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Thermodynamics basicsThermodynamics basics
A B C
Zeroth Law
If A and B and B and C are inthermal equil, then A and C arein thermal equil. [ie. At same T]
R Johnson/UAF/CEM/ME 2005
KE and PEKE and PE
KE = [1/2]mv2 ; PE = mgh
So, m = 1 kg, v = 100 m/s, h = 100 m,
KE = 0.5[104] kg[m2/s2 ] = 5000 J
via J = N-m and N = kg m /s2
PE = 1[9.8][100] kg m2 /s2 = 980 J
ThermalThermal, , chemical, chemical, nuclearnuclear energyenergy
1 kg water with T = 100 C changes internal energy by U = 420 kJ via mc T with c ~ 4.2 kJ/kg/K
1 kg liquid or gaseous fossil fuel has htg vl ~ 44MJ
If evap at P ~ 1 atm, U ~ 2 MJ
E = m c2 9 x 1016 J via c = 3 x 108 m /s
RJ 4.02
Energy amountsEnergy amounts
BMR ~ 70 W ~ 240 Btu/hr
Home htg in Fbks ~ 30 K Btu/hr in winter
Home electrical use ~ 1000 kWh/mo ~ 3.4 MBtu
or rate of ~ 1.4 kW or ~ 5 K Btu/hr
US energy use ~ 90 Q/yr with Q = 1015 Btu
10 kW/cap
RJ 4.02
Relative massesRelative masses
Mass of Water for 1000 Btu
0 1 2 3 4
Evap(1000 Btu/lbm)
Melting (144 Btu/lbm)
Thermal (1 deg F)
KE (224 ft/s)
PE (780 ft)
mo
de
log m (lbm) RJ 3/2000
PropertiesProperties
RJ UAF
saturated
SHliqT
s or v
P = const.
h = const
C P
Water: Crit Pt at P = 22 MPa, T = 374 oC
SH steamSH steam
RJ UAF
2500 3000 3500 4000 4500 50000
200
400
600
800
1000
1200T vs enthalpy at 0.3 MPa for SH steam
h [kJ/kg]
T [
C]
T = -2417 + 44.4ha + 0.084h + 9.29Pa = 0.5
RJ 9/2000 hstpcurvefit.m
Sat. steamSat. steam
Properties of Saturated Water
50 0.012 12.0 209.3 2234.2 209.3 2382.7 0.704 7.372
100 0.101 1.67 418.9 2087.6 419.0 2257.0 1.307 6.048
150 0.476 0.393 631.7 1927.9 632.2 2114.3 1.842 4.996
200 1.554 0.127
250 3.973 0.050 1080.4 1522.0 1085.4 1716.2 2.793 3.280
300 8.681 0.021
374 22.09 0.0031 2029 0 2099 0 4.430 0
T P vg uf ufg hf hfg s sfg
oC MPa m3/kg kJ/kg kJ/kg/K
50 0.012 12.0 209.3 2234.2 209.3 2382.7 0.704 7.372
100 0.101 1.67 418.9 2087.6 419.0 2257.0 1.307 6.048
150 0.476 0.393 631.7 1927.9 632.2 2114.3 1.842 4.996
200 1.554 0.127
250 3.973 0.050 1080.4 1522.0 1085.4 1716.2 2.793 3.280
300 8.681 0.021
374 22.09 0.0031 2029 0 2099 0 4.430 0
vvgg
vs. P vs. P
P = 1 2 4 6; vg = 0.194 0.0996 0.0498 0.0324
P = 0.02 0.05 0.10;
vg = 7.65
3.24 1.694
hhgg vs vs
PP
P = 0.02 0.05 0.1 0.2 0.5 1 ; hg = 2610 2646 2676 2707 2749 2778
FirstFirstConservation of Energy
E = Q - W
E
QW
mi me
eg. Q = 100 kJ
W = 60 kJ
E = 40 kJ
[for system]
Each term > 0 if by system
For a cycleFor a cycle
E = 0 so Qnet = W which leads to
W = QH - QL
2nd Law says all of QH can’t be
converted into W or < 100 %
Closed Closed transient transient
Q - W = m[u2 -u1]
eg. Let SH steam at P1 = 5 MPa and 900 oC cool at const v = 0.1076 m3 /kg to 450 oCSTs u1 = 3841 kJ/kg
STs P2 = 3 MPa and u2 = 3020 kJ/kg
So Q/ m = - 821 kJ/kg
T
v
Heat XferHeat Xfer
qdot
Qdot = -k [dT/dx]
k < 0.1 W/m/K for
good insulators
Required insulation thicknessRequired insulation thicknesskeep resting adult warm at T = -10 oC. ka = 0.026 W/[m oK] and Edot = - Qdot
Edot = 70 W = kA [T/x] with T= 40 oC. 70 W = 1.04 [A/x] W with A (m2 ) & x (m ) A ~ 2 m2 x ~ [2/70] m ~ 3 cm
2m Ta = - 10 oC30 oC
11stst Law for CV Law for CV
dEcv/dt = mdot [hin - hex ] + Qdot - Wdot
Ecv = mcv [u + 0.5v2 + gz]cv
We could also put KE & PE terms onRHS of 1st eqn.
