Student Handout 10 2014
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Transcript of Student Handout 10 2014
Macroscopic energy balance 1
CHEE 3363Spring 2014Handout 10
�Reading: Fox 4.8 and 4.9
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Learning objectives for lecture
1. State the control volume equation for energy.�
2. Apply the energy equation to calculate operating conditions for mechanical equipment.�
- Power�
- Rate of change of temperature�
- �
- Pressure drop�
- Rate of heat transer�
- Shaft work
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Macroscopic energy balance 1Start with conservation of energy:
where total system energy is given by:
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Recall: E is extensive (total energy of the system) and e is intensive
Q − W =dE
dt
∣
∣
∣
∣
sys
Esys =
∫
Msys
e dm =
∫
Vsys
eρdV
e = u +v2
2+ gz
(1) (2) (3)(1)
(2)
(3)
Macroscopic energy balance 2: RTT
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Apply the Reynolds transport theorem to Esys:
Recall: Reynolds Transport Theorem:
1: rate of change of any arbitrary extensive property of the system
2: time rate of change of arbitrary extensive property N, with η the corresponding intensive property
3: net rate of flux of extensive property N out through the control surface
dN
dt
∣
∣
∣
∣
sys
=∂
∂t
∫
CV
ηρ dV +
∫
CS
ηρv·dA
1 2 3
Q − W =dE
dt
∣
∣
∣
∣
sys
=∂
∂t
∫
CV
eρ dV +
∫
CS
eρv·dA
Macroscopic energy balance 3: workTypes of work done by control volume (remember W > 0 when work is done by CV on surroundings):
.
1: Shaft work: rate of work transferred out through CS by shaft
2: Work done by normal stresses:
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W = lim∆t→0
δW
∆t= lim
∆t→0
F · ds
∆t= F · v
W = Ws + Wnormal + Wshear + Wother
1 2 3 4
dFnormal·v = σnndA · v
Wnormal = −
∫
CS
σnndA · v = −
∫
CS
σnnv · dA
work done by CV through CS
dFshear = τdA Wshear = −
∫
CS
τ · vdA
3: Work done by shear stresses:
n·
Macroscopic energy balance 4: work3: Work done by normal stresses:
4: Other work: includes terms from electromagnetism� (we will generally neglect this term)
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Wshear = −
∫
CS
τ · vdA = −
∫
Ashaft
τ · vdA −
∫
Asurf
τ · vdA −
∫
Aport
τ · vdA
τ ·v = 0 Wshear = 0
dFshear = τdA Wshear = −
∫
CS
τ · vdA
already accounted
for in 1
0
v = 0 on surface
by choice of CS s.t. dA is perpendicular to velocity, can be made 0
and
n· n· n·
n·
n·
Macroscopic energy balance 5: all together
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Rearrange:
Q − Ws +
∫
Cs
σnnv · dA − Wshear − Wother =∂
∂t
∫
CV
eρ dV +
∫
CS
eρv · dA
Q − Ws − Wshear − Wother =∂
∂t
∫
CV
eρ dV +
∫
CS
eρv · dA −
∫
Cs
σnnv · dA
ρ =1
v
∫
CS
σnnv · dA =
∫
CS
σnnvρv · dA
Macroscopic energy balance 6: energy eqn
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σnn ≈ −pIgnore viscous effects and assume
Final energy equation (with substitution for enthalpy):
Q − Ws − Wshear − Wother =∂
∂t
∫
CV
eρ dV +
∫
CS
(e + pv)ρv · dA
Substitute into energy equation:Q − Ws − Wshear − Wother =
∂
∂t
∫
CV
eρ dV +
∫
CS
(e − σnnv)ρv · dA
Review: energy, enthalpy of ideal gasesFor an ideal gas:
energy
enthalpy
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u = u(T )cv =
du
dTu2 − u1 =
∫
T2
T1
cvdT = cv(T2 − T1)
cp =dh
dTh2 − h1 =
∫
T2
T1
cp dT = cp(T2 − T1)h = u +p
ρ
= u(T ) +RT
Mw
= h(T )
from ideal gas law
vf
Example: air compressor 1Given: Air at STP enters a compressor at velocity vi = 75 m/s and leaves at absolute pressure Pf = 200 kPa and temperature Tf = 345 K and velocity vfcooling water circulating around the compressor casing removes dQ/dm = 18 kJ/kg of air.Calculate: the power required by compressor.
Assumptions:
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ωsvi
Continuity:
..m
Example: air compressor 2
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vf
ωsvi
Check for solution:
Recall: power is energy/time!
.m
Equation
Example: Compressed air 1Given: Compressed air stored in bottle with volume V = 0.5 m3 at pressure p = 20 MPa and temperature T = 60°C. When a valve is
m = 0.05 kg/s..
Calculate: rate of change of temperature in bottle.
Apply continuity:Assumptions:
Apply 1st law:
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Example: Compressed air 2
cv:
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Check for solution:
Simplify the previous equation:
Example: pump 1
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Given: Centrifugal water pump with 0.1-m diameter inlet and 0.1-m
0.2m Hg vacuum and exit pressure 240 kPa. Inlet and outlet at same elevation. Measured power input 6.75 kW.Calculate
Apply continuity:Assumptions:
Basic equation:
Example: pump 2
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Given: Centrifugal water pump with 0.1-m diameter inlet and 0.1-m
0.2m Hg vacuum and exit pressure 240 kPa. Inlet and outlet at same elevation. Measured power input 6.75 kW.Calculate
Checks:
