Notes On Thermodynamics
Notes On Thermodynamics
Notes On Thermodynamics
gabriel, akb
(skimmed from thermodynamics an engineering approach - cengel & boles,
1. basic concepts
fluid — characterized by low resistance to flow, and tendency to assume the
shape of its container
closed system — no exchange of matter, mass does not exit the boundary
(energy like heat can)
properties of a system
intensive property — independent of the mass of a system (T, P, ρ)
extensive property — depend on the size/extent of the mass of a system
(total mass, V, total momentum)
notes on thermodynamics 1
V
specific property — extensive property per unit mass (specific volume v= m
)
state postulate
a system is a simple compressible system in the absence of electrical,
magnetic, gravitational, motion and surface tension effects.
the postulate requires that two specified properties of the system must be
independent to fix the state. two properties are independent if one can be
changed as the other remains constant.
processes + cycles
isothermal - T is constant
isobaric - P is constant
isometric - V is constant
zeroth law of
thermodynamics
two bodies are in thermal
equilibrium if both have the same
temperature even when not in
contact.
temperature conversion
T (K) = T (°C) + 273.15
T (R) = T (°F ) + 459.67
T (R) = 1.8 ⋅ T (K)
T (°F ) = 1.8 ⋅ T (°C) + 32
notes on thermodynamics 2
ΔT (K) = ΔT (°C)
ΔT (R) = ΔT (°F )
pressure
— the normal force exerted by a
fluid per unit area. in solids, this
is called normal stress.
P1 = P2
the difference between P1 and P3 ,
💡 try:
notes on thermodynamics 3
we can find hA by recalling that
VA MA /ρ MB /ρ
hA = = hB =
AA AA AB
MA /ρ MA g
PA = Patm + ρghA = Patm + ρg ⋅ ⇒ PA = Patm +
AA AA
MB g
PB = Patm + ρg(hB + h) ⇒ PB = Patm + ρgh +
AB
pure substance —
homogenous, invariable
chemical composition
saturation temperature —
temperature at which
vaporization occurs at a
given pressure
saturation pressure —
pressure at which
vaporization occurs at a
given temperature
notes on thermodynamics 4
💡 EXPERIMENT! CONCEPTUALIZE, UNDERSTAND.
Imagine a piston-
cylinder arrangement,
set up three times
with pressures
P1 , P2 , and P3 , where
P1 > P2 > P3 . Water is
being heated in the
tank, and the weight
of the piston is
negligible. The system
is in quasi-
equilibrium.
p3 reaches its saturation point faster than p2, p1, because at lower
pressure, the boiling point of water is lower. this is why at high
altitude, where the atmospheric pressure is lower, hikers find that water
heats up quicker.
quality (x)
quality x — ratio of the mass of vapor compared to the liquid. this quality
x is in the "dome" region of the T-v diagram.
for convention,
f - liquid
g - vapor
mg mg
x= =
mtotal mf + mg
the volume is the sum of the volume occupied by the vapor and liquid
V = Vf + Vg ⇒ vf mf + vg mg
notes on thermodynamics 5
V mf mg
v= = vf + vg
mT mT mT
v = vf (1 − x) + vg x
mg
mg
mT
= vf (
mT − mg
mT
) + vg
derived. if x= mT
, then v = vf (1 − x) + vg x
state postulate
for a pure substance, the state is defined by two independent, intensive
properties.
notes on thermodynamics 6
from the table, vf = 0.001452m3 /kg, vg = 0.01803m3 /kg
(in any situation, don't substitute values immediately. work out the algebra
first before putting in the values. it sharpens your algebraic skills!)
v = vf (1 − x) + vg x ⇒ v = vf − vf x + vg x
−vf x + vg x = v − vf ⇒ x(−vf + vg ) = v − vf
v − vf 0.003 − 0.001452
x= = ≈ 0.0934 = 9.34%
−vf + vg −0.001452 + 0.01803
RT a
P= − 2
v − b v − cbv + db2
wherein a, b, c, and d are parameters. if these parameters are 0, then we
end up with the ideal gas law.an ideal gas has a density ρ low enough that
its intermolecular distance is large.
