ME 10.311 Thermo I
Lect 12.docx
Lect.: 12 Title: Conservation of Energy Control
Volume – Part 1
Page 1 of 6
Delivery Notes:
Class L.O.:
1. Able to explain in words and mathematically the conservation of
energy (1st Law) for a control volume
2. Able to apply the 1st Law to steady-state and transient processes
so that unknown values of stored energy, work, or heat may be
calculated.
Class M.Map:
MAIN QUESTION? – How do we deal energy in a situation where mass is
flowing into and/or out of a control volume?
Topic 1: A General From of the 1st Law for CV’s
For a control mass.
dEcm
= Q −W
dt
E2 − E1 =1 Q2 −1 W2
Please know these
well.
ME 10.311 Thermo I
Lect 12.docx
•
Page 2 of 6
Slide – Energy and Flow (Arbitrary System)
Now consider mass
flow into and / or out of
cv
•
All mass flows enter or leave with a certain amount of stored energy:
1
e = u + V 2 + gZ
2
•
Å each term is energy
per unit mass, kJ/kg
When matter moves through a control surface – pressure is needed.
Either to push it INTO or push it OUT OF the CV.
W = F V = ∫ PV dA = PV = P v m
Å Pv = flow work /
mass
Rate of Flow Work By the CV → Pe ve me
Rate of Flow Work On the CV → Pv
i i mi
o Recall the “adding water to a city’s main water line.”
•
Pulling this all together in terms of total Work
W = Wcv − Pv
i i mi + Pe ve me
•
By convention, we can now re-write the original energy per unit mass
term as
1 2
e + Pv = u + Pv + V + gZ
2
h ≡ u + Pv
This refers to all other
forms of CV work
ME 10.311 Thermo I
Lect 12.docx
Page 3 of 6
Now let’s pull it together
Assume for the moment there is not heat transfer or work done and let’s
write the first law similar to the conservation of mass
dEcv
= ∑ mi e − ∑ me ee
dt
i
Å for general case of
multiple inlets &
outlets.
Now let’s add our energy per unit mass expression, assume one inlet
and one outlet.
dEcv
1
1
= mi (hi + Vi 2 + gZ i ) − me (he + Ve 2 + gZ e )
dt
2
2
Now let’s add control volume work and heat transfer
dEcv
1
1
= Q − W + mi (hi + Vi 2 + gZi ) − me (he + Ve 2 + gZe )
2
2
dt
To finish it off for general case, multiple outlets and inlets:
This summarizes the
control volume 1st law
Notice how work is
now located in two
terms.
dEcv
1
1
= Q − W + ∑ mi (hi + Vi 2 + gZ i ) − ∑ me (he + Ve 2 + gZ e )
2
2
dt
or
dEcv
= Q − W + ∑ mi htot ,i − ∑ me htot ,e
dt
Total enthalpy or
stagnation enthalpy
can be used:
h_tot = h + 1/2v^2 + gZ
h_stag = h + 1/2v^2
ME 10.311 Thermo I
Lect 12.docx
Page 4 of 6
So What?– How can we use this equation?
Topic 2: Steady-State Process From of the 1st Law
• Relevance – long term steady operation of
components like:
o
o
o
o
o
o
Å Components of heat
engines & heat pumps
turbines
compressors
nozzles
boilers
condensers
evaporators
• Our assumptions and their impact
1. The CV itself – the CV is not moving relative to the
coordinate frame of the device or group of devices of
interest
If I say the state of
matter is not changing
what am I saying …..
properties don’t
change!
Î all velocities are measured relative to control surface, no work
is associated with accelerated the CV.
2. Inside the CV – the state of matter as each point
inside the CV is not changing with time. (Changing
with space IS poss.)
Î
and
dmcv d
dρ
dV
= ( ρV ) =
=0
V +ρ
dt
dt
dt
dt
properties are not
changing.
also assumes the
volume is not changing
ME 10.311 Thermo I
Lect 12.docx
Page 5 of 6
dEcv
=0
dt
3. At the CS – the mass flux and state of matter does
not vary with time
Î Every quantity in the equations is independent of time.
•
Final Equation Form
∑m = ∑m
Q + ∑ m (h + V
i
cv
e
i
i
i
2
/ 2 + gZ i ) =∑ me ( he + Ve 2 / 2 + gZ e ) + Wcv
Î Vapor Compression System
Concept Question – TPS Option
Air at 500 K, 500 kPa is expanded to 100 kPa in two steady flow
cases. Case one is a throttle and case two is a turbine. Which has
the highest exit T? Why?
1. Throttle. Highest exit T. In the throttle flow no work is taken out,
no kinetic energy change occurs and we assume no heat transfer
takes place and no potential energy change. The energy equation
becomes constant h, which gives constant T since it is an ideal gas.
2. Turbine. Lowest exit T. In the turbine work is taken out on a
shaft so the fluid expands and P and T drops.
What happens when
there is only one
stream of matter
entering and leaving?
E_dot = Q_dot –
W_dot
E_dot = 0
W_dot = 0
h_in = h_out
(ideal gas)
ME 10.311 Thermo I
Lect 12.docx
Page 6 of 6
Example 1: Water Turbine
Å Draw side view
1
2
CV = inlet to outlet
Continuity : min = mex = m
Eq.6.131st Law :(h + V 2 / 2 + gz )in = (h + V 2 / 2 + gz )ex + wT
Assuming T ~ constant, P ~ const --> v constant, h constant.
Assuming V constant
wT = g ( zin − zex ) = 9.8m / s 2 ⋅
m=
200m
= 1.96kJ / kg
1000 j / kJ
WT 1300i103 kW
=
= 6.63i105 kg / s
wT
1.96kJ / kg
V = mv = 6.63i105 kg / s ⋅ 0.001001m3 / kg = 664 m3 / s
Example 2: Small Air Turbine
Home Work
Please see the
WIP