W
• For an infinitesimal change dU = δQ - δW
• As mentioned above, E = U + KE + PE, where KE and PE
are th bulk ki ti and gravitational potential energies of
the b lk kinetic d it ti l t ti l i f
the system, dU + d(KE) + d(PE) = δQ – δW
• KE = ½ mV2 and PE = mgZ
• dE = dU + d(KE) + d(PE) = δQ – δW
• E2 – E1 = U2 – U1 +m(V22 – V12) + mg(Z2 – Z1) = 1Q2 - 1W2
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4. Internal Energy U
• U is an extensive property (as are KE and PE)
• u = U/m is the specific internal energy
p gy
• Can be used to fix state of a phase
• In the two-phase say liquid-vapour region, u = (1-x)uf + xug
= uf + xufg, and u can be used with a table to determine the
quality ie., the relative amounts of the two phases
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5. Problem Analysis
• Identify system, sketch diagram, show forces, and heat and
work flows
• Identify (fix) the initial state and the final state
state,
• Characterize the process
• Sketch the process schematically
• Identify the thermodynamic model to be used
y y
• What is the analysis to be done?
• What is the solution technique?
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6. Adiabatic free expansion
Carry out a first law analysis of the process
• Initial state specified by, say v and T. Then u(T,v) = u1 also
known
• 1Q2 = 0 (adiabatic, 1W2 = 0 (rigid container), hence by I Law
U2 – U1 = 0. Also V2 = 2V1. Hence the final state is specified
by v2 = 2v1, and u2 = u1
What about other properties such as the temperature or
pressure? Yes, if we have the equation of state or other
equivalent information
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7. Adiabatic free expansion – ideal gas
• The final state is determined by V2 = 2V1, and U2 = U1.
• Joule expansion experiment: Joule carried out this
experiment,
experiment and found that there was no measurable
temperature change, in part because under the chosen
conditions, the gas closely approximated ideal behaviour.
• Conclusion: The internal energy of an ideal gas is a
function only of the temperature, u = u(T). This can be
proved from the EoS using the II Law as we will see
• Hence T2 = T1, and so P2 = P1/2
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8. Constant Pressure Expansion
For the isobaric conversion of saturated liquid water at 100º
C to saturated vapour at the same temperature, find the
heat required per unit mass.
• The control mass is 1kg of water, in a piston-cylinder, with
the pressure set at 101 3 kPa Draw a schematic sketch
101.3 kPa.
• All properties of the initial and final states are fixed, and
can be read from the steam tables. Sketch the processp
• The process is isobaric, sketch it in the P-v plane (or the T-
v plane)
• Th thermodynamic model i th eos i t b l f
The th d i d l is the in tabular form
• Analyse the system using the I Law
• Look up the internal energy values and the specific
values,
volumes. Compute the work done, and thence the heat
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9. Constant Pressure Expansion - Enthalpy
• 1w2 = Psat(vg – vf), with Psat = 101.325 kPa = P1 = P2
• u2 – u1 = ug – uf
• 1q2 = ug – uf + Psat(vg – vf) = (u + Pv)2 – (u + Pv)1 (true for
any constant pressure process for a control mass).
• Define the enthalpy H = U + PV, an extensive state
function with dimensions of energy
• 1q2 = h2 – h1 where h is the enthalpy per unit mass and is
mass,
tabulated
• It is useful to define the enthalpy H, since the combination
U + PV appears naturally here. It will also arise in the
discussion of flow work
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10. Back to the example
• Refer to the steam tables for the properties of
saturated water at 100 C
C.
Psat (kPa) vf (m3/kg) vg (m3/kg) vfg uf (kJ/kg) ug (kJ/kg) ufg
101.325 0.001044 1.67290 1.67185 418.91 2506.50 2087.58
Values of u with reference to ug at 0.01 C taken as 0
0 01
Find 1Q2 = ∆(u + Pv) = hg – hf= hfg
At 100 C, hf = 419.02, hg = 2676.05 and hfg = 2257.03 kJ/kg
The th l
Th enthalpy per unit mass i an i t
it is intensive property, and
i t d
may be used as one of the variables to fix the state of a
phase, or to locate the position on a tie-line in the two
phase region
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11. Specific Heat
• Specific Heat C = 1/m (δQ/δT)
• Since Q path dependent, or δQ inexact, must specify the
path, or conditions
• Specific heat at constant volume Cv = 1/m (δQ/δT)V, which
for a simple compressible substance may be written as Cv =
1/m (∂U/∂T)V = (∂u/∂T)v
• Specific heat at constant pressure CP = 1/m (δQ/δT)P,
which for a simple compressible substance may be written
as CP = 1/m (∂H/∂T)P = (∂h/∂T)P
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12. Ideal Gas – Specific Heats
• Air at 1 bar and 300 K is heated (a) at constant volume,
and (b) at constant pressure, to a final temperature of 420
K. In each case, calculate the work, heat, internal energy
change, and enthalpy change
Aside: Since u = u(T) for an ideal gas h = u + Pv = u + RT =
gas,
h(T). Also Cv = (∂u/∂T)v and (∂h/∂T)P are functions only of
the temperature
CP – Cv = R
Cv and CP are
functions of T
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13. Ideal Gas Example (contd.)
