1. BITS Pil i
Pilani
Pilani Campus
Lecture 8 – Fi t Law for Control Mass
L t First L f C t l M

2. Work
• Work is done by a system if the sole effect on the
surroundings could be the lifting of a weight in the
gravitational field
• δW = Fsds, 1W2 = ∫ δW
• W i t k as positive if the corresponding energy t
is taken iti th di transfer
f
is from the system to the surroundings, and negative if the
transfer is from the surroundings to the system
g y
• W = ∫δW is path dependent, and δW is inexact
• Work at the moving boundary of a simple compressible
substance (PdV work, work of expansion/compression), W =
b t k k f i / i )
∫δW = ∫PdV if quasiequilibrium, W = ∫δW = ∫PextdV if not
• Polytropic process PVn = constant
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3. Other types of work
• Stretching of a wire under tension δW = -TdL where T is
the tension, and dL the infinitesimal extension. In the elastic
region, with the stress σ = T/A = Ee, where E is the modulus
of elasticity and e the strain defined by de = dL/L0, δW = -
AEeL0de
• Expansion of a surface δW = -S dA
S is the surface tension, and dA the
infinitesimal change in area
• Work done in transporting charge dZ across potential
difference Φ i δW = - ΦdZ
diff is
• Shaft work
• Flow work
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4. Internal Energy U
• Joule experiment – For a closed system (control mass) in
an adiabatic enclosure, W = -∆U where U is the internal
energy, an extensive function of the state of a system
(extension of the work-energy theorem
• Internal energy – sum total of energy of inter molecular
inter-molecular
interaction, molecular translational kinetic energy, and intra-
molecular contributions such as electronic, vibrational, and
rotational
• Very generally, one may replace the internal energy above
by E the total energy which will include the kinetic energy of
the system as a whole, and the potential energy in the
gravitational field E = U + KE + PE
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5. Heat Q
• State of control mass can be changed without the
performance of work, specifically by placing it in thermal
contact with another object at a different temperature
• ∆U = Q (when W = 0), Q is called heat
•TTransfer of energy across boundary surface that occurs
f f b d f th t
solely due to a difference in temperature between two
bodies in thermal contact
• Energy transfer from the body at higher temperature to
that at lower temperature
• U it i J same as f W U or E T k as positive when
Unit is J, for W, U, E. Taken iti h
transfer of energy is from the surroundings to the system
• Q = ∫δQ is path dependent, and δQ is inexact
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6. Work and Heat
• Both heat and work are transient phenomena, not
properties
• Both are modes of energy transfer across system
boundary when system undergoes a process
• Both are path dependent, and the infinitesimals δQ and
dependent
δW are inexact differentials
•What can we say about the internal energy change of a
control mass in a process in which work and heat are both
non-zero?
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7. 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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8. 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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9. 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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10. 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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11. Adiabatic free expansion – ideal gas
• The final state is determined by V2 = 2V1, and U2 = U1.
• Joule expansion experiment: Joule carried out this
experiment, and found that there was no measurable
temperature change, in part because under th chosen
t t h i tb d the h
conditions, the gas closely approximated ideal behavior.
• 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, but for now we will
take it as an additional postulate from observation
observation.
• Hence T2 = T1, and so P2 = P1/2
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12. 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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13. 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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14. 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 = Δ(u + Pv) = hg – hf= hfg
At 100 C, hf = 419.02, hg = 2676.05 and hfg = 2257.03 kJ/kg
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