MECH-325: Introduction to Energy Systems
2004-05 Catalog Data:
MECH-325 Introduction to Energy Systems Credit: 4 (4-0-0-4)
Prerequisite: MATH-102 and Junior Standing Co-requisite: None
Description: This introductory course is designed for non-mechanical engineering students. It integrates the fundamental principles of thermodynamics, fluid mechanics and heat transfer. Basic principles of thermodynamics are followed by properties of pure substances, application of the first and the second laws of thermodynamics to thermal systems, application of conservation of mass, momentum and energy to fluid systems, principles of conduction, convection and thermal radiation. Topics such as thermodynamic cycles, internal and external flows, refrigeration, aerodynamic lift and drag, pump performance, free and forced convection, and heat exchangers are covered. Practical applications of the principles discussed are emphasized. A PEM Fuel Cell and a Cylinder Convection experiment are incorporated into this course.
Textbook: Introduction to Thermal Systems Engineering: Thermodynamics, Fluid Mechanics, and Heat Transfer, Michael J. Moran, Howard N. Shapiro, Bruce R. Munson, and David P. DeWitt, 2002, Wiley
Reference: Introduction to Thermal Sciences: Thermodynamics Fluid Dynamics Heat Transfer, 2nd ed., Frank W. Schmidt, Robert E. Henderson and Carl H. Wolgemuth, 1993, Wiley
Coordinator: Ahmad Pourmovahed
Course Learning Objectives: Upon completion of this course, the student will be able to: 1. Apply the fundamental principles of thermodynamics, fluid mechanics and heat
transfer to a variety of thermal-fluid systems. 2. Apply measurement techniques and experimental methods to energy systems. 3. Apply team skills. 4. Conduct experiments, analyze and interpret data. 5. Communicate effectively. Prerequisites by topic: 1. Elementary calculus 2. Basic computer skills Topics covered: Week Topic 1 Thermodynamic Concepts and Definitions 2 Properties of Pure Substances 3 System Analysis: First and Second Laws
Dr. A. Pourmovahed, Kettering University 1 Introduction to Energy Systems
4 Control Volume Analysis 5 External Flows, Lift and Drag 6 Internal Flows, Piping System Losses 7 Conduction Heat Transfer 8 Forced and Natural Convection Heat Transfer 9 Heat Exchangers and Thermal Radiation 10 PEM Fuel Cell and Cylinder Convection Laboratory Experiments 11 Laboratory Presentations and Final Exam Schedule: Two 120-minute sessions per week Computer usage: Any Spreadsheet and Word Processing Software Laboratory Experiments: Two laboratory experiments performed by the students at the
Energy Systems Laboratory during Week 10. Prepared by: Ahmad Pourmovahed, Professor of Mechanical Engineering, 11/25/2003
Dr. A. Pourmovahed, Kettering University 2 Introduction to Energy Systems
THERMODYNAMICS A science which deals with energy and its transformation and with relationships between the properties of substances. System: Any object, quantity of matter, or region of space selected for study. Closed System: No mass crosses the boundary (only energy does).
Q
GasW
Open System (Control Volume): Mass (as well as energy) flows through the boundary.
W
o u tmi nm
Q
Surroundings: Everything external to the system.
Dr. A. Pourmovahed, Kettering University 3 Introduction to Energy Systems
Property: A quantity which serves to describe a system.
Intensive Property: Independent of system mass e.g., , , , , , ,p T v u h sρ .
Extensive Property: Depends on system mass e.g., V m . , , , , ,U H S E Process: An event which takes the system from one state to another. Isothermal: Constant temperature (T = const.). Adiabatic: No heat transfer (Q = 0). Isometric: Constant volume (V = const.). Isobaric: Constant pressure (p = const.). Reversible: When reversed, leaves no change in system or surroundings. Isentropic: Constant entropy (S = const.) (reversible and adiabatic).
Laws of Thermodynamics:
Oth Law: Two systems in thermal equilibrium with a third system will be in thermal equilibrium with each other.
1st Law: During a given process, the net heat input minus the net work
output equals the change in total energy.
2nd Law: A device operating in cycles cannot produce work continually while exchanging heat with a single reservoir.
3rd Law: The entropy of a pure crystalline substance is zero at 0o K.
Entropy: .rev
Qd ST
δ ≡
Comments about the 2nd Law • A decrease in entropy is not possible for an isolated system.
Dr. A. Pourmovahed, Kettering University 4 Introduction to Energy Systems
• All systems, left to themselves, will reach an equilibrium state such that the entropy function reaches a maximum.
• Entropy is not conserved in real processes. • The entropy production depends on the magnitude of irreversibilities in the
system • The increase in entropy corresponds to a decrease in the quality of energy. Ideal Gas Law: pV mRT=
( )pv RT p RTρ= = uRRM
=
.8.314 1.986 1545 f
u o o
ft lbkJ BtuRkg mole K lb mole R lb mole R
= = = o⋅ ⋅ ⋅
Absolute Temperature: ( ) ( ) 460o oT R T F= +
T K ( ) ( ) 273o oT C= +
Enthalpy: H U pV≡ +
ph u pv uρ
≡ + = +
Quasistatic Compression Work: 1 2
21
pdVW→
= ∫
Introduction to Energy Systems
1P
V
Area = 1 2W→
2
Dr. A. Pourmovahed, Kettering University 5
p, V
The Phase Rule: 2F C= + − P F = Number of intensive properties required to fix the intensive state of each phase. C = Number of components. P = Number of phases.
Specific Heats:
Vv
ucT
∂≡ ∂
dudT
for an ideal gas
pP
hcT
∂≡ ∂
d hd T
for an ideal gas
1st Law for a Closed System:
E Q W∆ = − (with negligible kinetic and potential energy changes, E∆ = U∆ )
Dr. A. Pourmovahed, Kettering University
Q
E W
6 Introduction to Energy Systems
1st Law for an Open System: (W excludes flow work)
Q
W
o u tminmE
2
2i n o u ts y s t e m
d E vQ W m h g zd t −
= − + ∑ + +
LQ
HQ
HeatEngine netW
HT
LT
Carnot Efficiency:
net H LW Q Q= − (1st Law)
H H
L L
Q TQ T
=
,max 1 Lth
H
TT
η = −
2nd Law for Closed System:
2
2 1 1
producedSQS Szero or positiveT
δ− = +∫
Dr. A. Pourmovahed, Kettering University 7 Introduction to Energy Systems
2nd Law for Open System:
producedAreain outsystem
d S Q m s Sd t T −
= + +
∑∫∫
Isolated System: 2 1S S⟩
Ideal Gas Equations:
pv RT= p vc c R− =
vdu c dT= 1.0p
v
ck
c≡ ⟩
.kpv const= isentropic process constant cp, cv
. ,p vconst c c :
2 2
1 1p
T ps c ln R ln cT p
∆ = − =
variable c c , :p v
0 02 1s s s R∆ = − −
( )0s from table Incompressible Substance:
2
1
Ts c lnT
∆ =
Q mc T= ∆
Dr. A. Pourmovahed, Kettering University 8
1 1
2 2 1
1 1 2
k kkT p v
T p v
−−
= =
2
1v
Tln R lnT
+ 2
1
vv
2
1
plnp
( ).const c
( ).const c
Introduction to Energy Systems
Pure Substance:
, mass of vaporquality xmixture mass
≡
( )0x saturated liquid=
( )1x saturated vapor=
( )0 1x saturated mixture⟨ ⟨
f fv v x v= + g
g
g
g
P m
f fu u xu= +
f fh h x h= +
f fs s x s= +
T
superheatedvapor
v
liquid
p
mixtureT
c.p.
Dr. A. Pourmovahed, Kettering University 9
Vapor
Liquid
Introduction to Energy Systems
evapor.
Triple Pt.
elting
vapor
sublimation
T
liquid
solid
P
v or s
c.p.
c.p.
