basic energy concepts

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.

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.

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.

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

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]

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

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]

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

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

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 = constC P

Water: Crit

Pt at P = 22 MPa, T = 374 oC

Sat. steamSat. steam

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

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 %

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

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

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

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

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

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 ↑

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

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.

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 PropertiesProperties

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 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

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

Heat EnginesHeat Engines

RJ UAF

Receive heat at high T and reject to low T

QH

QL

W

η

= W/ QH

~ 30 %

ApplicationsApplications

RJ UAF

Diesel electric generators, gas turbines power plants

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

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 ]

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.

Energy, Energy, ExergyExergy, and Entropy, and Entropy

Masanori Shukuya

& Abdelaziz

Hammache, 2002

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

Quality of EnergyQuality of Energy

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

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

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