Open sys in SSSF Qcv + mi hi = me he + Wcv
i
e
QW
Qcv = - 200 kW
mi = me = 3 kg/s
hi = 3000 kJ/kg; he = 2600 kJ/kg
Wcv = 3[3000 – 2600] – 200 = 1000 kW
Note: dot omitted over Q, W, m
ThrottlingThrottling
hi = he X
Eg. Steam at 4 MPa and 700 C exits valve at 0.5 MPa
h = 3906 kJ/kg so Te = 691 oC via
i e
T = 600 + [(3906 – 3702)/(3926 – 3702)] 100
With [ ] = 0.91
s from 7.62 to 8.47because of irreversibilities
T
s
P
SteamSteam Properties Properties
BoilerBoileri
e
T
s
Now take sat liq at 4 MPaentering boiler and heat to 800 C
hi = hf = 1087 and he = 4142 kJ/kg
So q = he - hi = 3055 kJ/kg
Note: Tsat = 250 C
Heating of Heating of LiquidLiquid
Q = m cp [T2 - T1 ]
Above true whether P const or not as heating occurs
For water, cp = c = 1 Btu/lbm/ oF = 4.2 kJ/kg/ K
US Solar US Solar InsolationInsolation
http://www.eren.doe.gov/erec/factsheets/solrwatr.pdf
Solar water heaterSolar water heater
NREL photo
Solar Hot WaterSolar Hot Water
http://www.solaraccess.com/news/story?storyid=5271
$ 375 K cost and annual savings of $ 67 K100 panels heat 800 gph
Vancouver Int.Airport
Solar water heatingSolar water heating solarwaterFbks.m NREL Fbks data; vls are monthly averages [kWh/m^2/day] for March - Nov. for surface tilted at lat angle of 64 deg
mo = 2:10; effic1 = [0.5 0.6 0.7*ones(1,4) 0.6 0.5 0.4];Sinsoltilt9 = [2.4, 4.7, 5.6, 5.3, 5.2, 4.9, 4.2, 3.4, 2.0];% avg = 3.3 for yr & if mult by 3600, we convert to kJ/m^2/day
Tl = 10; Th = 40; rhow = 1; % kg/liter Cpw = 4.2; effic = 0.7; Toi = [Th Tl]; % kJ/kg/K Volw = effic*3600.*Sinsoltilt9/Cpw/rhow/(Th - Tl);Volw1 = 3600.*effic1.*Sinsoltilt9/Cpw/rhow/(Th - Tl);
figure(2); plot(mo,Volw,mo,Volw1,'linewidth',2); grid on; title('Warm water prepared','fontsize',16); % etc
2 3 4 5 6 7 8 9 1020
40
60
80
100
120Warm water prepared
month of year
lite
rs/d
ay
collector tilt at latitude angle = 64 deg
Tout Tin = 40 10 [deg C], effic = 0.7
solar heated water per m2 collectorgreen has effic from 40% in Nov to 70% in summer
Fbks NREL solar data
Solar Air Heater 1Solar Air Heater 1
Solar air heater at UAF
PV panel
Data fromData from Fieldpoint Fieldpoint
Panel output voltage
fan currentamps
Pyranometer W/m^2
Outlet T [deg F]
InletT [deg F]
11:11 4.254 0.04 0.0 -0.5 -1.4
12:11 4.234 0.076 153.9 19.5 1.9
12:41 4.122 0.116 38.5 10.2 2.2
12:56 12.036 0.776 307.8 18.6 4.6
13:26 3.776 0.136 384.7 32.6 3.2
13:41 10.596 0.656 38.5 11.7 2.5
14:11 2.53 0.152 384.7 49.0 5.3
14:56 4.216 0.04 38.5 14.3 2.3
16:41 1.068 0 0.0 2.9 0.1
1/12/03
UAF EnergyCenter
http://pug.engr.uaf.edu/data_html\011203_ecenter.html
Solar Air HeaterSolar Air Heater
~ 8 hr period represented
Solar air heater UAF 2.18.03
y = 0.311x + 4.9705
R2 = 0.8645
0
100
200
300