Mollier Diagram
s
h
T
P
The Rankine Cycle: The Rankine cycle is the ideal cycle for a steam power plant. It consists of the following processes: 1→ 2 isentropic compression (ideal pump) 2→ 3 isobaric heat addition (boiler) 3→ 4 isentropic expansion (ideal turbine) 4→ 1 isobaric heat rejection (condenser)
4
3
1
3
C
P
B
4
2
2T
Example 1:
Dr. A. Pourmovahed, Kettering University 10
1
s
Introduction to Ene
T
rgy Systems
Heat from a combustion process is supplied to an ideal Rankine cycle using steam as the working fluid. Steam leaves the boiler at 800°F; 600 psia and turbine exhaust pressure is 1 psia. Determine the thermal efficiency of this cycle. Solution:
3 3600 800p psia and T F= = ° From superheated steam tables:
3
3
1407.551.6343
Btulbm
Btulbm R
hs
== i
( )4
4 3
1 , .p psia assume saturated mixtures s ideal turbine
==
4 4f fgs s x s= +
( )41.6343 0.13266 1.8453x= +
4 0.814x =
( )( )4 4 69.74 0.814 1036 912.8 Btulbmf fgh h x h= + = + =
1 @1 69.74 Btulbmfh h psia≅ =
( ) ( )( )2 1 0.016136 600 1 144 / 778 1.8 Btulbmp fw v p p≅ − = − =
2 1 71.5 Btulbmph h w= + =
3 4 494.75 BtulbmTw h h= − =
493 Btulbmnet T Pw w w= − =
3 2 1336 BtulbmBq h h= − =
4 1 843 Btulbmcq h h= − =
Note: net B cw q= − q Thermal Efficiency:
Dr. A. Pourmovahed, Kettering University 11 Introduction to Energy Systems
493 36.9%
1336net
B
wq
η = = =
Example 2: Consider heating a one-mile high building. Steam for the heating system enters a pipe at ground level as dry saturated vapor at 30 psia. On the top floor, the pressure in the pipe is 14.7 psia. The heat transfer from the steam as it flows up the pipe is 50 Btu/lbm. What is the state of the steam at the top floor?
First Law: 2
2in outsystem
d E vQ W m h g zd t −
= − + ∑ + +
( ) (1 2 1 20 0q h h g z z= − + − + − )
( )2 1 1 2h h q g z z= + + −
1 @30 1164.3g Psiam
Btuh h lb= =
50
m
Btuq lb= −
( ) ( )2 1164.3 50 32.2 5280 / 32.2 778.2 1107.5
m
Btuh lb= − + − × =
2 14.7 180.15 , 970.4f fgm m
Btu Btup psia h hlb lb
= ⇒ = =
2 2f fh h x h= + g
( ) 22
95.6%1107.5 180.15 970.4 xxsaturated mixture
== + ⇒
Dr. A. Pourmovahed, Kettering University 12 Introduction to Energy Systems
Example 3: A 2000 lb. automobile traveling at 60 mph stops suddenly. Its kinetic energy (309 Btu) is converted to heat in the brakes raising their temperature from 80oF to 600oF. The
specific heat of the brake material is 0.1.o
m
Btulb R
. The ambient temperature is 520oR.
a. What is the change in the entropy of the system (the automobile)? b. How much of the original kinetic energy of the vehicle is now unavailable?
a Mass of the brakes, Qmc T
=∆
( )309 5.94
0.1 600 80 mm l= =−
b
2
1
460 6005.94 0.1 0.401460 80 o
T BS mc n nT R
tu+∆ = = × ==
+
0.401 osystem
BtuS R∆ =
b . 0 ( )surrS notime for heat transfer∆ =
. 0.401 0 0.401univ o
BtuSR
∆ = + =
0 .. univUnavail Energy Irreversibility T S= = ∆
( )520 0.401 208.5 Btu= =
Example 4: One lbm of air is contained in a rigid, insulated tank at 500oR, 14.7 psia. There is a work transfer of 146 Btu to the contents, paddling the air.
Neglect the mass of the container and the paddle wheel. Assume 0.17.v o
m
Btuclb R
= . The
ambient temperature is 500oR. Determine the destroyed available energy.
Dr. A. Pourmovahed, Kettering University 13 Introduction to Energy Systems
Air
0, 146Q W Bt= = − u
1st Law: ( )0 146 146U Q W Btu∆ = − = − − =
( )2 1 ( )vU mc T T ideal gas∆ = −
2 1v
UT Tmc∆
= +
2146500 1358.8
1 0.17oT R= + =
×
22 1
1 1v
TS S m c ln R lnT v
− = +
2v
2
1
1358.81 0.17 0.16996500system v o
T BS m c ln lnT R
∆ = = × =tu
. 0surrS∆ =
0 .. . univAvail Destroyed Irrev T S= = ∆
( )500 0.16996 0 85 Btu= + =
Dr. A. Pourmovahed, Kettering University 14 Introduction to Energy Systems
FLUID MECHANICS A branch of engineering concerned with forces and energy generated by fluids at rest or in motion. Fluid: A substance that moves and deforms continuously as long as a shear stress is applied. Liquids: Form a free surface in a gravitational field; relatively incompressible. Gases: No free surface; highly compressible compared to liquids.
Isothermal Bulk Modulus: E V PVV
T≡ −
∂∂
EV ≅ ×2 104 atm for liquid water @ 70o F.
( = 20 atm 0.1% reduction in volume) ∆P ⇒
EV = P for any ideal gas (at constant temperature)
(95% reduction in volume if PP
2
1
20= )
For a Fluid at Rest: - Pressure, P, does not change horizontally. - For a change in elevation by z∆ , P decreases by ρ.g. z∆ . - Pressure is independent of orientation.
Density: mass per unit volume, Vm
=ρ
Specific Weight: weight per unit volume, γ ρ= g
Specific Gravity: S 0H2
ρρ
≡ (13.6 for Hg)
Viscosity: Shear stress/rate of deformation
τ=µ
dydu
v
fluidy
x
u
( )h/vτ
=µ h
Dr. A. Pourmovahed, Kettering University 15 Introduction to Energy Systems
Kinematic Viscosity : ρµ
≡ν
- The viscosity, µ , increases with the temperature for gases. - The viscosity, µ , of a liquid decreases with the temperature.
- The viscosity, µ , increases slowly with the Pressure, for gases and most liquids.
- The kinematic viscosity, ν, varies significantly with both P and T for gases. The Speed of Sound:
For an ideal gas, c kR= T k = 1.40 for air
R = 53.35 Rlbm
lb.ft f
− for air
c = 1129 ft/s for air at 70O F. c is higher in “incompressible” substances (liquids and solids) than in gases.
Dr. A. Pourmovahed, Kettering University 16 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 17 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 17 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 18 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 19 Introduction to Energy Systems
Fluid Statics: Study of forces in fluids at rest. Pressure variation in vertical direction:
γ−=ρ−= gdzdp
In a fluid at rest, the pressure varies only in the direction of gravity.
z
h
Pa
Gas ghPP .liq
essurePrGage
agas ρ=−
liquid
Pa = Atmospheric Pressure (can be measured with a
barometer).
U-Tube Manometer
At sea-level Pa = 101.35 kPa = 761 mm Hg
= 29.96 in Hg
Buoyancy: B = net upward force on submerged or floating object
∇ Air
Liquid
B
B = Weight of displaced Liquid + weight of displaced Air
B
∇Liquid
Air
B = Weight of displaced Liquid
Example:
What percentage of the total volume of an iceberg floats above the water surface?
ρ icemlb
ft=57 2 3. ; ρ water
mlbft
=62 4 3.
Weight of iceberg = Buoyant force ρ ρice iceberg water submerged∀ = ∀ ∀∀ = = =submerged
iceberg
ice
water
ρρ
572624 917%.
. . (8.3% above free surface)
Example A block of concrete weighs 100 lbf in air and “weighs” only 60 when immersed in
fresh water
lbf
( .γ = 624 3lb
ftf ) . What is the average specific weight of the concrete?
Dr. A. Pourmovahed, Kettering University 20 Introduction to Energy Systems
flb60'W =
flb100W =
flb40B =
B
W
W∇
B = Weight of the displaced water. B = γ water concrete∀
∀ = =concrete ft40
62 40 641 3
..
γ concreteconcrete
fW lbft
=∀
= =100
0 641156 0 3.
.
Fluid Dynamics: Study of fluids in motion
Volumetric Flow Rate:
Q A V= ⋅ m3
s or
fts
cfs3
=
V = Average velocity
Mass Flow Rate:
m AV Q= =ρ ρ
m has units of: lbm
s or
kgs
Reynolds Number:
νhh
eDVDV
R =µ
ρ≡
Dr. A. Pourmovahed, Kettering University 21 Introduction to Energy Systems
V = Average velocity (in pipe). Dh = Hydraulic Diameter
PerimeterWetted)AreaSectionalCross(4
D h =
Dh = D for a circular pipe. For Pipe flow: ⇒ laminar flow Re ⟨ 2200 ⇒ critical flow 2200 4000⟨ ⟨Re
Re ⟩ 4000 ⇒ turbulent flow
Parabolic
laminar
Turbulent core
Laminar sublayer
turbulent
Fundamental Equations:
Conversation of Mass (Continuity):
outm inm m
c.v.
∂∂mt
m mc v
in out = −
. .
