First Law
Sept 2005: changes reflect text: Woud
Sept 2006: added examples
first law: during any cycle a system undergoes, the cyclic integral of the heat is proportional to the cyclic
integral of the work
pg 83 van Wylen & Sonntag Fundamentals of Classical Thermodynamics 3rd Edition SI Version
⌠
⌠
⎮ 1 dQ = ⎮ 1 dW
⌡
⌡
first law for cycle
(5.2)
The net energy interaction between a system and its environment is zero for a cycle executed by the system.
pg 2 Cravalho and Smith
⌠
⌠
⎮ 1 dQ − ⎮ 1 dW = 0
⌡
⌡
where integral are cyclic and dQ = δQ dW = δW
plot data
Pressure Volume plot for Processes
2.5
p
2
1.5
1
0.5
0.5
1
1.5
2
2.5
v
A process
B process
C process
⌠
⌠
⎮ 1 dQ = ⎮ 1 dW
⌡
⌡
apply first law to cycle A B
apply first law to cycle A C
subtract A C from A B
rearrange ...
1
1
2
2
⌠
⌠
⎮ 1 dQ_WB = ⎮ 1 dQ_WC
⌡
⌡
dE = δQ − δW
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2
1
2
1
1
2
1
2
⌠
⌠
⌠
⌠
⎮ 1 dQA + ⎮ 1 dQB = ⎮ 1 dWA + ⎮ 1 dWB
⌡
⌡
⌡
⌡
2
1
2
1
1
2
1
2
⌠
⌠
⌠
⌠
⎮ 1 dQA + ⎮ 1 dQC = ⎮ 1 dWA + ⎮ 1 dWC
⌡
⌡
⌡
⌡
1
1
1
1
2
2
2
2
⌠
⌠
⌠
⌠
⎮ 1 dQB − ⎮ 1 dQC = ⎮ 1 dWB − ⎮ 1 dWC
⌡
⌡
⌡
⌡
i.e. δQ - δW is a point function ... only dependent
upon the end points => define as ...
energy, point function
1
first law for system (Woud: Closed system) - change of state
rearrange and integrate ...
Q1_2
δQ = dE + δW
Q1_2 = E2 − E1 + W1_2
E1
N.B. Woud starts with rate
equation and obtains this
assuming steady state
is the heat transferred TO system
E2
W1_2
are intial and final values of
energy of system and ...
(5.5)
is work done BY the system
energy E consists of internal energy + kinetic energy + potential energy
E = U + KE + PE
dE = dU + dKE + dPE
Closed System
(5.4)
δQ = dE + δW = dU + dKE + dPE + δW
and first law can be restated ...
d
U = Q_dot − W_dot
dt
dU = δQ − δW
m_dote = m_doti = 0
(W 2.3)
d and δ: VW&S: page 62 Woud page 11
d = differential of point functions state variables
δ = differential of path functions - amount depends on
path/process: diminutive
see discussion of cyclic process below
difference between d and δ
cycle may be considered a closed system; initial state and final state are identical, For example (detailed
discussion later)
set up limits and calculations
p-v plot of Brayton cycle
6
pressure
4
2
0
0
1
2
3
4
5
volume
adiabatic compression
heat addition
adiabatic expansion in turbine
heat rejection
closed (cycle)
d
U = 0 = Q_dot − W_dot
dt
Q_dot = W_dot
Qcycle = Wcycle
(W 2.6)
this is where we started above using
different approach to first law
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2
example 5.3
Sonntag example 5.3: vessel with volume 5 m 3 contains 0.05 m 3 of saturated liquid water and 4.95 m 3 of
saturated water vapor at 0.01 MPa. Heat is added until the vessel is filled with saturated vapor. Determine Q.
