ME 200 L15: ME 200 L15:Conservation of Energy: Control Volumes 2.5, 4.4-4.5 HW 5 cancelled; HW 6...
15
ME 200 L15: ME 200 L15: Conservation of Energy: Control Volumes 2.5 , 4.4-4.5 HW 5 cancelled; HW 6 assigned: Due 02/26/14 https://engineering.purdue.edu/ME200/ ThermoMentor © Program Launched Spring 2014 MWF 1030-1120 AM J. P. Gore [email protected] Gatewood Wing 3166, 765 494 0061 Office Hours: MWF 1130-1230 TAs: Robert Kapaku [email protected] Dong Han [email protected]
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Transcript of ME 200 L15: ME 200 L15:Conservation of Energy: Control Volumes 2.5, 4.4-4.5 HW 5 cancelled; HW 6...
ME 200 L15:ME 200 L15: Conservation of Energy: Control Volumes 2.5, 4.4-4.5
HW 5 cancelled; HW 6 assigned: Due 02/26/14 https://engineering.purdue.edu/ME200/
ThermoMentor© Program Launched
Spring 2014 MWF 1030-1120 AM
J. P. Gore [email protected]
Gatewood Wing 3166, 765 494 0061Office Hours: MWF 1130-1230
TAs: Robert Kapaku [email protected] Dong Han [email protected]
2
In this Lecture …
• Allowing flow of mass (with associated energy) into and out of a system to make the “closed system,” an “open system.”
• Heat and work transfer are common to both closed and open systems.
• In flow and out flow of mass occurs with associated energy transport.
• In flow and out flow of mass requires energy expenditure called “Flow Work.”
• Keeping track of energy that comes in and the energy that goes out to ensure what stays in.
Energy Rate Balance/Energy Conservation/1st Law of Thermo
time rate of change of the energy
contained within the control volume
at time t
net rate at whichenergy is beingtransferred in
by heat transferat time t
net rate at whichenergy is beingtransferred out
by work attime t
net rate of energytransfer into thecontrol volumeaccompanying
mass flow
eeiiCV gzp
umgzp
umWQdt
dE)
2
V()
2
V(
22cv
Exit Flow workInlet Flow work
Energy Rate Balance/Energy Conservation/1st Law of Thermo
time rate of change of the energy
contained within the control volume
at time t
net rate at whichenergy is beingtransferred in
by heat transferat time t
net rate at whichenergy is beingtransferred out
by work attime t
net rate of energytransfer into thecontrol volumeaccompanying
mass flow
eeiiCV gzp
umgzp
umWQdt
dE)
2
V()
2
V(
22cv
Exit Flow work=0Inlet Flow work=0
=0=0
Control Mass
Mass
=0=0
CM
dEQ W
dt
CMCM
Does not matter
=0=0
Does not matter
Energy Rate Balance/Energy Conservation/1st Law of Thermo
time rate of change of the energy
contained within the control volume
at time t
net rate at whichenergy is beingtransferred in
by heat transferat time t
net rate at whichenergy is beingtransferred out
by work attime t
net rate of energytransfer into thecontrol volumeaccompanying
mass flow
eeiiCV gzp
umgzp
umWQdt
dE)
2
V()
2
V(
22cv
Exit Flow workInlet Flow work
Evaluating Work for a Control Volume
)()( iiieeeCV vpmvpmWW (Eq. 4.12)
The expression for work is
cvW accounts for boundary work (associated with rotating shafts, displacement of the boundary, and electrical effects)
where►
)( eee vpm is the flow work at exit e.►
)( iii vpm is the flow work at inlet i.►
7
Total Energy
• For a non-flowing fluid:
• For a flowing fluid, we must add the fluid’s potential for doing flow work:
• The energy contained in a flowing fluid is:
pekeue
p
pekepue pekehe
Now you see the utility of h!
7
gzV
ue 2
2
8
First Law for Open Systems
8
dE/dt = net rate of Q
net rate of W
- +
Rate of energy addition due to inflow
Rate of energy loss due to outflow
-
9
First Law for Open Systems
• What is our convention for Q and W?
• What assumptions are inherent in this form of the 1st Law?
e
ee
eei
ii
iicvcv
cv gzV
hmgzV
hmWQdt
dE
22
22
9
10
First Law for Open Systems
• We also have steady-flow forms of mass and energy conservation. We use this form most often in ME 200.
e
eii mm
i
ii
iie
ee
eecvcv gz
Vhmgz
VhmWQ
22
22
10
11
Common Assumptions
• In addition, there are common assumptions for each term:– Insulated (adiabatic)– Negligible ΔKE– Negligible ΔPE– No work– 1-inlet, 1-outlet
2 2
2 2
e i
cv cv e e e i i ie i
V VQ W m h gz m h gz
11
12
Common Assumptions
• We can still use previous assumptions such as– Incompressible (for liquids)
– ideal gas – absence of Q and/or W
types
12
(if applicable)
13
Example
Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state.
• Find– Q = ? in kW
• Sketch
• Assumptions– The control volume is at steady
state.– ΔWcv = Δke = Δpe = 0
• Basic Equations
1 2
h2 = 1700 kJ/kgh1 = 3000 kJ/kgm1 = 0.5 kg/ssteam
e
ee
eei
ii
iicvcv
cv gzV
hmgzV
hmWQdt
dE
22
22
13
e
eii
cv mmdt
dm ;
14
Example
Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state.
Find– Q = ? in kW
System (mass flowing through pipe)
1 2
h2 = 1700 kJ/kgh1 = 3000 kJ/kgm1 = 0.5 kg/ssteam
14
cvQ
cvQ
15
Example
Assumptions
• Wcv = 0
• ΔKE = 0
• ΔPE = 0
• Steady State, Steady Flow Operation
Basic Equations
•Solution
e
ee
eei
ii
iicvcv
cv gzV
hmgzV
hmWQdt
dE
22
22
cv 2 1Q m h h
skJ
kW
kg
kJskgQcv 1
1300017005.0
kWQcv 65015
e
eii
cv mmdt
dm mmm 21