Accumulation of mass in the c.v. (zero in steady flow). Bernoulli’s Equation:
Dr. A. Pourmovahed, Kettering University 22 Introduction to Energy Systems
For steady, frictionless flow of an incompressible fluid along a streamline:
gzv21P 2 ρρ ++ = const.
or:
.constzg2
vP 2
=++γ
2
Pγ
= Pressure head
g2v2
= Vel. Head 1
z = Elevation head Sum of all three = Total head The Momentum Equation:
∑−−
−
−
+⋅∂∂
=
timeunitperinout.momx
inoutx
timeunitper.v.cin.momxofonAccumulati
.v.cxx )vm()vm(t
F
Sum of all external forces on the c.v. including pressure forces, body forces, and shear and contact forces (x dir.)
The Energy Equation:
Q
W
o u tminm
E
Dr. A. Pourmovahed, Kettering University 23 Introduction to Energy Systems
∂∂Et Q W m h v gz
c vS
in out
= − + + +
−∑
. .
2
2
Zero if steady Applications of Bernoulli’s Equation
Emptying a Large Tank
A A1 2>>
0v1≅
P P2 1 0= =
∇ 1
2
A2
h
Bernoulli’s equation: 2
222
1
211 gz
2vP
gz2
vP++
ρ=++
ρ
2
22
1 gz2
v0gz00 ++=++
( ) gh2zzg2v 2122 =−=
gh2v2=
2 2 2 2Q A v A gh and m Qρ= = = Venturi Meter
Dr. A. Pourmovahed, Kettering University 24 Introduction to Energy Systems
2
222
1
211 gz
2vP
gz2
vP++
ρ=++
ρ
21 zz =
Continuity: 21
212211 v
AA
vvAvA =⇒ρ=ρ
−=
−=
ρ− 2
1
222
22
2
1
22221
AA
12
v2
vAA
2vPP
( )2
1
2
212
AA1
PP2v
−
−ρ
=
( )2
1
2
212
AA1
PP2AQ
−
−ρ
=
For compressible fluids, such as air, correction factors must be applied to the above equation. Pitot Tube
Dr. A. Pourmovahed, Kettering University 25 Introduction to Energy Systems
A simple device used to measure fluid velocity.
PP0
V
2
v
P0
P
1
22
22
11
21 gzP
2vgzP
2v
+ρ
+=+ρ
+
PP,PP,0)zz(g,0v 102212 ===−=
ρ=
ρ+ 0
21 PP2v
ρ−
=PP2v 0
1
or: 20 v
21PP ρ=−
Dyn. Press.
Stag. Press.
Stat. Press.
Applications of the Momentum Equation:
Thrust of a rocket:
Example:
A rocket is tested on a test stand. Fuel is fed in at a rate of 1 kg/s and liquid oxygen (LOX) at 10 kg/s. The flexible couplings and supporting rollers offer negligible horizontal force. Exhaust gases exit at atmospheric pressure with a velocity of 2000 m/s. Find the rocket thrust. Assume steady-state operation.
Dr. A. Pourmovahed, Kettering University 26 Introduction to Energy Systems
Conservation of Mass:
∂∂mt m m
c vin out
= −
. . = 0
m mout in= = 1 + 10 = 11 kg/s
x-Momentum Equation:
( )F m v vx xout in= + −∑ 0 x
Fx∑ = 11 kgs × 2000 m
s = 22000 N
Force on c.v. = + 22 kN
Force on the thrust stand = - 22 kN
Rocket Thrust = 22 kN
Jet Deflected by Vane:
A jet is deflected by a stationary vane. If the jet speed is 100 ft/s and its diameter is 1 in., what force is exerted on the vane by the jet?
Bernoulli’s: V1 = V2
V2x = 100 Cos (60°) = 50 ft/s
Dr. A. Pourmovahed, Kettering University 27 Introduction to Energy Systems
V2y = -100 Sin (60°) = -86.6 ft/s
V1x = 100 ft/s
V1y = 0
The Momentum Equation
ΣFx = 0 + ( )m v vx xout in−
ΣFy = 0 + ( )m v vy yout in−
Forces on the c.v. (water):
Fx = ( )m v vx x2 1−
m = ρA1V1 = 62.4 lbft
m3
π4
112 100
22
×ft ft
s
m = 34.03 lbm/s
Fx = 34.03 lbsm (50-100) ft
slb
lb fts
f
m322 2.= -52.9 lbf
Fy = ( )m v vy y2 1−
Fy = 34.03 (-86.6 - 0) 1322. = -91.6 lbf
Forces on the Vane
Fx = + 52.9 lbf
Fy = + 91.6 lbf
Force on a Bend
No Friction
Pipe Diameter = 2 cm
Fluid is water
Dr. A. Pourmovahed, Kettering University 28 Introduction to Energy Systems
v1 = v2 = QA
m s
m= ×
−1 5 10
42
100
3 3
22
.π
/ = 4.775 m/s
A = 3.142 × 10-4 m2
V1x = -4.775 m/s V1y = 0
V2y = -4.775 m/s V2x = 0
ΣFx = - PxwfF 1A1 = ρ Q (0 - V1x)
xwfF = P1A1 - ρ QV1x = 500 × 103 × 3.142 × 10-4 - 1000 × 1.5 × 10-3 (-4.775)
= 164.3 N
ΣFy = + PywfF 2A2 = ρ Q (v2y - v1y)
ywfF = 1000 × 1.5 × 10-3 (-4.775 - 0) - 500 × 103 × 3.142 × 10-4 = -164.3 N
Force on a U-Tube
Neglect Friction, P2 = Atmospheric
A1 V1 = A2V2
V2 = AA V V1
21 14=
V2 = 20.8 ft/s
P2 = 0, z1 = z2
2
222
1
211 gz
2vPgz
2vP
++ρ
=++ρ
P1 = P2 + ( ) ( )ρ2
6242 208 52 1
32174 14422
12
12 2v v P− ⇒ = −
×. . . . = 2.73 psi
m = ρA1v1 = 62.4 × π4
212
2
× 5.2 = 7.079 lbm/s
ΣFx = F + Pwfx 1A1 = m (v2x - v1x)
Dr. A. Pourmovahed, Kettering University 29 Introduction to Energy Systems
Fwfx= m (v2x - v1x) - P1A1
Fwfx= 7.079 (-20.8 - 5.2) 1
32174. - 2.73 π4 (2)2 = -14.3 lbf
Force on the U tube = + 14.3 lbf
The K.E. Correction Factor
Corrects for nonuniform velocities at inlets and outlets.
v
v A vdAA
= ∫1
v = 12 (centerline velocity) for laminar flow in a circular pipe
α =
∫
1 3
Avv dA
A
uniform flow
most turbulent flows
laminar flow in a circular pipe
α = 10
10520
...
≈
General Energy Equation for a Piping System
Pump Turbine
2
1
wP wT
steady flow ρ = constant
m
P v
g z h h h P vg zP T L
11
12
12
222
22 2γ α γ α+ + + − − = + +
h wmg
wQP
P≡ =γ
P Pump Head
Dr. A. Pourmovahed, Kettering University 30 Introduction to Energy Systems
h wmg
wQT
T T≡ =γ
Turbine Head
gmQ
guuh 12
L −−
≡ Head Loss
hL > 0 2nd Law
Total Head = P vg zγ α+ +2
2
Pressure Head Velocity
Head
ElevationHead
Head Loss for an Abrupt Expansion:
V1
hm = ( )v vg
1 22
2−
Head Loss for a pipe discharging into a large reservoir:
hm = vg2
2 (entire velocity head is lost.)
Dr. A. Pourmovahed, Kettering University 31 Introduction to Energy Systems
Example:
Find pump power for the system shown.
1
2
P
2hL = 1.5 vg22
2
Q = 0.25 m3/s
P1 = 0, v1 ≅ 0, α2 ≅ 1.0
P vg z h h h P v
g zP T L1
112
12
222
22 2γ α γ α+ + + − − = + +
( )v Q
A22 2
025
4 030354= = =.
..
π m/s
hP = ( )P vg z z hL
2 22
2 12γ + + − + 15 222
. vg
hP = P vg z z2 22
2 12 5 2γ + + −.
( )hP = × +×
100 109810 25
3542 981
3 2
..