400
0 200 400 600 800 1000
pyranometer [W/m^2] [Ac ~ 0.8 m^2]
W o
f h
ot
air
del
iver
ed
Solar Air HeaterSolar Air Heater
Heat EnginesHeat Engines
RJ UAF
Receive heat at high T and reject to low T
QH
QL
W
= W/ QH ~ 30 %
Carnot CycleCarnot Cycle
T
s
1
2 3
4
TH
TL
= 1 - TL /TH Is upper bound for heat engine
QH = TH [s3 – s2]
QL = TL [s4 – s1]
= [QH - QL] /QH
W = QH - QL
QH /TH = QL /TL
Tdq
= 0QH
QL
ApplicationsApplications
RJ UAF
Diesel electric generators, gas turbines power plants
http://www.gensetcentral.com/pdf/js170uc.pdf
Example - DEGExample - DEG
RJ UAF
6 gph fuel
138 K Btu/gal
QH = 818 K Btu/hr = 242 kW and = 33 %
Wel = 81 kW with rest of output as heat flux
up exhaust and rejected to jacket water and ambient
Wel
QL
[ 13.5 kWh/gal ]
Second LawSecond Law
Processes only proceed in certain drcts:
Clausius:
Can't build cyclic device whose sole
effect is heat xfer from cold to hotter
body.
EntropyEntropy
T
dqs rev
gensys sT
dqs
or
Heat xfer thru finite T is irreversible
Clausius InequalityClausius Inequality
Consider Rankine Cycle
B
C1
3
4
P
T
2
P2 = P3 = 1 MPaT2 = 100 CT3 = 350 C
SH vap
Sat liq
x = 1
P1 = P4 = 100 kPa and sat so T = 100 C
Tdq < = 0
T
s
1
2
3
4
cond
boilerturb
180 C
100 C
350 C2a
1.30 7.36
s3 = 7.30
Now calc Now calc
Tdq = q23 /T23 + q41 / T34
q23 = h3 - h2 = 3051 – 417 = 2634 kJ/kg
q41 = h1 - h4 = - 2258 kJ/kg
q41 / T34 = - 2258/373 = -6.05 kJ/kg/K
Adiabatic from 1..2 and 3..4
To calc. 3
2T
dqSplit into three parts
In sat regime, = hfg /Tsat = 2015/453 = 4.45
In SH regime, [h3 - hg ]/Tavg = [3051 - 2778]/538
0.51 so
3
2T
dq= 5.79
Tdq = 5.79 - 6.05 = - 0.26
Note : wnet = q23 - q14 = 515 kJ/kg
In CL regime, =[ h2a – h2 ]/Tavg = [762 – 418]/413 = 0.83
Open sys in SSSF
Sgen = m [se - si] – Q/T
I = To Sg
Wrev = W + I
i e
Q
SSSFSSSF
Example - STExample - ST
RJ UAF
hi = 3344; he = 2769
W = 8[hi - he ] - 300
= 4. 3 MW
8 kg/s at 3 MPa
450 oC
0.2 MPa150 oC
W
Q = - 300 kW
Ex 7-15 C&B
Ex. [cont.]Ex. [cont.]
RJ UAF
si = 7.083 se = 7.280 kJ/kg/K so
Sg = 8[se - si] + 300/298 = 2.56 kW/oK
I = To Sg = 765 kW
Wrev = W + I = 5.07 MW
2 = W/Wrev = 85 %
If 1 kg water heated isothermally and reversibly at 100 oC from sat liquid to sat vapor, 1st Law
Q = m[u2 – u1 ] + P[v2 - v1 ] = m[h2 – h1]
= 1[h2 - h1 ] = 1[hfg ] = 2257 kJ
S = Q/T = 2257/373 = 6.048 kJ/K
= m sfg
T
s or v
1 2
Environmental ImpactsEnvironmental Impacts
> 1272 grams of fossil fuel and chemicals used to produce a 32-bit DRAM memory chip [mass of 2 grams]
Another 400 grams of fossil fuel required to produce electricity to operate it over its lifetime.
This doesn’t even account for ultimate disposal
cf ratio of 2:1 for automobile
Environmental Science & Technology / January 1, 2003