energy analysis of closed systems chapter 4. recall that a closed system does not include mass...
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Energy Analysis of Closed Systems
Chapter 4
Recall that a closed system does not include mass transfer
Heat can get in or out Work can get in or out Matter does not cross
the system boundaries
Mechanical WorkThere are many kinds of mechanical workThe most important for us will be moving
boundary work Wb
Sometimes called PdV work
The primary form of work involved in automobile engines is moving boundary work produced by piston cylinder devices
The mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move.
Boundary Work
W W FdsF
AAds
PdV
b b
z z zz
1
2
1
2
1
2
1
2
The area under the curve is the work
The work depends on the process path
Consider some special cases
Constant volumeConstant pressureIsothermal for an ideal gasPolytropic
Constant Volume
P
V
1
2
W PdVb z12 0
Constant Pressure
P
V
1 2
W
W PdV P dV P V Vb z z1
2
1
2
2 1b g
Isothermal for an ideal gas
P
V
1
2
PmRT
V
W PdVmRT
VdV
mRTV
V
b
FHG
IKJ
z z1
2
1
2
2
1
ln
Polytropic
P
V
1
2
PV n constant
W PdVConst
VdV
PV PV
n
PVV
V
b n
FHG
IKJ
z z1
2
1
2
2 2 1 1
2
1
1,
ln ,
n 1
= n = 1
n=1
PV1= constant equivalent to the isothermal case for an ideal gas
PV= mRT
2
1
lnb
VW mRT
V
2
1
lnV
PVV
P
V
Lets go through the polytropic case integration step-wise
CPV n
nV
CP nCV
2
1
2
1dVCVPdVW n
b
1
1
n
VCW
n
b
2
1
1
11
12
n
VVC
nn
But…
nV
CP
22 And…. nV
CP
11
1
11
12
n
VVC
nn
Wb
So…
1122 VPVPWb 1 n
nn V
CV
V
CV
1
1
2
2
1 n
So far, we have not assumed an ideal gas in this derivation, if we do, then….
mRTPV
n
mRTmRTWb
112
n
TTmRWb
112
Energy Balance for a Closed System (Chapter 2)
E E Ein out system How can energy get in and out of a closed system?
Heat and Work
Total Energy entering the system
Total Energy leaving the system
The change in total energy of the system
- =
Rate form
/in out systemE E E dE dt
Q W Enet net system
E = U + KE + PE
E U KE PE
0 0If the system isn’t moving
Q W U KE PEnet net
In a cyclic process, one where you end up back where you started:
0E0 WQ
net in net outQ W
You convert heat to work, or vise versa
The net work is the area inside the figure
1
2
Pre
ssure
Volume
Wnet
If we can calculate the work done for each step in this process, we can find the net work produced or consumed by the system
Calculating Properties
Chapter 4b
We know it takes more energy to warm up some materials than others
For example, it takes about ten times as much energy to warm up a pound of water, as it does to warm up the same mass of iron.
Specific Heats – Cp and Cv
Also called the heat capacityEnergy required to raise the
temperature of a unit mass one degree
UnitskJ/(kg 0C) or kJ/(kg K) cal/(g 0C) or cal/(g K) Btu/(lbm 0F) or Btu/(lbm 0R)
Consider a stationary constant volume system
E=U+KE +PE
ddU
UU
mCvdT
duCvdT
vv T
uC
Q-W=ΔU
Q
First Law
Consider a stationary constant pressure system
It takes more energy to warm up a constant pressure system, because the system boundaries expand
You need to provide the energy to increase the internal energy do the work required to move the system
boundary
Consider a stationary constant pressure system
E=U+KE +PE
Q=ΔU+PΔVΔH
UU
mCpdTdhCpdT
pp
hC
T
Q-W=ΔU
Q
Q-PΔV=ΔU
dH
Cp is always bigger than Cv
pp T
hC
h includes the internal
energy and the work required to expand the system boundaries
Cp and Cv are properties
Both are expressed in terms of u or h, and T, which are properties
Because they are properties, they are independent of the process!!
The constant volume or constant pressure process defines how they are measured, but they can be used in lots of applications
Ideal Gases
RTPv
Tuu
Joule determined that internal energy for an ideal gas is only a function of temperature
Pvuh
RTPv
RTuh Which means that h is also only a function of temperature for ideal gases!!
For non ideal gases both h and u vary with the state
Specific Heats vary with temperature – but only with temperature –for an ideal gas
Note that the Noble gases have constant specific heats
Why is water on this chart?
dTTCdu v )(
dTTCuuu v2
112
dTTCdh p )(
dTTChhh p2
112
If Cv and Cp are functions of temperature – how can we integrate to find U and H?
Use an average value, and let the heat capacity be a constant
2
1 12,12 TTCdTTCuuu avevv
2
1 12,12 TTCdTTChhh avepp
That only works, if the value of heat capacity changes linearly in the range you are interested in.
OK Approximation
Crummy Approximation
Sometimes the best you can do is the room temperature value
What if you need a better approximation?
All of these functions have been modeled using the form
Cp = a + bT + cT2 + dT3
The values of the constants are in the appendix of our book – Table A-2c
2
1
322
1)( dTdTcTbTadTCh p
2 2 3 3 4 42 1 2 1 2 1
2 1( )2 3 4
b T T c T T d T Th a T T
This is a pain in the neck!! Only do it if you really need to be very accurate!!
Isn’t there a better way?
Use the Ideal Gas Tables
Table A-17 pg 910 (air)Both u and h are only functions of T –
not pressureRelative values have been tabulated
for many ideal gasesIf the gas isn’t ideal – then it’s a
function of both T and P and these tables don’t work!!
Cp is modeled in the Appendix as a function of temperature – so you could calculate h, but what if you want to calculate u? You’d need Cv
There is no corresponding Cv table !!
Cp = Cv + R
Three Ways to Calculate u
v
p
C
Ck
Specific Heat Ratio
k does not vary as strongly with temperature as the heat capacity
k = 1.4 for diatomic gases (like air)
k = 1.667 for noble gases
Use in Chapter 7
Solids and Liquids
Treat as incompressible fluids
Cp = Cv = C
du C dT CdTV
u C T C T T ( )2 1
h u Pv
dh du Pdv vdP
h u v P C T v P
But dv is 0 if the system is incompressible
0
small
Summary Boundary work
Looked at four different special cases Constant volume Constant pressure Constant temperature Polytropic
2
1bW PdV
Summary
First Law for a closed system
Q W U KE PEnet net
Summary
Defined Constant pressure heat capacity Constant temperature heat capacity
By performing a first law analysis of a closed system
pp T
hC
vv
uC
T
SummaryThe three ways to calculate changes in
Internal energy (u)Enthalpy (h)
Look up properties at state one and two in the tables
Assume constant values of specific heat, then integrate
Use the curve fit equation for specific heat, then integrate