. + (10 - 6) = 15.79 m
wP = γQhP = 9810 × 0.25 × 15.79 = 38.72 kw (minimum power required)
Dr. A. Pourmovahed, Kettering University 32 Introduction to Energy Systems
Example
zA = 30 m
zB = 32 m
zC = 27 m
zD = 26 m
- Head loss from inlet to B = 34 of velocity head
- Head loss from B to D = 14 of velocity head
Find Q and PB
Remember hL = vg2
2 at Point D
Total head loss from A to C = 2 vg2
2
P vg z
vg
P vg zA A
Ac c
cγ γ+ + − = + +
2 2 2
2 2 2 2
vg
2
= zA - zC = 3m ⇒ v = ×981 3. = 5.42 m/s
Q = Av = π4 (0.3)2 × 5.42 = 0.384 m
s3
P vg z
vg
P vg zA A
AB B
Bγ γ+ + − = + +
2 2 2
234 2 2
PB = γ (zA - zB) - γ ( 1.75vg
2
2 )
PB = 9,810 [(30-32) - 1.75 32 ] = -45,371 Pa = -45.4 kPa, gage
Head Loss Calculations v D
L
Head Loss due to Pipe Friction:
hf ≡ f LD
vg
2
2
f = friction factor
Dr. A. Pourmovahed, Kettering University 33 Introduction to Energy Systems
For Laminar Flow (Re < 2200): f = 64Re
Re ≡ ρvDµ
For Turbulent Flow:
f = f (Re, ∈D ) (Moody’s Diagram)
∈D = Relative Pipe Roughness
Minor Losses (Valves, bends, fittings, etc…)
hm = Kvg
2
2 (K ≡ Loss Coefficient)
hL = hf + hm Total Head Loss
Dr. A. Pourmovahed, Kettering University 34 Introduction to Energy Systems
Loss Coefficients for Valves and Fittings* Fitting or Valve K Standard 45° elbow 0.35 Standard 90° elbow 0.75 Long-radius 90° elbow 0.45 Coupling 0.04 Union 0.04 Gate valve Open 0.20 ¾ Open 0.90 ½ Open 4.51 ¼ Open 24.0 Globe valve Open 6.4 ½ Open 9.5 Tee (along run, line flow) 0.4 Tee (branch flow) 1.5
Roughness of Pipe Materials Type ∈ (mm) ∈ (ft) Glass Smooth Smooth Asphalted cast iron 0.12 0.0004 Galvanized iron 0.15 0.0005 Cast iron 0.26 0.00085 Wood stave 0.18-0.90 0.0006-0.003 Concrete 0.30-3.0 0.001-0.01 Riveted steel 1.0-10 0.003-0.03 Drawn tubing 0.0015 0.000005
Dr. A. Pourmovahed, Kettering University 35 Introduction to Energy Systems
Loss Coefficients for Sudden
Contractions* A2/Al K 0.1 0.37 0.2 0.35 0.3 0.32 0.4 0.27 0.5 0.22 0.6 0.17 0.7 0.10 0.8 0.06 0.9 0.02
Dr. A. Pourmovahed, Kettering University 36 Introduction to Energy Systems
Example a) Is the flow laminar?
∇µ = ⋅.015 P sa
ρ = 1200 3kg
m
D = 4 mm
v ms= 1
(1)
3 mfully-developed flow
(2)
constant - head tank
P1
b) Find P1.
a) R vDe = = × ×ρ
µ1200 1 0004
0015.
.
Re = 320 < 2200 ⇒ laminar flow
b) f = 64 64320Re
= = 0.2000
hf = f ( )LD
vg2 2
2 02 30004
12 981= ×. . . = 7.65 m
2 2
1 1 2
2 2f
P v P vhg g
α αγ γ
+ − = + 2
with α = 2.
v1 = v2 , P2 = 0 ⇒ P1 = γ h f = ρ gh f = 1200 × 9.81 × 7.65
= 90,000 Pa
= 90 k Pa, gage
Dr. A. Pourmovahed, Kettering University 37 Introduction to Energy Systems
Example Determine the required reservoir depth to maintain water flow rate of 0.03 m3/s through the pipe.
ρ = 998
kg/m3 γ = 9790 N/m3 ν = 1.00 × 10-6 m2/s
( )v Q
Ams= = =003
075679
2
.
..
π4
Re = vDν = ×
× −679 0075100 10 6. ..
= 5.09 × 105 >> 2200 ⇒ Turbulent Flow
f = 0.0131 (Moody Diagram)
Ke = 0.50 Square-edged inlet
P vg z f L
Dvg K
vg
P vg ze
1 12
1
2 22 2
2
22 2 2 2γ γ+ + − − = + +
z1 - z2 = d
d = vg
2
2 (1 + f LD + Ke)
d = 67981 1 00131 100
0075 052.
. . . .× +
2 9 = 44.6 m +
∇
d
(1)
z
20°CSmooth Pipe D = 75 mm
L = 100 m
(2)
17.47
WaterTables
Dr. A. Pourmovahed, Kettering University 38 Introduction to Energy Systems
Example (Pipes in Parallel) What size pump will be necessary in the 3-inch pipe so that 5 ft3/s of water will flow through each pipe? Neglect losses in bends and elbows. Assume that water is at 70°F and all pipes are smooth. 6” Pipe: v Q
Afts= =
=5
46
12
25462π.
Re = ρµvD 10 c (1) 10(2) fs
cfs
5 cfs
5 cfsPump
D = 3”, L = 100’
D = 6″, L = 100′
Re = 623 2546 6
12202 10 321745
. .
. .
× ×
× ×−
Re = 1.2 × 106 (Turbulent Flow)
f = 0.011 (Moody chart, smooth pipe)
P z f LD
vg
P z11
22
22γ γ+ − = +
P1 - P2 = f LD
vg
2
2 ⋅ γ
P1 - P2 = 0.011 ( )( )100 25462 3226
12
2..× 62.3
= 1381 lb = 9.59 Psi ft
f2
3” Pipe: v = 4 (25.46) = 101.84 ft/s
Re = 2 (1.2 × 106) = 2.4 × 106
f = 0.0099
P z f LD
vg h P zP
11
22
22γ γ+ − + = +
Dr. A. Pourmovahed, Kettering University 39 Introduction to Energy Systems
( )h f LD
vg P PP = + −2
2 121γ
( )( )hP = × +00099 100 101842 322
16233
12
2
..
. . (-1381)
= 616.1 ft
. .W Qh lbft
fts ftP P
f= = × ×γ 62 3 5 61613
3
= 191914 fts = 349 HP lbf.
(Minimum Power Required)
Dimensional Analysis Dimensional analysis is a method for reducing the number and complexity of experimental variables which affect a given physical phenomenon (a compacting technique). Consider the drag on a sphere:
D,
V
ρ µ→
The drag force, F, is a result of pressure forces and viscous shear. It is very difficult to find F analytically (due to turbulence, boundary-layer separation, etc.). Thus, F must be determined experimentally.
( ), , ,F f D V ρ µ=
Generally, it takes about 10 experimental points to define a curve. To find the effect of D, we need to run 10 experiments. For each D, we need 10 values of V, 10 values of ρ , and 10 values of µ . This means 10,000 experiments are needed to get complete data. If we plot F vs. V with D as a parameter, we will have 100 sheets of curves. Dimensional analysis will significantly reduce the time and the cost of the experimental investigation. It also allows us to present the data in a compact form.
Dr. A. Pourmovahed, Kettering University 40 Introduction to Energy Systems
The Principle of Dimensional Homogeneity (PDH)
If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous; i.e., each of its additive terms will have the same dimensions.
/x yA B e C= +
A, B, and C must have the same dimensions; x and y also must have the same dimensions. Dimensional Variables are quantities that actually vary during a given case and would be plotted against each other to show the data. They all have dimensions, and all can be nondimensionalized as a dimensional analysis technique ( , , , ,D V Fρ µ in our example). Dimensionless Variables ( ∏ ’s) are dimensionless and are constructed by combining dimensional variables. The Buckingham PI Theorem (1914) If a physical process satisfies the PDH and involves n dimensional variables, it can be reduced to a relation between only k dimensionless variables or ∏ ’s. The reduction
equals the maximum number of variables which do not form a ∏ among themselves and is always less than or equal to the number of dimensions describing the variables.
j n k= −
Procedure to find ∏ ’s 1. List and count the n variables involved in the problem. If any important variables
are missing, dimensional analysis will fail. 2. List the dimensions of each variable. Use or MLT FLTθ θ . 3. Find j. Initially guess j equal to the number of different dimensions present.
Look for j variables that do not form a ∏ product. If no luck, reduce j by 1 and look again.
4. Select j variables that do not form a ∏ product. These variables will appear in
every group when you are done. ∏
Dr. A. Pourmovahed, Kettering University 41 Introduction to Energy Systems
5. Add one additional variable to your j variables and form a power product. Algebraically find the exponents that make the product dimensionless.
Example 1: Drag on a sphere
( ), , ,F f D V ρ µ=
Step 1: 5n = , , , ,F D V ρ µ Step 2: 2:F MLT −
:D L V L 1: T −
3: MLρ − 1 1: ML Tµ − − Step 3: ( )3 , ,j M L= T
( ), , Cannot form V Dρ ∏ Step 4: Select , ,V Dρ Step 5: , , ,V Dρ µ 5 3 2n j− = − = , , ,V D Fρ 2 ∏ groups
3 0a b c d a a b b c d d dV D M L L T L M L T M L Tρ µ − − − −= = 0 0
0
111
: 0M a d+ = : 3L a b c d− + + − = : 0T b d− − =
Select any one: 1d = − a = b = c = 1 /V Dρ µ∏ = Reynolds Number
3 2e f g h e e f f g h h hV D F M L L T L M L T M L Tρ − − −= = 0 0 0
0
: 0M e h+ = : 3L e f g h− + + + =
Dr. A. Pourmovahed, Kettering University 42 Introduction to Energy Systems
: 2T f h− − = 0
122
Select: 1h = e = − f = − g = −
2 2 2
FV Dρ
∏ = Force Coefficient
There are only two ∏ groups.
