basic energy concepts
TRANSCRIPT
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Basic Energy Concepts Basic Energy Concepts –– IntroIntro R JohnsonUAF -
2008
Power different from work or energy
Proper use of units critical
Both supply and demand sides important
Proper utilization of resources requires interdisciplinaryskills.
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UnitsUnits
Fundamental ones include mass, length, time leading to
F has units of m L/t2
and Wk = F d and P = Wk/t
Eg. 1 N = kg m/s2
and 1 lbf
= 32.2 lbm
ft/s2
J = N m ; W = J/s, J = coul-V, MW = 106
W
kWh = 3412 Btu etc.
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ThermodynamicsThermodynamicsDeals with energy and
its conversion
Four fundamental laws with
First dealing with conservation of energy and
Second dealing with concept that energy has quality in addition to quantity
Efficiency sometimes defined as work outdivided by heat in.
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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
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First and Second LawsFirst and Second Laws
1st
Law examples: Solar energy into PV cell =
electrical energy out + heat generated or
Fuel energy into heat engine = work out plus heat released to coolant plus out exhaust
2nd
Law example: Not all solar energy into PV cell
can be converted into electrical work
Nor can all fuel into a heat engine plus
Heat doesn’t spontaneously flow uphill
[nor does water]
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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
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Heat CapacityHeat Capacity
Can use to calc amt of heat xferred
via
q = Δh = c ΔT where c = heat capacity
c ~ 4.2 kJ/kg/K = 1 Btu/lbm/F for water
cp
~ 1 kJ/kg/K
for air [note subscript now]
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ThermalThermal, , chemical, chemical, nuclearnuclear energyenergy1 kg water with ΔT = 100 C
changes internal energy
by ΔU = 420 kJ via mc ΔT with c ~ 4.2 kJ/kg/K
1 kg liquid or gaseous fossil fuel has htg
vl
~ 44MJ
If evap
at P ~ 1 atm, ΔU ~ 2 MJ
E = m c2
⇒
9 x 1016
J via c = 3 x 108
m /s
RJ 4.02
90 B MJ
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Energy amountsEnergy amountsBMR ~ 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 ~ 100 Q/yr with Q = 1015
Btu ⇒
11 kW/cap
RJ 4.08
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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)
mod
e
log m (lbm) RJ 3/2000
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PropertiesProperties
RJ UAF
saturated
SHliqT
s or v
P = const.
h = constC P
Water: Crit
Pt at P = 22 MPa, T = 374 oC
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Sat. steamSat. steam
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FirstFirstConservation of Energy
ΔE = Q -
W
E
QW
mi me
eg. Q = 100 kJ
W = 60 kJ
ΔE = 40 kJ
[for system]
Each term > 0 if by
system
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For a cycleFor a cycle
ΔE = 0 so Qnet
= W which leads to
W = QH
- QL
2nd
Law says all of QH
can’t be
converted into W or η
< 100 %
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Carnot Cycle
For heat engine [brown ↑], heat added qh = area under 2-3 and heat rejected qL = area under 1-4. Via 1st Law, w = qh - qL
and η = w/qh since η = [what one wants]/[what one pays for]. ηC = 1 – TL/Th since , qL/qh = TL /Th . For heat pump or refrig. [blue ↑], work is added as the working fluid goes from 4 to 3 and heat flows out from 3 to 2 with heat coming in from 1 to 4. From 1st law, energy in = energy out so w + qL = qh .COPHP = qh/w. For Carnot cycle, COPHP = Th /[Th - TL ]. COPref = TL /[Th - TL ].
1
23
4
T w = area of enclosure
s
qh
qL
w
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Closed Closed transient transient
Q -
W = m[u2
-u1
]
eg. Let SH steam at P1
= 5 MPa
and 900 oCcool at const v = 0.1076 m3
/kg to 450 oC
STs
⇒ u1
= 3841 kJ/kg
STs
⇒ P2
= 3 MPa
and u2
= 3020
kJ/kg
So Q/ m = -
821 kJ/kg
T
v
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Heat TransferHeat Transfer
Three modes are conduction, convection, andradiation.
1st
two require presence of a physical medium
Latter is how suns energy reaches earth
Conduction relates to R value [thermal resistance]
Convection relates to wind chill
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Conductive Heat Conductive Heat XferXfer
qdot
qdot
= -k [dT/dx] is Fourier’s Law
k < 0.1 W/m/K for good insulators
dT/dx
is Temperature gradient
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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 [ΔT/Δx] with ΔT= 40 oC.
70 W = 1.04 [A/Δx] W with A (m2
) & Δx (m )
A ~ 2 m2 ⇒ Δx ~ [2/70] m ~ 3 cm
2m Ta
= -
10 oC30 oC
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ConvectiveConvective
qdot
= h ΔT with h 25 to 200 W/m2
/K
for gases and 100 to 20K for liquidsin forced convection
Tb
u Ta
h ↑
as u ↑
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RadiationRadiation
qdot = ε σ T4 with σ
= 5.67 x 10-8 W/m2/K4
Sun emits as black body at T = 5762 K so
qdot = 56.7 MW/m2
cf. T = 300 K ⇒ qdot = 460 W/m2
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11stst Law for CVLaw 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.
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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
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ThrottlingThrottling
hi
= he XEg. 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
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SteamSteam PropertiesProperties
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BoilerBoileri
eT
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
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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
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US Solar US Solar InsolationInsolation
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http://www.eren.doe.gov/erec/factsheets/solrwatr.pdf
Solar water heaterSolar water heater
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NREL photo
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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
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Solar water heatingSolar water heatingsolarwaterFbks.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/literCpw
= 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
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2 3 4 5 6 7 8 9 1020
40
60
80
100
120Warm water prepared
month of year
liter
s/da
y
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
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Solar Air Heater 1Solar Air Heater 1
Solar air heater at UAF
PV panel
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Heat EnginesHeat Engines
RJ UAF
Receive heat at high T and reject to low T
QH
QL
W
η
= W/ QH
~ 30 %
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ApplicationsApplications
RJ UAF
Diesel electric generators, gas turbines power plants
http://www.gensetcentral.com/pdf/js170uc.pdf
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Example Example -- DEGDEG
RJ UAF
6 gph fuel
138 K Btu/gal
QH
= 818 K Btu/hr = 242 kW and η
= 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 ]
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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.
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Energy, Energy, ExergyExergy, and Entropy, and Entropy
Masanori Shukuya
& Abdelaziz
Hammache, 2002
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Energy and Energy and exergyexergy
“Current focus on energy conservation, as a strategy, is at best incomplete and at worst wholly incorrect. As it is converted from one form to another, energy is neither lost nor destroyed. It does, however, "lose a certain quality which can be described as its ability to do work." [1] Since it is the ability of energy to do work which gives energy its value to society, we should strive to conserve available work”
[exergy] “, not energy.
“
Simpson and Kay, 1989
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Quality of EnergyQuality of Energy
Wall, 1977, http://exergy.se/ftp/paper1.pdf
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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
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Solar water heaterSolar water heater
Consider 676 W/m2 incident solar rad
with collector
α
= 0.9 and ε
= 0.1 [selective absorber]
qdotnet-rad
= α
Gsolar
-
ε σ [Ts4
- Tsky
4
]
Ts
= 320 K and Tsky
= 260 K ⇒
qdotnet-rad
= 575 W/m2
If instead, α
= ε
= 0.9, qdotnet-rad
= 307
Ex 12.5 C & B –
Ht xfer