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### Transcript of Basic Energy Concepts

Basic Energy Concepts IntroPower different from work or energy Proper use of units critical Both supply and demand sides important

R Johnson UAF - 2008

Proper utilization of resources requires interdisciplinary skills.

UnitsFundamental 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.

ThermodynamicsDeals 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 out divided by heat in.

Thermodynamics basicsZeroth LawA B C

If A and B and B and C are in thermal equil, then A and C are in thermal equil. [ie. At same T]R Johnson/UAF/CEM/ME 2005

First and Second Laws1st 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 Heat doesnt spontaneously flow uphill [nor does water] plus

KE and PEKE = [1/2]mv2 ; PE = mgh

So, m = 1 kg, v = 100 m/s, h = 100 m, KE = 0.5[104] kg[m2/s2 ] via J = N-m = 5000 J

and N = kg m /s2

PE = 1[9.8][100] kg m2 /s2 = 980 J

Heat CapacityCan use to calc amt of heat xferred via q = h = c T 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]

Thermal, chemical, nuclear energy1 kg water with T = 100 C changes internal energy by U = 420 kJ via mc T with c ~ 4.2 kJ/kg/K If evap at P ~ 1 atm, U ~ 2 MJ 1 kg liquid or gaseous fossil fuel has htg vl ~ 44MJ E = m c2 9 x 1016 J90 B MJ RJ 4.02

via c = 3 x 108 m /s

Energy 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

Relative massesMass of Water for 1000 Btu PE (780 ft) mode KE (224 ft/s) Thermal (1 deg F) Melting (144 Btu/lbm) Evap(1000 Btu/lbm) 0 1 2 3 4

log m (lbm) RJ 3/2000

Propertiesh = const CP T liq saturated s or v Water: Crit Pt at P = 22 MPa, T = 374 oC SH P = const.

RJ UAF

Sat. steam

Conservation of Energy

First

E = Q - W[for system]

Q W

eg. Q = 100 kJ W = 60 kJ E = 40 kJ

Emi me

Each term > 0 if by system

For a cycle E = 0 so Qnet = W which leads to W = Q H - QL 2nd Law says all of QH cant be converted into W or < 100 %

Carnot Cycle

qh2

3

T

w = area of enclosure

w

1

qL s

4

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 = w/qh since = [what one wants]/[what one pays for]. C = 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 ].

Closed transient Q - W = m[u2 -u1]

T v

eg. Let SH steam at P1 = 5 MPa and 900 oC cool at const v = 0.1076 m3 /kg to 450 oC STs u1 = 3841 kJ/kg STs P2 = 3 MPa and u2 = 3020 kJ/kgSo Q/

m = - 821 kJ/kg

Heat TransferThree modes are conduction, convection, and radiation. 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

Conductive Heat Xfer qdot = -k [dT/dx] is Fouriers Law k < 0.1 W/m/K for good insulators dT/dx is Temperature gradient

qdot

Required 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 cm30 oC 2m Ta = - 10 oC

Convectiveqdot = h T with h 25 to 200 W/m2 /K for gases and 100 to 20K for liquids in forced convection u Ta Tb h as u

Radiationqdot = T4 with = 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

1st Law for CVdEcv/dt = mdot [hin - hex ] + Qdot - Wdot Ecv = mcv [u + 0.5v2 + gz]cv We could also put KE & PE terms on RHS of 1st eqn.

Open sys in SSSF Qcv + mi hi = me he + Wcv Qcv = - 200 kW mi = me = 3 kg/s

i

Q

W e

hi = 3000 kJ/kg; he = 2600 kJ/kg Wcv = 3[3000 2600] 200 = 1000 kWNote: dot omitted over Q, W, m

Throttling

h i = he

i

X

e

Eg. Steam at 4 MPa and 700 C exits valve at 0.5 MPa h = 3906 kJ/kg so Te = 691 oC via T = 600 + [(3906 3702)/(3926 3702)] 100 With [ ] = 0.91T P s

s from 7.62 to 8.47 because of irreversibilities

Steam PropertiesCengel & Boles

BoilerT Now take sat liq at 4 MPa entering boiler and heat to 800 C s i

e

hi = hf = 1087 and he = 4142 kJ/kg So q = he - hi = 3055 kJ/kgNote: Tsat = 250 C

Heating of LiquidQ = 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 Insolation

Solar water heater

http://www.eren.doe.gov/erec/factsheets/solrwatr.pdf

NREL photo

Solar Hot Water

Vancouver Int. Airport

\$ 375 K cost and annual savings of \$ 67 K 100 panels heat 800 gphhttp://www.solaraccess.com/news/story?storyid=5271

Solar 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/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

Warm water prepared120

100

collector tilt at latitude angle = 64 degTout Tin = 40 10 [deg C], effic = 0.7

liters/day

80

solar heated water per m2 collectorgreen has effic from 40% in Nov to 70% in summer

60

40

20

2

3

4

5

6

7

8

9

10

month of year

Fbks NREL solar data

Solar Air Heater 1

PV panel

Solar air heater at UAF

Heat EnginesReceive heat at high T and reject to low T

QL QH W

= W/ QH

~ 30 %RJ UAF

ApplicationsDiesel electric generators, gas turbines power plants

http://www.gensetcentral.com/pdf/js170uc.pdfRJ UAF

Example - DEG6 gph fuel138 K Btu/gal

Wel QL

QH = 818 K Btu/hr = 242 kW and = 33 % Wel = 81 kW with rest of output as heat fluxup exhaust and rejected to jacket water and ambient

[ 13.5 kWh/gal ]RJ UAF

Second LawProcesses only proceed in certain drcts:

Clausius: Can't build cyclic device whose sole effect is heat xfer from cold to hotter body.

Energy, Exergy, and Entropy

Masanori Shukuya & Abdelaziz Hammache, 2002

Energy and exergyCurrent 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

Quality of Energy

Wall, 1977, http://exergy.se/ftp/paper1.pdf

Environmental 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 doesnt even account for ultimate disposal cf ratio of 2:1 for automobileEnvironmental Science & Technology / January 1, 2003

Solar water heater

Consider 676 W/m2 incident solar rad with collector = 0.9 and = 0.1 [selective absorber] qdotnet-rad = Gsolar - [Ts4 - Tsky4 ]

Ts = 320 K and Tsky = 260 K qdotnet-rad = 575 W/m2 If instead, = = 0.9, qdotnet-rad = 307Ex 12.5 C & B Ht xfer