thermodynamics lecture 10
TRANSCRIPT
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Lecture 10 First Law for Control MassLecture 10 – First Law for Control Mass
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Problem 1
A piston/cylinder contains 50 kgof water at 200 kPa with avolume of 0.1 m3. Stops in thecylinder restricts the enclosedyvolume to 0.5 m3, as shown infig. The water is now heated to200oC Find the final pressure200oC. Find the final pressure,volume and the work done bythe water.
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Example 2A piston-cylinder arrangement has a linearspring and the outside atmosphere actingg gon the piston. It contains water at 3 MPaand 400oC with a volume of 0.1 m3. If thepiston is at the bottom the spring exerts apiston is at the bottom, the spring exerts aforce such that a pressure of 200 kPainside is required to balance the forces. The
t l til thsystem now cools until the pressurereaches 1 MPa. Find the heat transfer forthe process.
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Example 2:SolutionInitial state determined by P and v, it is superheated. Final state fixed by P (given) and v again. How do we find v?P = 200 kPa + ksV/A2, where ks is the spring constant, and A the area of cross-section. This is since the pressure to just balance when piston is at bottom is given as 200 kPa Sincebalance when piston is at bottom is given as 200 kPa. Since P1 = 3 MPa and V1 = 0.1 m3 given, ks/A2 = 2.8x104 kJ/m3, so V2 = 0.02857 m3, v2 = 0.02840 m3/kg, saturated mixture with x2 = 0.14107
State P (kPa) T (C) V (m3) v (m3/kg) Mass M (kg) x u (kJ/kg) U(kJ)
1 3000 400 0.1 0.09936 1.006 ‐ 2950
2 1000 143.6 0.02857 0.02840 1.006 0.14107 1019
W = ∫PdV = 200 (∆V) + 2 8x104(V 2 V 2)/2 = 143 kJ
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1W2 = ∫PdV = 200 (∆V) + 2.8x104(V22 – V1
2)/2 = -143 kJ1Q2 = ∆U + 1W2 = -1925 -143 = -2068 kJ
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Problem 3
An insulated piston cylinder devicecontains 5 L of saturated liquidcontains 5 L of saturated liquidwater at a constant pressure of175kPa. Water is stirred by a peddlewheel while current of 8 A flows for45 min through a resistor place inthe water. If 50% of liquid (by mass)q ( y )is evaporated during this constantpressure process and the peddlework amounts to 300kJ determinework amounts to 300kJ, determinethe voltage of the source. Also showthe on P-v diagram.
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Fi 4 2 E l f k i th b d f t
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Fig 4.2 Example of work crossing the boundary of a system because of electric current flow across the system boundary
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An insulated piston cylinder device contains 5 L of saturated liquid water at a constant pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this constant pressure process and the peddle work amounts to 300kJ, determine the voltage of the source. Also show the on P-v diagram.
Solution:Assumptions1 The cylinder is stationary and thus the kinetic and potential
energy changes are zero.2 The cylinder is well-insulated and thus heat transfer is2 The cylinder is well insulated and thus heat transfer is
negligible.3 The compression or expansion process is quasi-equilibrium.
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An insulated piston cylinder device contains 5 L of saturated liquid water at a constant pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this constant pressure process and the peddle work amounts to 300kJ, determine the voltage of the source. Also show the on P-v diagram.
• We take the contents of the cylinder as the system.• This is a closed system since no mass enters or leaves• This is a closed system since no mass enters or leaves.• The energy balance for this stationary closed system can be
expressed as
)()( 21
22 ZZVVmUUWQ +−
+ )(2
)(12
12122121 ZZmgUUWQ −++−=−
1221 UUW −=−
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An insulated piston cylinder device contains 5 L of saturated liquid water at a constant pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this constant pressure process and the peddle work amounts to 300kJ, determine the voltage of the source. Also show the on P-v diagram.
UWWW =−+ outb,inpw,ine, ∆
)hh(mW)tI()hh(mWW
12
12
−=+
−=+
ip
inpw,ine,
,p ,,
∆V )hh(mW)tI( 12+ inpw,∆V
/kJ/kg486.97 kPa 175
3kPa 175 ==
⎬⎫= @f11 hhP
kPa, 175/kgm0.001057 liquidsat. 3
kPa 175
==
==⎭⎬
22
@f1
5.0xPvv
( )
k4 7304m0.005
kJ/kg1593.7553
=×+=+=
1
fg2f2 57.22135.097.486hxhh
V
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kg4.7304/kgm0.001057 3 ===
1
1mv
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An insulated piston cylinder device contains 5 L of saturated liquid water at a constant pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min through a resistor place in the water. If 50% of liquid (by mass) is evaporated during this constant pressure process and the peddle work amounts to 300kJ, determine the voltage of the source. Also show the on P-v diagram.
SUBSTITUTING
=−=+
kJ kg486.97)kJ/55kg)(1593.7(4.7304kJ)(300
VV
536.4935tItI
∆∆
V 228.497=⎟⎟⎠
⎞⎜⎜⎝
⎛×
=kJ/s 1VA 1000
s) 60A)(45 (8kJ 4935.536
VP
1 2
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v
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First law as a rate equation
WQPEKEU δδ −=∆+∆+∆
tW
tQ
tPE
tKE
tU
δδ
δδ
δδδ−=
∆+
∆+
∆
dtdU
tU
t =∆δδ 0lim
dttδ
KEdKE )()(lim =∆
dttt lim0 δδ
PEdPE )()(lim =∆
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dttt lim 0 =δδ
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First law as a rate equation
QQ &δlim QtQ
t =δδ 0lim
Wδ Wt
Wt
&=δδ
δ 0lim
WQdtPEd
dtKEd
dtdU && −=++
)()(
WQdE &&=
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WQdt
−=
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FIRST LAW ANALYSIS FOR AFIRST-LAW ANALYSIS FOR A CONTROL VOLUME
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FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUME
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Conservation of mass and Control Volume
Control volumes:
Control Volume
Mass can cross the boundaries, and so we must keep track
of the amount of mass entering and leaving the controlof the amount of mass entering and leaving the control
volume.
Conservation of Mass Principle
Net mass transfer to or from a system during a process is
equal to the net change in total mass of the system
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during that process.
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Conservation of mass and Control VolumeControl Volume
∑∑VCdm&& ∑∑ −= ei
VC mmdt
..
Total mass inside the control volume
∫ ∫ +++=== .......)/1(.. CBAVC mmmdVvdVm ρ
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Conservation of mass and Control VolumeControl Volume
Th l fl t iThe volume flow rate is
∫ ∨=∨= dAAV local&
The mass flow rate becomes
∫The mass flow rate becomes
AdAVVm local ∨=
∨=== ∫ )(
&&& ρ
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vdA
vvVm avg ==== ∫ )(ρ
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FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUME
For a Fixed mass
12 EE −
For a Fixed mass
= 21Q − 21W12 21Q 21
The instantaneous rate equation isThe instantaneous rate equation is
dE MC Q& W&dtMC .. = Q − W
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