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 energyenergy1 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 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 ~ 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)

mod

e

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 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

SteamSteam Properties Properties

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

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

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

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 + 4.9705R2 = 0.8645

0100200300400

0 200 400 600 800 1000

pyranometer [W/m^2] [Ac ~ 0.8 m^2]

W o

f hot

air

deliv

ered

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 = 0

QH

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

Tdqs rev

gensys sTdqs

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

4cond

boilerturb180 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

dq Split 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 MPa450 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