3
3
V := 5m
State 1:
6
Vvap := 4.95m
MPa := 10 Pa
3
Vliq := 0.05m
p := 0.1MPa
3
m
v f := 0.001043
constant volume and
mass => constant v
3
steam tables at p =
0.1 MPa
v g := 1.694
m
State 2: V2 = V
v = vg
u g := 2506.1
kg
Q1_2 = E2 − E1 + W1_2
first law:
Vliq
vf
V = m ⋅v n
n
n
mass1_liq = 47.9386 kg
x 1 :=
x = quality( of_steam)
Vap H2O
page 616
kg Sonntag
kJ
u fg := 2088.7
kg
Q1_2 = U2 − U1
kJ
kg
Liq
H2O
ΔV = 0
Δz = 0
E=U
Vf
massf =
vf
mass1_vap :=
mass1_vap
Vvap
mass1_vap = 2.9221 kg
v g
x 1 = 0.0575
mass1_vap + mass1_liq
4
U1 := mass1_liq⋅u f + mass1_vap⋅u g
1 Q2
kJ
u = ug
W1_2 = 0
have volume, determine mass of each
mass1_liq :=
u f := 417.36
kg
3
kJ := 10 ⋅J
U1 = 2.7331 × 10 kJ
x%1 := x 1 ⋅100
x%1 = 5.7453
intensive property = not dependent on
mass (x, v, u, ρ)
extensive does (U, V, mass)
or ... using average specific properties
u 1 := u f + x 1 ⋅u fg
u 1 = 537.3611
kJ
kg
(
U11 := u 1 ⋅ mass1_vap + mass1_liq
)
4
U11 = 2.7331 × 10 kJ
which can be shown by ...
U1 = mass1_liq⋅u f + mass1_vap⋅u g
u g = u f + u fg
mass1_liq⋅u f + mass1_vap⋅u g
u 1a =
mass1_liq + mass1_vap
(
mass1_liq⋅ u f + mass1_vap⋅ u f + u fg
mass1_liq + mass1_vap
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u fg = 2.0887 × 10
3 kJ
kg
3 kJ
u g − u f = 2.0887 × 10
substitute for u g
) = (mass1_liq + mass1_vap)⋅ uf + mass1_vap⋅ ufg = u
mass1_liq + mass1_vap
3
f + x 1 ⋅u fg
kg
state 2: need 2 properties, e.g. quality = 100%, can
calculate v (specific volume). T and p will be rising as heat
is added.
mass_total := mass1_liq + mass1_vap
v 2 :=
3
V
v 2 = 0.0983
mass_total
m
and ... is saturated vapor so we need to look up
(interpolate) steam tables for v g = v2
kg
my_interp( x2 ,
x1, y2, y1, vx) := y1 +
vx − x1
x2 − x1
⋅(y2 − y1)
an interpolation statement
v is value between x1 and x 2
result is y at v
could use mcd function linterp(vx,vy,x) where:
vx is a vector of real data values in ascending order.
vy is a vector of real data values having the same number of elements as vx . x is the value of the independent variable at which to interpolate a result. For best results, this
should be in the range encompassed by the values of vx .
using Table A.1.1
T
vg
ug
3
values at 1
values at 2
T1 := 210
v 1g := 0.10441
u 1g := 2599.5
m
kg
3
T2 := 215
v 2g := 0.09479⋅
pg
kJ
u 2g := 2601.1
m
kg
p 1g := 1.9062MPa
kg
kJ
p 2g := 2.104MPa
kg
vx := v 2
interpolated values at
(
)
T2 := my_interp v 2g , v 1g , T2 , T1 , vx
(
)
p 2 = 2.0317 MPa
(
)
u 2 = 2.6005 × 10
p 2 := my_interp v 2g , v 1g , p 2g , p 1g , vx
u 2 := my_interp v 2g , v 1g , u 2g , u 1g , vx
total internal energy at state 2:
3 kJ
kg
5
U2 := mass_total⋅ u 2
Q1_2 := U2 − U1
heat added ...
T2 = 213.1717
U2 = 1.3226 × 10 kJ
Q1_2 = 104933 kJ
same result can be obtained using Table A.1.2 Pressure Tables
my_interp( x2 , x1, y2, y1, vx) := y1 +
p
ug
vg
3
values at 1
p 1g := 2.0MPa
v 1g := 0.09963
m
v 2g := 0.08875⋅
m
kg
u 1g := 2600.3
kJ
u 2g := 2602.0
kJ
3
values at 2
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p 2g := 2.25MPa
kg
4
kg
kg
vx − x1
x2 − x1
T
⋅(y2 − y1)
T1 := 212.42
T2 := 218.45
vx := v 2
interpolated values at
(
)
T2 := my_interp v 2g , v 1g , T2 , T1 , vx
T2 = 213.1529
(
)
p 2 = 2.0304 MPa
(
)
u 2 = 2.6005 × 10
p 2 := my_interp v 2g , v 1g , p 2g , p 1g , vx
u 2 := my_interp v 2g , v 1g , u 2g , u 1g , vx
Q1_2 := U2 − U1
heat added ...
kg
5
U2 := mass_total⋅ u 2
total internal energy at state 2:
3 kJ
U2 = 1.3226 × 10 kJ
Q1_2 = 104933 kJ
example 5.3
first law as a rate equation
Q1_2 = E2 − E1 + W1_2 = U2 − U1 + KE2 − KE1 + PE2 − PE1 + W1_2
from above ...