• (a) ∆h = ∫Cp0dT
Especially over small ranges of T, often Cp0 taken to be
constant, in which case the integral is ≈ Cp0(T2 – T1) = 125
kJ/kg
If Cp0 not assumed constant, it is usually expressed as an
expansion in T, Cp0 = C0 + C1θ + C2θ2 + C3θ3 where θ = 10-
3T(K) Using the values of the constants from the table, find
∆h = 133.2 kJ/kg
The enthalpiesTare also evaluated and tabulated from which
T2 1
∆h = ∫Cp0dT - ∫Cp0dT = 125.1 kJ/kg
T0 T‐0
0
∆u = ∆h - P ∆v = ∆h - R ∆T
w=0
q = ∆u
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14. Ideal Gas Example (contd.)
• (b) ∆u and ∆h are both the same as in (a) since both u and
h are functions of T alone, and the initial and final
temperatures are the same in both cases.
q = ∆h as this is a process at constant pressure
w = q – ∆ = R ∆T
∆u
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15. Liquids and Solids
• Estimate the change in internal energy u and enthalpy h
for liquid water when it is taken from the saturated liquid at
20 C to (a) 20 C and 500 kPa, (b) 40 C and 2000 kPa
Solids, and liquids away from the critical point, are highly
incompressible,
incompressible and so dv ≈ 0 is an excellent approximation
approximation.
Also, the heat capacities at constant pressure and at
constant volume are virtually the same, say C.
Hence du ≈ C dT, and dh = du + d(Pv) ≈ C dT + vdP
Taking C and v as constant, one has
∆u
∆ ≈ C ∆T and ∆h ≈ C ∆T + v ∆P
∆T, d
(a) ∆u ≈ 0, ∆h ≈ vf ∆P ≈ 0.5 kJ/kg
(b) ∆u ≈ C ∆T = 84 kJ/kg, ∆h ≈ C ∆T + vf ∆P = 86 kJ/kg
kJ/kg
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16. Example
• A constant pressure piston-cylinder assembly
contains 0 20 kg water as saturated vapour at 400 kPa
0.20 kPa.
It is now cooled so that the water occupies half of the
original volume. Fi d Q i th process
i i l l Find in the
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17. Solution
• Initial state characterized by pressure and specific volume.
Process is isobaric, so final state at same pressure, and v2
= v1/2. The steam table will be used. By comparing v2 to vf
(0.0010840, and vg 90.46138) at given P, determine that the
final state is a mixture of saturated liquid and vapour with x2
vapour,
= (v2 – vf)/vfg, and then calculate u2 = uf + x2ufg
State
St t P (kP )
(kPa) T (C)
T (C) v (m3/k )
( /kg) V ( 3)
V (m x u (kJ/kg)
(kJ/k ) U (kJ)
U (kJ)
1 400 143.6 0.46246 0.092492 1 2553.55
2 400 143.6 0.23123 0.046246 0.49882 1576.6
From the table above, ∆u = -977 kJ/kg, or ∆U = -195.4 kJ
1W2 = P(V2 – V1) = -18.5 kJ
18 5
1Q2 = ∆U + 1W2 = -213.9 kJ
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18. Example
• A piston-cylinder arrangement has a linear spring and the
outside atmosphere acting on the pistion. It contains water
at 3 MPa and 400 C with a volume of 0.1 m3. If the piston is
at the bottom, the spring exerts a force such that a pressure
of 200 kPa inside is required to balance the forces The
forces.
system now cools until the pressure reaches 1 MPa. Find
the heat transfer for the process.
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19. Solution
Initial state determined by P and v, it is superheated. Final
state fixed by P (given) and v again. How do we find v?
P = 200 kPa + ksV/A2, where ks is the spring constant, and A
the area of cross-section. This is since the pressure to just
balance when piston is at bottom is given as 200 kPa Since
kPa.
P1 = 3 MPa and V1 = 0.1 m3 given, ks/A2 = 2.8x104 kJ/m3, so
V2 = 0.02857 m3, v2 = 0.02840 m3/kg, saturated mixture with
x2 = 0.14107
State P (kPa) T (C) V (m3) v (m3/kg) Mass M (kg) x u (kJ/kg) U(kJ)
1 3000 400 0.1 0.09936 1.006 ‐ 2950
2 1000 143.6 0.02857 0.02840 1.006 0.14107 1019
1W2 = ∫PdV = 200 (∆V) + 2 8x104(V22 – V12)/2 = -143 kJ
2.8x10 143
1Q2 = ∆U + 1W2 = -1925 -143 = -2068 kJ
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