( )1 2f∏ = ∏
( )2 2
F f R eV Dρ
=
This will result in a single plot which requires about 10 experiments. Example 2: Laminar Flow in a Circular Pipe For laminar flow through a circular pipe, the volumetric flow rate Q is a function only of the tube radius R, the fluid viscosity µ , and the pressure drop per unit pipe length, dp/dx. Using the theorem, find an appropriate dimensionless ∏relationship.
1. , , , 4d pQ f R nd x
µ
= =
2. R L
1 1M L Tµ − − 2 2d p M L T
d x− −
3 1Q L T − 3. Guess 3j =
, , and cannot form a groupd pRd x
µ ∏
Dr. A. Pourmovahed, Kettering University 43 Introduction to Energy Systems
2 2 0 0c
a b a b b b c c cd p 0R L M L T M L T M L Td x
µ − − − − = =
: 2L a b c− − = 0
0 0 no :M b c+ = c = ∏ group : 2T b c− − = 0 0
0b =
a = 3j = 4. only one ∏ group 1n j− =
1 0d pQ R M L Td x
γα βµ
=
0 0
0
0
3 1 2 2 0 0L T L M L T M L T M L Tα β β β γ γ γ− − − − − =
: 3 2L α β γ+ − − = 1γ = − : 0M β γ+ = 1β = + : 1 2T 0β γ− − − = 4α = −
14
const.Qd pRd x
µ∏ = =
Laminar-flow theory shows that 1 8π
∏ =
Dr. A. Pourmovahed, Kettering University 44 Introduction to Energy Systems
Aerodynamic Drag
v0
separation
wake
Laminar B.L.
Stag.Pt.
v0
separation
Turbulent B.L.Re > 5 × 105
wake
For streamlined objects (such as airfoils), skin friction is a major contributor to drag. For blunt objects (cylinders, spheres, etc…) pressure drag is the primary component. The Drag Coefficient
CD ≡ Fv A
D
P12 0
2ρ
Ap = Projected Frontal Area
FD = Drag Force
Dr. A. Pourmovahed, Kettering University 45 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 46 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 47 Introduction to Energy Systems
Example
Estimate the added power (kw) required for the car when the cartop carrier is used and the car is driven at 80 km/h in a 20-km/h head wind over that required when the carrier is not used under the same conditions.
V0 = 80 + 20 = 100 km/h (Velocity of Air relative to the car) Assume Rectangular Plate
2.05.1
b=
l = 7.5 ⇒ CD ≅ 1.25 (Table)
FD = 12 D
20Cvρ (b ⋅ l )
FD = 12 × 1 ⋅ 2 100 10
36003 2
×
× 1.25 × (0.2 × 1.5) = 173.6 N
Power = FD ⋅ V = 173.6 × 80 103600
11000
3× × = 3.86 kW
Note: at 1atm, 20°C, ρair = 1.2 kg/m3
Example
A 3000-lb automobile with a projected frontal area of 20.0 ft2, a drag coefficient of 0.40, and a rolling resistance coefficient of 0.10 is traveling at 65 mph. Determine the % savings in gas mileage if the speed is reduced to 55 mph. The air temperature is 60°F. Energy = Force x Distance Energy required per distance of travel = force, F F = Rolling Resistance + Aerodynamics Drag
3000 × 0.10 12
2vρ ⋅ CD AP
Dr. A. Pourmovahed, Kettering University 48 Introduction to Energy Systems
F = 300 + 12 × 0.0763 V2 × 0.40 × 20 / 32.174
ρ F = 300 + 0.00949 V2 (V in ft/s, F in lbf)
V = 65 mph = 95.33 ft/s ⇒ F = 386.24 lbf
V = 55 mph = 80.67 ft/s ⇒ F = 361.76 lbf
Energy savings = 386 24 36176386 24
0 063. ..
.−=
( 6.3 % Savings )
Example (Drag and Buoyancy)
A sphere 2cm in diameter rises in oil at a constant velocity of 3 cm/s. What is the
specific weight of the sphere if the oil density is 900 kg/m3
and its viscosity is 0.096 Pa. s?
Sphere ρs
W
D
fluidρ
B
D = Drag Force
B = Buoyant Force
W = Weight of sphere
D = 2D
20 d
4Cv π
ρ21
B = Weight of Displaced Fluid = 4 3 π d g2
3
ρ
W = 43 2
3
π d gs
ρ
At constant velocity: W = B - D
43 2
3
π d gs
ρ = 2
D20
3
d4
Cv21g
2d
34 π
ρ−ρ
π
ρ ρ ρsDg g V C
d= − 34
02
Re = ρ =V 900 0.03 0.020 d CDµ× × = ⇒ ≅0096 563 7 0. . .
( )γ s sg
Nm
≡ = × −×
=ρ 900 981 34
900 003 7 0002 8616
2
3.. .
.
Dr. A. Pourmovahed, Kettering University 49 Introduction to Energy Systems
Example
A 60 mph (88 ft/s) wind blows past the water tower shown. Estimate the bending-
moment at the tower base. T = 60°F
V0 = 88 fts
Ds = 40'
Dc = 15'
b = 50 ft
b2
Ds
Fc
Dc
b b + Ds2
Fs
R V De
ss
= = ××
= ×−0
4788 40
1 58 102 23 10
ν ..
R V De
cc
= = ××
= ×−0
4688 15
1 58 108 35 10
ν ..
CDs≅ 03.
CDc≅ 07.
PD20 ACV
21F ρ= ρ = 0 slug00237 3.
ft
Fs = ( ) ( )12 000237 88 03 4 40 34603
2 2
22
4
2
2
× × ×. .slugft
fts
ft lb
D
f
s
π
π
=
Fc = ( )12 000237 88 07 50 15 48182× × × × × =
⋅
. ( ) .b D
f
c
lb
M = Fs b D F bsc+
+2 2
M = 3460 50 402 4818 50
2+
+ = 362,650 ft.lbf
Note: slug = 32.2 lbm
Dr. A. Pourmovahed, Kettering University 50 Introduction to Energy Systems
HEAT TRANSFER
Energy transfer due to a temperature difference.
Three heat transfer mechanisms:
Conduction: Transfer of energy due to interactions between particles.
Convection: Energy transfer by the bulk motion of the fluid.
Radiation: Energy emitted by matter through electromagnetic waves.
Fourier’s Law of Conduction:
x
T2
T1
gradient eTemperatur
Area sectional-Cross tyConductivi Thermal
RateTransfer Heat
=
===
−=
dxdTAkq
dxdTkAq
x
x
Dr. A. Pourmovahed, Kettering University 51 Introduction to Energy Systems
Newton’s Law of Cooling (Convection)
Ts
T∞
Velocity distribution
q’’
U∞
Temperature distribution
q’’ = h(Ts - T∞)
Heated Surface
q’’ ≡ Heat Flux (Heat transfer rate per unit area)
h ≡ Convection Heat Transfer Coefficient
Ts ≡ Surface Temperature
T∞ ≡ Fluid Temperature far from the surface
Stefan-Boltzmann Law (Radiation):
q’’ = ε σ Ts4
Ts = Absolute Surface Temperature
ε = Surface Emissivity
q’’ = Heat Flux Surface at Ts
σ = Stefan-Boltzmann Constant
4281067.5
KmW⋅
×= −
Dr. A. Pourmovahed, Kettering University 52 Introduction to Energy Systems
Thermal Resistance (steady state): 0 x x = L
Plane Wall Steady-State
L
T1
qx
( )
kALR
RVV
i
kAL
TTq
TTLkAq
LTT
LTT
dxdT
dxdTq
dxdTkAq
cond
x
x
x
x
=∴
−=
−=
−=
−−=
−=
=⇒=
−=
:is wall)(plane Resistance Conduction
Law) s(Ohm' with Compare
constant constant
21
21
21
2112
T2
=
kALR
T1 qx T2
Dr. A. Pourmovahed, Kettering University 53 Introduction to Energy Systems
x
Fluid flow Convection
T∞ ( )
hAR
hA
TTq
TThAq
conv
sx
sx
1 :is Convectionfor Resistance
1
=
−=
−=
∞
∞
Ts
The Composite Wall
A
kA
LA
B
kB
LB
C
kC
LC
T∞,1 T1 Series T3 T∞,4 T2
h1
T4 h4
Hot Fluid
Cold Fluid
Ak
L
A
A T1 T2 T3 T4
Ak
L
B
B
Ak
L
C
C
Ah1
1
Ah4
1
qx
T∞,1 T∞,4
∑=
=n
iitotal RR
1
Total Thermal Resistance:
Dr. A. Pourmovahed, Kettering University 54 Introduction to Energy Systems
ARU
RTT
q
AhAkL
AkL
AkL
AhR
total
totalx
C
C
B
B
A
Atotal
1:as tCoefficienTransfer Heat Overall theDefine
11
4,1,
41
≡
−=
++++=
∞∞
Then
qx = U A ∆T
where ∆T = T∞,1 - T∞,4.
