Thermodynamics lecture 10

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Lecture 10 First Law for Control MassLecture 10 – First Law for Control Mass

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

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.


Fi 4 2 E l f k i th b d f t


Fig 4.2 Example of work crossing the boundary of a system because of electric current flow across the system boundary

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.



• 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 −=−



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


kg4.7304/kgm0.001057 3 ===

1

1mv


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


v

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 =∆


dttt lim 0 =δδ

First law as a rate equation

QQ &δlim QtQ

t =δδ 0lim

Wδ Wt

Wt

&=δδ

δ 0lim

WQdtPEd

dtKEd

dtdU && −=++

)()(

WQdE &&=


WQdt

−=

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FIRST LAW ANALYSIS FOR AFIRST-LAW ANALYSIS FOR A CONTROL VOLUME

FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUME


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


during that process.

Conservation of mass and Control VolumeControl Volume

∑∑VCdm&& ∑∑ −= ei

VC mmdt

..

Total mass inside the control volume

∫ ∫ +++=== .......)/1(.. CBAVC mmmdVvdVm ρ


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


vdA

vvVm avg ==== ∫ )(ρ

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


Thermodynamics lecture 10

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