δQ = δU + δKE + δPE + δW
in small time interval δt
δQ
divide by δt
δt
=
δU
δt
+
δKE
δt
+
δPE
δt
+
δW
δt
d
d
d
d
d
d
d
Q = U + KE + PE + W = E + W
dt
dt
dt
dt
dt
dt
dt
first law as a rate equation
(5.31 and 5.32)
first law as a rate equation - for a control volume (Woud: system boundary)
Q1_2 = E2 − E1 + W1_2
=>
δQ
δt
=
E2 − E1
δt
+ δW
(5.38)
= energy in control volume at time t
Et
Et_δt
= energy in control volume at time t +dt
E1 := Et + ei⋅δmi
E2 := Et_δt + ee⋅δme
= the energy of the system at time t
= the energy of the system at time t +dt
system consists of control
volume and differential entities
δmi each with ei, vi, Ti, pi
where i = input
and differential entities δme
each with ee , ve , Te , pe where
e = output
E2 − E1 → Et_δt + ee⋅δme − Et − ei⋅δmi
ee⋅δme − ei⋅δmi
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E2 − E1 = Et_δt − Et + ee⋅δme − ei⋅δmi
(5.39)
represents flow of energy across boundary during δt as a result of
δmi and δme crossing the control surface
5
now consider work associated with masses δmi and δme
work done ON mass δmi is ...
p i⋅v i⋅δmi
ON as work must be done to make it enter
system
work done BY mass δme is ...
p e⋅v e⋅δme
BY as leaving represents work done
work done BY system in δt is then ...
δQ
divide by δt and substitute into first law ... (5.38) and combining and rearranging
δQ
+
δt
δmi
δt
(
Et_δt − Et
)
⋅ ei + p i⋅v i =
δt
2
e + p ⋅v = u + pv +
V
2
δme
+
δt
(
)
⋅ ee + p e⋅v e +
V
2
δt
δWc_v
=
E2 − E1
δt
H = U + p⋅ V
+ g⋅ z
+ δW
(5.38)
(5.42)
δt
2
+ g⋅ z = h +
(5.41)
δW + p e⋅v e⋅δme − p i⋅v i⋅δmi
(5.43)
h = u + p⋅ v
enthalpy defined - is
a property (5.12)
therefore ...
2
2
⎛⎜
⎞⎟ E
⎞⎟ δW
Vi
Ve
δme ⎛⎜
t_δt − Et
c_v
+
⋅ ⎜ hi +
+ g ⋅ zi⎟ =
+
⋅ ⎜ he +
+ g ⋅ ze⎟ +
2
2
δt ⎝
δt ⎝
δt
δt
δt
⎠
⎠
δQ
δmi
δmi
δme
and
(5.44)
are summed over all inputs and outputs ...
example 5.4
Sonntag example 5.4 a cylinder fitted with a piston has volume 0.1m3 and contains 0.5 kg steam at 0.4 MPa.
Heat is transferred until the temperature is 300 deg_C while pressure is constant
What are the heat and work for this process?