41
111
hkL
kL
kL
h
U
C
C
B
B
A
A ++++=
Dr. A. Pourmovahed, Kettering University 55 Introduction to Energy Systems
Radial Systems
r3r 2
r1 h3
T∞,3
h1, T∞,1
BA
r1 = internal radius of pipe
r2 – r1 = thickness of pipe wall (A)
r3 – r2 = thickness of insulation layer (B)
T∞,1 = fluid temperature inside the pipe
T∞,3 = fluid temperature outside the pipe
( )
Lrr
B
23
k2/ln
π( )Lrh 11 21π
( )Lrh 33 21π
( )Lrr
A
12
k2/ln
πqr
T∞,3 T∞,1
L = Pipe Length
totalr R
TTq 3,1, ∞∞ −
=
(Radial Heat Transfer Rate)
Dr. A. Pourmovahed, Kettering University 56 Introduction to Energy Systems
Example
Determine the heat loss per foot from a 3 inch steel schedule 40 pipe (3.07
inch I.D., 3.50 inch O.D., ks = 25 Btu/hr-ft-°F) covered with a ½ inch
thickness of insulation (ki = 0.11 Btu/hr-ft-°F). The pipe transports a fluid at
300°F with h1 = 40 Btu/hr-ft2-°F and is exposed to ambient air at 80°F with
h3 = 4.0 Btu/hr-ft2-°F.
( )
1
2
123 2
1 2
3 2
1
1 1
2 12
1.535"1.750"
" 2.250"
25
0.11
40
4.0
1 1internal flow 0.03111.5352 /40 2 1
12
ln( / ) ln(1.75steel2
s
i
s
rrr r
Btukhr ft F
Btukhr ft F
Btuhhr ft F
Btuhhr ft F
FRh rL Btu hr
r rRk L
π π
π
=== + =
=⋅ ⋅
=⋅ ⋅
=⋅ ⋅
=⋅ ⋅
= = =
= =
( )
3 23
4
3 3
0 /1.535) 0.000832 25 1 /
ln( / ) ln(2.250 /1.750)insulation 0.36362 2 0.11 1 /
1 1external to pipe 0.21222.2502 /4.0 2 1
12
i
FBtu hr
r r FRk L Btu hr
FRh r L Btu h
π
π π
π π
=× ×
= = = × ×
= = =
r3r 2
r1 h3
T∞,3=80° F
h1
T∞, 1= 300° F
steel insulation
r
Dr. A. Pourmovahed, Kettering University 57 Introduction to Energy Systems
hrBtu
RTT
q
hrBtuFRRRRR
totalr
total
0.3626077.0
80300/
6077.0
3,1,
4321
=−
=−
=
=+++=
∞∞
Without Insulation
R1 and R2 will not change, R3 = 0
( )
hrBtuq
hrBtuFRRRRR
hrBtuF
LrhR
r
total
0.7223047.0
80300/
3047.0
/2728.0
112750.120.4
121
4321
234
=−
=
=+++=
=
==ππ
Removing the insulation nearly doubles the heat loss.
Dr. A. Pourmovahed, Kettering University 58 Introduction to Energy Systems
Convection Heat Transfer
- Forced Convection: Flow caused by external means, e.g. fans, pumps,
wind.
- Free (Natural) Convection: Flow induced by buoyant forces.
( )2
3
L
L
Gr Number, Grashof
Nu Number,Nusselt
Pr Number, Prandtl
Re Number, Reynolds
νβ
ανµ
ν
LTTgk
hLkc
VL
s
P
∞−≡
≡
=≡
≡
Dimensionless Numbers:
L = length; V = fluid velocity
ν = kinematic viscosity = µ/ρ
cp = specific heat; k = thermal conductivity
α = thermal diffusivity ≡ k/(ρ cp)
β = coefficient of thermal expansion (β = 1/T for an ideal gas)
h = heat transfer coefficient
Dr. A. Pourmovahed, Kettering University 59 Introduction to Energy Systems
Convective Heat Flux:
U∞
For a flat plate:
( ) 31
54
Pr871Re037.0 −= LLNu
All fluid properties are evaluated at Tf.
q’’ = h(Ts - T∞)
e)Temperatur (Film 2
10Re ,60Pr6.0
values)average are & (
Re
8
∞
∞
+≡
≤<<
≡
≡
TTT
hNukLhNu
LU
sf
L
LL
L ν
Fluid Solidq’’
Ts
L T∞
Circular Cylinder in Cross Flow:
41
PrPrPrRe
=
s
nmDD CNu
U∞ T∞
Ts
6
Re
( & are average values)
0.7 Pr 500, 1 Re 10
D
D D
D
U D
hDNu Nu hk
ν∞≡
≡
< < < <
All properties at T∞ except for Prs (@ Ts).
Dr. A. Pourmovahed, Kettering University 60 Introduction to Energy Systems
Constants for Cross-flow over a Cylinder
ReD C m
1 – 40 0.75 0.4
40 – 1000 0.51 0.5
1000 – 2x105 0.26 0.6
2x105 - 106 0.076 0.7
If Pr ≤ 10, n = 0.37.
If Pr > 10, n = 0.36.
Flow over a Sphere:
( ) 41
4.032
21
PrRe06.0Re4.02
++= ∞
sDDDNu
µµ
).T (@ for except Tat properties All
2.30.1
106.7Re5.3380Pr71.0
s
4
s
s
D
µµµ
∞
∞ <<
×<<
<<Ts D
T∞
U∞
Dr. A. Pourmovahed, Kettering University 61 Introduction to Energy Systems
Internal Pipe Flow
Laminar Flow:
- Constant wall temperature ⇒ NuD = 3.66
- Constant wall heat flux ⇒ NuD = 4.36
Both for fully-developed flow, Pr > 0.6.
Turbulent Flow:
31
54
PrRe023.0 DDNu =
Fully-developed ⇒
160Pr7.0 ,10Re ,10 4 ≤≤>>
DDL
For Liquid Metals:
( ) 827.0PrRe0185.082.4 DDNu +=
425 10PrRe10 ,1005.9Re3600 <⋅<×<< DD
( ) 8.0PrRe025.00.5 DDNu +=
(T
(T
urbulent, constant heat flux)
urbulent, constant wall temperature)
100PrRe >⋅D
Dr. A. Pourmovahed, Kettering University 62 Introduction to Energy Systems
Noncircular Tubes
Hydraulic Diameter: PAD C
h
4≡
AC = Cross-sectional Area
P = Wetted Perimeter
Re , h hVD hDNuk
ρµ
≡ =
For turbulent flows:
Use circular tube correlations.
For laminar flows, use Table below:
Dr. A. Pourmovahed, Kettering University 63 Introduction to Energy Systems
Concentric Tube Annulus
0' DDDh −=
hi
D’
Di
hoAdiabatic Wall
Do
kDhNu h0
0 ≡
If the flow is laminar, use table below: D0/D’ Nu0
0 ----
0.05 17.46
0.10 11.56
0.25 7.37
0.50 5.74
1.0 4.86
Dr. A. Pourmovahed, Kettering University 64 Introduction to Energy Systems
Heat Exchangers
Devices used to exchange heat between two fluids that are at different temperatures and separated by a solid wall.