P v = RT
all three properties P, v and T are intensive. for the density ρ to be low
enough for Ideal Gas, then the Temperature must be high or the Pressure must
be low.
notes on thermodynamics 7
💡 try: a 1m^3 rigid tank initially constains air at 1Mpa, 400K and is
connected to an air line. the valve is opened and air flows to the
tank until it reaches 5 MPa, at which point the volve closes and the
final temperature is 450K. Find a) initial and final mass of air in
the tank and b) if the tank eventually cools to room temperature,
300k, what is the pressure in the tank then?
P V = mRT
P1 V
m1 = = 8.71kg
RT1
P2 V
m2 = = 38.71kg
RT2
b)
T3
P3 = P2 ( ) = 3.33 MP a
T2
notes on thermodynamics 8
2
W =∫ F ⋅ dx
1
F = P Ak^
dz = dzk^
2 2
W12 = ∫ F ⋅ dz = ∫ P Adz
1 1
2
W12 = ∫ P dV
1
W12 = P (V2 − V1 )
2 2
notes on thermodynamics 9
💡
2 2
W12 = ∫ P dV = P ∫
2
work out the integration by dV = P ⋅ V ∣∣1
1 1
yourself!
W12 = P (V2 − V1 ) = P ΔV
2
W12 = ∫ δW =
W2 − W1
1
2
Q12 = ∫ δQ =
Q2 − Q1
1
ΔV is an exact differential.
💡 Heat and work are not thermodynamic properties. They can also cross
the system boundary.
2
W12 = ∫ P dV
1
W12 = 0
notes on thermodynamics 10
case B: piston is moving, P is constant
2
W12 = P ∫ dV = P (V2 − V1 )
1
constant
P V n = constant ⇒ P =
Vn
plugging this into the Work equation
2 2
const. dV
W12 = ∫ n
dV = constant ∫ n
1 V 1 V
if n = 1, then const. = P1 V1 = P2 V2 if n =
1, const. = P1 V1 = P2 V2
n n
V2 V −n ∣2 const. 1−n
W12 = const. ln ∣∣ ∣∣ W12 = const. ∣ ⇒ (V −
V1 −n + 1 1 1−n 2
plugging in the constant into the plugging in the constant into the
2 2 dV
work equation, W12 = constant ∫1 dV
Vn
work equation, W12 = constant ∫1 Vn
V2 V2 P2 V2 − P1 V1
W12 = P1 V1 ln = P2 V2 ln W12 =
V1 V1 1−n
notes on thermodynamics 11
that
v1 = 0.157m3 /kg
if V2 = 1.41V1 , then
@T = 20∘ C, vg = 0.14922 — the largest possible value can have and STILL be
saturated. Because v2 > vg , then the ammonia is a superheated vapor.
because P is a straight line, we can assume that the pressure can be the
midpoint between 1 and 2.
W12 = −204.868 kJ
notes on thermodynamics 12
sign conventions
W12 is positive when done by the
system
rates
Ẇ = power = work/unit time = Watts
Q̇ = heat transfer done per unit
time = Watts
notes on thermodynamics 13
v1 = 0.424m3 /kg
W12 = 4.91kJ
Win = Qout ⇒ ∮ δW = ∮ δQ
2
Because ∫1 (δQ − δW ) is path independent, this must be a property of the
system. That property is the ENERGY (E) of the system.
2
∫ (δQ − δW ) = ΔE − E2 − E1
1
or
δQ − δW = dE ⇒ Q12 − W12 = ΔE
Energy (E) is the sum of all energies of the system, aka the total energy.
This is an extensive property.
E = U + ke + pe
notes on thermodynamics 14
v1 = V1
m1
= 0.2m3 /kg
@ 0.4MPa, vg < v1 < vf , so it is a
mixture.
At state 2, we can conclude that it
is superheated because T2 >
Tsat @ .4MP a
from table, v2 = 0.655m3 /kg
W12 = P m(v2 − v1 ) = 91 kJ
Q12 = U2 − U1 + W12
ΔE = E2 − E1 = Q12 − W12
Q12 = m(u2 − u1 ) + W12 = 772.27 kJ
neglecting all energies apart from
internal energy,
notes on thermodynamics 15