3
V1 := 0.1m
T2 := 300 deg_C
quality =
mtot := 0.5kg
p := 0.4MPa
mass_of_vapor
total_mass
v f := 0.001084
m
kg
3
v g := 0.4625
v 1 = v f + x ⋅ v fg
constant pressure
Q1_2 = E2 − E1 + W1_2
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1−x=
an intensive property
=x
3
I think by definition, steam = water + vapor at quality = x
3
V1
m
=>
v 1 :=
v 1 = 0.2
mtot
kg
3
m
kg
x :=
v fg := v g − v f
v1 − vf
v fg = 0.4614
m
kg
x = 0.4311
v fg
(
)
(
W1_2 = p ⋅ V2 − V1 = p ⋅ mtot⋅ v 2 − v 1
E2 − E1 = U2 − U1
6
)
as ΔV^2 and Δz = 0
mass_of_liquid
total_mass
(
E2 − E1 = U2 − U1 = mtot⋅ u 2 − u 1
(
)
(
)
)
(
)
Q1_2 = E2 − E1 + W1_2 = mtot⋅ u 2 − u 1 + p ⋅ mtot⋅ v 2 − v 1 = mtot⎡u 2 + p ⋅ v 2 − u 1 + p ⋅ v 1 ⎤
⎣
⎦
h f := 604.74
kJ
kg
h fg := 2133.8
(
Q1_2 := mtot⋅ h 2 − h 1
kJ
kg
)
h 1 := h f + x ⋅ h fg
h 1 = 1.5246 × 10
3 kJ
kg
m
Q1_2 = 771.0904 kJ
3
Table A.1.2 Saturated Steam Pressure Table
v 1 = 0.2
kg
(
W1_2 := p ⋅ mtot⋅ v 2 − v 1
)
W1_2 = 90.96 kJ
Q1_2 = E2 − E1 + W1_2 = U2 − U1 + W1_2
(
U2 − U1 = Q1_2 − W1_2 = Q1_2 − p ⋅ mtot⋅ v 2 −
v 1
(
ΔU := Q1_2 − p ⋅ mtot⋅ v 2 − v 1
)
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9/21/2006
)
ΔU = 680.1304 kJ
example 5.4
7
kJ
kg
Table A.1.2 Saturated
Steam Pressure Table
3
v 2 := 0.6548
h 2 := 3066.8
m
kg
first law as a rate equation - for a control volume
d
Qc_v +
dt
∑
n
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Vi
d
m_dot
⋅
h
+
+
g
⋅
z
⎢
i⎜ i
i⎟⎥ = Ec_v +
2
⎣
⎝
⎠⎦ dt
∑
n
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
(5.45)
d
m_dot
⋅
h
+
+
g
⋅
z
⎢
e⎜ e
e⎟⎥ + Wc_v
2
⎣
⎝
⎠⎦ dt
this is where Woud starts
W_dot
system boundary
m_dote, ve, he
Q_dot
z = elevation
Q_dot = heat_flow
i = inlet
W_dot = work_flow
ze
e = exit
m_dot = mass_flow
v = velocity
m_doti, vi, hi
m, U
m = mass_system
h = specific_enthalpy
U = internal_energy_system
zi
First law: change of energy within the system equals the heat flow into the system, minus the work flow delivered
by the system, plus the difference in the enthalpy, H, kinetic energy E kin and potential energy Epot of the entering
and exiting mass flows.
assuming energy
E = U + Ekin + Epot
and ...
Ekin = Epot = 0
⎛⎜
⎞⎟
⎛⎜
⎞⎟
Ve
Vi
d
U = Q_dot − W_dot + m_doti⋅ ⎜ h i +
+ g ⋅ zi⎟ − m_dote⋅ ⎜ h e +
+ g ⋅ z e⎟
2
2
dt
⎝
⎠
⎝
⎠
2
enthalpy = H = U + p ⋅ V
2
enthalpy = h = U + p ⋅ v
E=U
N.B. dot => rate not d( )/dt
W (2.1)
a defined property
steady state, steady flow process ... Woud: open systems steady state (stationary)
assumptions ...
1. control volume does not move relative to the coordinate frame
2. the mass in the control volume does not vary with time
3. the mass flux and the state of mass at each discrete area of flow on the control surface do not vary with time
and .. the rates at which heat and work cross the control surface remain constant.
d
mc_v = 0
dt
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d
Ec_v = 0
dt
8
steady state, steady flow process ...
m_dot = flow_rate
∑
m_dotin = ∑ m_doten
n
d
Qc_v +
dt
∑
n
(5.46)
n
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Vi
⎢m_dotin⋅ ⎜ h i + 2 + g ⋅ zi⎟⎥ =
⎣
⎝
⎠⎦
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
d
⎢m_doten⋅ ⎜ he + 2 + g⋅ ze⎟⎥ + dt Wc_v
⎣
⎝
⎠⎦
∑
n
2
2
⎡⎢
⎤⎥
Vi − Ve
W_dot = Q_dot + m_dot⋅ ⎢h i − h e +
+ g ⋅ ( zi − ze)⎥
2
⎣
⎦
q=
d
Qc_v
dt
w=
m_dot
(5.47)
(W 2.8) see text for examples of
application to steam turbine,
boiler or heat exchanger, nozzle
and throttle
d
Wc_v
dt
are heat transfer and work per unit mass
flow (5.51)
m_dot
steady state steady flow ... - single flow stream
Vi
2
2
q + hi +
+ g ⋅ zi = h e +
2
Ve
2
+ g ⋅ ze
+ w
(5.50)
uniform state, uniform flow process USUF
1. control volume remains constant relative to the coordinate frame
2. state of mass within the control volume may change with time, but at any instant of time is uniform
throughout the entire control volume - I define this as f(t) but not of space
3. the state of mass crossing each of the areas of flow on the control surface is constant with time although the
mass flow rates may be changing
∑ m_doten − ∑ m_dotin = 0
d
mc_v +
dt
at time t, continuity equation ...