Dr. A. Pourmovahed, Kettering University 65 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 66 Introduction to Energy Systems
The Overall Heat Transfer Coefficient, U
ooo
ofi
o
i
if
iiooii AhAR
kLDD
AR
AhAUAUUA1
2
ln1111 ''
,'', ++
++===π
Rf’’ = fouling factor (See Table below)
Dr. A. Pourmovahed, Kettering University 67 Introduction to Energy Systems
Heat Exchanger Analysis
mass flow rate constant pressure specific heat
hot fluid cold fluid inlet outlet
Btu heat transfer rate from hot fluid to cold fluid (W or )s overall heat transfer coefficien
P
P
mc
C m chcio
q
U
•
•
==
= ⋅====
=
=
( ) ( )
lm
, ,,, , ,
t
heat transfer area heat exchanger length
T or log-mean temperature difference
h c c o c iP cP h h i h o
AL
LMTD
q m c T T m c T TCC
ch
• •
==
∆ =
= − = −
Dr. A. Pourmovahed, Kettering University 68 Introduction to Energy Systems
Parallel-Flow, Concentric Tube Th,o
Th,i
Th,o
Tc,i
Tc,o
∆T2 ∆T1
Tc,i Tc,o Th,i
,1 ,
,2 ,
2 1
2
1ln
o o i i
c ih i
c oh o
lm
lm
o o
i i
U A U A
T T TT T T
T TT TT
q UA TA D LA D L
ππ=
∆ = −∆ = −
∆ − ∆∆ ≡ ∆∆
= ⋅∆==
Dr. A. Pourmovahed, Kettering University 69 Introduction to Energy Systems
Counter-Flow, Concentric Tube
Th,i
Th,o
Tc,o Tc,i
Th,i
Th,o
Tc,o
Tc,i
∆T2
∆T1
,1 ,
,2 ,
2 1
2
1ln
c oh i
c ih o
lm
lm
T T TT T T
T TT TT
q UA T
∆ = −∆ = −
∆ − ∆∆ ≡ ∆∆
= ⋅∆
Dr. A. Pourmovahed, Kettering University 70 Introduction to Energy Systems
Multipass and Cross-Flow Heat Exchangers
q = U A F ∆Tlm,CF
∆Tlm,CF = Counter-Flow LMTD
F = Correction Factor (See Figures 11.10-11.13)
∆T1 = Th,i - Tc,o
∆T2 = Th,o - Tc,i
In the following figures, t is always assigned to the tube-side fluid. It
does not matter if the hot fluid or the cold fluid flows through the shell
or the tubes.
Dr. A. Pourmovahed, Kettering University 71 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 72 Introduction to Energy Systems
Dr. A. Pourmovahed, Kettering University 73 Introduction to Energy Systems
Example 1 (Calculation of U)
Hot water at 90° C flows on the inside of a 2.5 cm I.D. steel tube
(wall thickness = 0.8 mm, k = 15 W/m.K) at a velocity of 4 m/s. This
tube forms the inside of a double-pipe heat exchanger. The outer pipe
has a 3.75 cm I.D. Engine oil at 20° C flows in the annular space at 7
m/s. Determine the overall heat transfer coefficient based on the
outside diameter of the inner tube.
water
oil
4 m/s
7 m/s
90 C
20 C
Dr. A. Pourmovahed, Kettering University 74 Introduction to Energy Systems
Properties of Water at 90° C:
23001005.31016.3
025.04965Re
96.1Pr
676.0
1016.3
965
5
4
3
4
3
>×=
⋅×
××==
=⋅
=
⋅×=
=
−
−
smkg
msm
mkg
DV
CmWk
smkg
mkg
µρ
µ
ρ
Turbulent Pipe Flow:
( ) ( )( ) ( )
KmWh
DkNu
Dkh
Nu
i
i
⋅=
×===
=
2
33.054533.05
4
33.054
1850
96.11005.3023.0025.0676.0PrRe023.0
PrRe023.0
Dr. A. Pourmovahed, Kettering University 75 Introduction to Energy Systems
Properties of Engine Oil at 20° C:
( )
2
2
0.0009
0.145
Pr 10400' 0.0375 0.0266 0.0109
7 0.0109Re 85 Laminar Flow0.00090.02665 (from Table) 0.71' 0.0375
0.1455 670.0109
oh
h
oo
ooh
ms
Wk m C
D D D mVD
DNu Dk Wh Nu D m
ν
ν
=
=⋅
== − = − =
×= = =
≅ =
= = =
( )
1
1
,
ln1 1 1 No Fouling2
ln1 1 1
2
ln1 1 1
2
0.0266ln1 1 10.02500.0266 1850 0.025 2 15 67 0.0266
o o i i
o
i
o o o oi i
o
i
o o o oi i
o
io
o o oi i
o
CA D L A D L
DD
U A h A kL h ADD
U D h D k h D
DDU D h D k h D
U
π π
π
−
−
⋅= =
= + +
= + +
= + +
= + +× × ×
264oWU m C=⋅
=
Dr. A. Pourmovahed, Kettering University 76 Introduction to Energy Systems
Example 2
A transmission oil cooler is used to cool hot transmission oil from 160
to 60 °C using engine coolant at 25 °C. The oil and coolant mass flow
rates are both 2 kg/s. The outside diameter of the inner tube is 50 mm.
The oil flows in the inner tube and Uo = 2500 W/m2.K. How long
must this counter-flow heat exchanger be?
( )( )
( )( )
,
,
, , ,
, ,,
,
160 60Average oil temperature 110 38322260
Assuming an average temperature of 40 for water:
4179
2 2260 160 60 452,000
452,000 2 4179 25
P h
P c
h P h h i h o
c c o c iP c
c o
C K
Jc kg KC
Jc kg Kq m c T T
q W
q m c T T
T
•
•
+= = =
=⋅
=⋅
= −
= × − =
= −
= × −
( ) ( ) ( )
,
,1 ,
,2 ,
79160 79 8160 25 35
81 35 54.881ln 35452,000 21 must coil up
0.050 2500 54.8
c o
c oh i
c ih o
lm
o o lm
T CT T T CT T T C
T C
q U D L T L mππ
=∆ = − = − =∆ = − = − =
−∆ = =
= ∆ ⇒ = =×
h
c
∆T2
∆T1
Dr. A. Pourmovahed, Kettering University 77 Introduction to Energy Systems
Example 3
Water at the rate of 30,000 lbm/hr is heated from 100 to 130°F in a shell-
and-tube heat exchanger. On the shell side, one pass is used with water as
the heating fluid, 15,000 lbm/hr entering at 200°F. The overall heat transfer
coefficient is 250 Btu/(hr-ft2-°F) and the average water velocity in 3/4 inch
I.D. tubes is 1.2 ft/s. The heat exchanger length must not exceed 8 feet.
Determine the number of tube passes, the number tubes per pass and the
length of the tubes.
U=250 Btu/(hr-ft2-°F)
30,000 lbm/hr 100°F
15,000 lbm/hr 200°F
130°F
Dr. A. Pourmovahed, Kettering University 78 Introduction to Energy Systems
( )( )
( )( )
( ) ( )( )
( )acceptablenot 5.936342342
342
6.5312
4/3250000,900
tubesofnumber n
6.53
4070ln
4070
ln
40100140
70130200
& flow,-counterFor
3636002.1
124/34.62
000,3044
4
1402001000,15000,900
000,9001001301000,30
2
1
21
2
1
,,2,,1
22
2
,,
,,,
,,,
ftLLn
ftLn
Ln
TDLnUq
F
TT
TTT
FT
FT
TTTTTT
VDmn
nVDnVAm
FTT
TTcmq
hrBtuq
TTcmq
lm
lm
icohocih
c
cc
ohoh
ohihhPh
icoccPc
==⇒=⋅
=⋅
⋅×=
=∆=
=−
=
∆∆
∆−∆=∆
=−=∆
=−=∆
−=∆−=∆
=
×
×
×==
==
=⇒−×=
−=
=−×=
−=
•
•
•
•
π
π
πρπ
πρρ
Use 2 tube passes (see Figure 11.10).
( ) ( )
( ) ( )ftL
TnFUDqL
TFDLnUq
F
ttTTR
tTttP
lm
lm
io
oi
ii
io
40.56.533688.0250
124/32
000,9002
passes 2 2
88.02
100130140200
30.0100200100130
=
×
=
∆⋅⋅⋅=
×∆⋅=
≅⇒
=−−
=−−
=
=−−
=−−
=
π
π
π
Dr. A. Pourmovahed, Kettering University 79 Introduction to Energy Systems
Radiation
Energy emitted by matter through electromagnetic waves. Radiation propagates at the speed of light: 83 10 /c m= × s The wavelength λ is related to the frequency υ by:
cλυ
=
λ is usually expressed in micrometers:
61 10m mµ −≡
Thermal radiation lies in the range of 0.1 to 100 mµ . The visible portion is very narrow (0.4 to 0.7 mµ ).
Spectrum of electromagnetic radiation.
Dr. A. Pourmovahed, Kettering University 80 Introduction to Energy Systems
Blackbody: An ideal surface. • A blackbody absorbs all incident radiation. • For a prescribed T and λ , no surface can emit more energy than a blackbody. • A blackbody is a diffuse emitter. Radiation emitted by a blackbody is independent of
direction. The Stefan Boltzman Law:
4bE Tσ=
bE = Blackbody emissive power (energy emitted per unit time and per unit area, 2
Wm
).
T = Absolute temperature. 8
2 45.669 10.