n
n
t
integrating over time gives change in mass of the control volume
⌠
⎮ d m
dt = m2_c_v − m1_c_v
⎮ dt c_v
⌡
0
mass entering and leaving
⌠
⎮
⎮
⎮
⌡
t
∑
∑
n
n
m_doti dt =
n
0
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9/21/2006
mi
n
⌠
⎮
⎮
⎮
⌡
t
0
9
∑
m_doten dt = ∑ men
n
n
m2_c_v − m1_c_v +
continuity for USUF process ...
∑ men − ∑ min = 0
n
(5.53)
n
apply first law at time t (5.45)
d
Qc_v +
dt
∑
n
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Vi
d
⋅
m_dot
⋅
h
+
+
g
z
⎢
i⎜ i
i⎟⎥ = Ec_v +
2
⎣
⎝
⎠⎦ dt
∑
n
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
(5.45)
d
⋅
m_dot
⋅
h
+
g
z
+
+ Wc_v
⎢
⎟
⎥
e⎜ e
e
2
⎣
⎝
⎠⎦ dt
since at time t c.v. is uniform ...
d
Qc_v +
dt
∑
n
2
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
⎡ ⎛
⎞⎟⎤⎥
Vi
Vc_v
d ⎢ ⎜
=
m_dot
⋅
h
+
+
g
⋅
z
m
⋅
u
+
+
g
⋅
z
⎢
⎢ ⎜ c_v
i⎜ i
i⎟⎥
c_v⎟⎥ ...
2
2
⎣
⎝
⎠⎦ dt ⎣ ⎝
⎠⎦
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
d
m_dot
⋅
h
+
+
g
⋅
z
+
⎢
e⎜ e
e⎟⎥ + Wc_v
2
⎣
⎝
⎠⎦ dt
∑
now integrate over time t
t
⌠
⎮ d Q
dt = Qc_v
⎮ dt c_v
⌡
0
n
⌠
⎮
⎮
⎮
⎮
⌡
t
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Vi
⎢m_doti⋅ ⎜ h i + 2 + g ⋅ zi⎟⎥ dt =
⎣
⎝
⎠⎦
∑
∑
n
n
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥
Vi
⎢mi⋅ ⎜ hi + 2 + g⋅zi⎟⎥
⎣ ⎝
⎠⎦
0
t
⌠
2
2
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥
⎛⎜
⎞⎟
⎛⎜
⎞⎟
⎮
V2
Vc_v
V1
⎮ d m⋅ u
dt
m
u
+
+
g
⋅
z
=
⋅
+
+
g
⋅
z
−
m
⋅
u
+
+
g
⋅
z
⎢ ⎜
2⎜ 2
c_v⎟⎥
2⎟
1⎜ 1
1⎟
⎮ dt ⎣ ⎝ c_v
2
2
2
⎠⎦
⎝
⎠
⎝
⎠
⌡
0
⌠
⎮
⎮
⎮
⎮
⌡
t
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
⎢m_dote⋅ ⎜ h e + 2 + g ⋅ ze⎟⎥ dt =
⎣
⎝
⎠⎦
∑
∑
n
n
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥
Ve
⎢me⋅ ⎜ he + 2 + g⋅ ze⎟⎥
⎣ ⎝
⎠⎦
t
⌠
⎮ d W
dt = Wc_v
⎮ dt c_v
⌡
0
0
uniform state, uniform flow process USUF
Qc_v +
∑
n
9/21/2006
9/21/2006
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥
Vi
⎢min⋅ ⎜ h i + 2 + g ⋅ zi⎟⎥ =
⎣ ⎝
⎠⎦
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
⎢men⋅ ⎜ h e + 2 + g ⋅ ze⎟⎥ ...
⎣
⎝
⎠⎦
∑
n
2
2
⎛⎜
⎞⎟
⎛⎜
⎞⎟
V2
V1
+ m2 ⋅ ⎜ u 2 +
+ g ⋅ z2⎟ − m1 ⋅ ⎜ u 1 +
+ g ⋅ z1⎟ + Wc_v
2
2
⎝
⎠
⎝
⎠
10
(5.54)