Wm K
σ −= ×
Radiation Properties: Reflectivity, ρ = fraction reflected Absorptivity, α = fraction absorbed Transmissivity, τ = fraction transmitted
1ρ α τ+ + = Most solid bodies do not transmit thermal
Absorption
ReflectionIncidentRadiation
Transmission
radiation:
0 1τ ρ α= ⇒ + =
Emissivity: Ratio of the emissive power of a body to that of a blackbody at the same temperature.
b
EE
∈ ≡
Dr. A. Pourmovahed, Kettering University 81 Introduction to Energy Systems
Kirchhoff’s Law:
α∈ =
In general, , , , and ρ α τ ∈ are all functions of the wavelength, λ . Monochromatic Emissive Power: Eλ The emissive power per unit wavelength. Monochromatic Emissivity:
λ∈
b
EE
λλ
λ
∈ ≡
bE λ is the monochromatic emissive power of a blackbody at the same λ and T.
For a graybody, λ∈ is independent of λ .
λ∈ = ∈
Spectral Distribution of Blackbody Emission (Plank Distribution):
( )5
12
2 .exp / 1bc WE
m mc Tλλ
µλ
− = −
4
81 2
.3.7420 10 W mcmµ
= ×
4
2 1.4388 10 .c mµ= × K
Dr. A. Pourmovahed, Kettering University 82 Introduction to Energy Systems
Spectral blackbody emissive power.
Wiens Displacement Law:
bE λ
λo
4Area = Tσ
T = const.
( )max . 2897.6 .T mλ µ= K
maxλ is the wavelength at which bE λ is a
maximum. The emissive power: 4
b boE E dλ λ σ
∞= =∫ T
Dr. A. Pourmovahed, Kettering University 83 Introduction to Energy Systems
The View Factor: The view factor is the fraction of the radiation leaving surface i, which is intercepted by surface j:
i jF
2dA
2A
2θR
1dA1A
1n1θ
2n
Reciprocity Relation: For N surfaces forming an enclosure:
i i j j jA F A F= i
11
i j
N
Fj
==Σ
1 12 2 21A F A F=
Dr. A. Pourmovahed, Kettering University 84 Introduction to Energy Systems
View factors for two-dimensional geometries.
Dr. A. Pourmovahed, Kettering University 85 Introduction to Energy Systems
View factors for two-dimensional geometries.
Dr. A. Pourmovahed, Kettering University 86 Introduction to Energy Systems
View factors for three-dimensional geometries.
Dr. A. Pourmovahed, Kettering University 87 Introduction to Energy Systems
View factor for aligned parallel rectangles.
View factor for coaxial parallel disks.
Dr. A. Pourmovahed, Kettering University 88 Introduction to Energy Systems
View factor for perpendicular rectangles with a common edge.
Areas used to illustrate view factor relations.
Dr. A. Pourmovahed, Kettering University 89 Introduction to Energy Systems
Blackbody Radiation Exchange:
i j i bi iq A E→ = jF
iF
j i j b j jq A E→ = Net Radiation Exchange:
i j i j j iq q q→ →= −
4bE Tσ=
j
j
AT
,i iA T
( )4 4i j i i j i jq A F T Tσ= −
If is a part of N blackbodies forming an enclosure, the iA net radiative energy leaving is:
iA
iJ
iAiG
jG
jJ
jA
( )4 4
1i i i j i
N
q A F Tj
σ= −=Σ jT
Radiation Between Nonblackbodies: All surfaces are assumed to be gray.
λ∈ = ∈ G = irradiation total radiation incident upon a surface per unit time and per unit area.
≡
J = radiosity total radiation that leaves a surface per unit time and per unit area.
≡
Assume that the surfaces do not transmit radiation:
0; 1;τ ρ α α= + = ∈ ≡
Dr. A. Pourmovahed, Kettering University 90 Introduction to Energy Systems
1ρ = − ∈
Assumptions:
• Each surface is isothermal, opaque, diffuse, and gray. • Radiosity and irradiation are uniform for each surface. • The medium within the enclosure is nonparticipating.
Radiation exchange in an enclosure of diffuse-gray surfaces with
a nonparticipating medium.
iq = net rate at which radiation leaves surface i. = The rate at which energy would have to be transferred to surface i by other means
to maintain it at a constant temperature.
( )i i iq A J G= − i
Dr. A. Pourmovahed, Kettering University 91 Introduction to Energy Systems
1bi i
ii
i i
E Jq
A
−=
− ∈ ∈
iq = net radiation heat transfer from the surface.
the driving potentialbi iE J− =
1
surface radiative resistancei
i iA− ∈
=∈
( ) 111
bi i i j
ii i j
i i
NE J J J
A FjA
−
−
− −=
− ∈=
∈
Σ
( ) 1
1
i ji
i i j
NJ J
qA Fj
−
−=
=Σ
Dr. A. Pourmovahed, Kettering University 92 Introduction to Energy Systems
Network representation of radiative exchange between surface i
and the remaining surfaces of an enclosure.
RADIATION SHIELDS
Low emissivity (high reflectivity) materials used to reduce the net radiation transfer between two surfaces. Assumptions • Each surface is isothermal, opaque, diffuse, and gray. • Radiosity and irradiation are uniform for each surface. • The medium within the enclosure is nonparticipating. Since the surfaces do not transmit radiation,
0; 1;τ ρ α α= + = ∈ ≡ ; 1ρ = − ∈
Dr. A. Pourmovahed, Kettering University 93 Introduction to Energy Systems
q 1 1 1, ,A T ε 2 2 2, ,A T ε
12
1 2
J2 Eb2
1 12
1A F
2
2 2
1A
−∈∈
1
1 1
1A
−∈∈
J1Eb1
q1
Radiation Exchange Between Large Parallel Planes with No Shield
Dr. A. Pourmovahed, Kettering University 94 Introduction to Energy Systems
1 1 1, ,A T ε 2 2 2, ,A T ε3 3,A T
3,1ε 3,2ε
Radiation Shield
q13 q32
1 2
3,1 3,2
1
1 1
1A
−∈∈
J1Eb1
1 13
1A F
J3,1
3,1
3,1 3
1A
−∈
∈
Eb3 J3,2 J2 Eb2
3,2
3,2 3
1A
−∈
∈ 3 32
1A F
2
2 2
1A
−∈∈
q1
Radiation Exchange Between Large Parallel Planes with One Shield
The Heat Transfer Rate With no shield:
Dr. A. Pourmovahed, Kettering University 95 Introduction to Energy Systems
1 2
12 01 2
1 1 1 12 2 2
( )1 11
b bE E
q
A A F A
ε ε
ε ε
−=
− −+ +
With 1 2
A A A== and 12
1F = , we get:
1 2
0
4 4
12
1 2
( )( )
1 1 1
A T Tq
σ
ε ε
−=
+ −
With one shield:
1 2
12 13 ,1 3 ,21 2
1 1 1 13 3 ,1 3 3 ,2 3 3 32 2 2
( )1 11 11 1
b bE E
q
A A F A A A F
ε ε
A
ε ε
ε ε ε ε
−=
− −− −+ + + + +
With 1 2
A A A== and 13 32
1F F= = , we get:
Dr. A. Pourmovahed, Kettering University 96 Introduction to Energy Systems
1 2
4 4
12 13 ,1 3 ,2
1 2 3 ,1 3 ,2
( )( )
1 11 1
A T Tq
σε ε
ε ε ε ε
−=
− −+ + +
Note that for a radiation shield to be effective, 3 ,1
ε and 3 ,2
ε should be small (high
reflectivity). For N radiation shields, when all the emissivities are equal,
12 12 0
1( ) ( )1
Nq q
N=
+
Where is the radiation heat transfer rate with no shields. 12 0
( )q
Dr. A. Pourmovahed, Kettering University 97 Introduction to Energy Systems
Example Two very large parallel plates with emissivities 0.3 and 0.8 exchange heat by radiation. Find the percentage reduction in heat transfer when a polished aluminum shield with an emissivity of 0.04 is placed between them. No Shield:
1 2
0 1
4 44 4
12
( )( ) 0.2791 ( )
1 1 10.3 0.8
A T Tq A
σσ
−= =
+ −2
T T−
With one Shield:
1 2
1 1 2
4 44 4
12
( )( ) 0.0190 ( )
1 1 1 0.04 1 0.04
0.3 0.8 0.04 0.04
A T Tq A
σσ
−= =
− −+ + +
T T−
93.2% reduction in q.
Dr. A. Pourmovahed, Kettering University 98 Introduction to Energy Systems
Introduction to Fuel Cells References: Larminie, J. and Dicks, A. (2003), Fuel Cell Systems Explained, 2nd ed., Wiley. Sonntag, R.E., Borgnakke, C. and Van Wylen, G.J. (2003), Fundamentals of Thermodynamics, 6th ed., Wiley. TVN Systems (2004), http://www.tvnsystems.com Nomenclature: At total electrode area A/F air-fuel ratio E reversible open circuit voltage e charge on one electron (1.602 x 10-19 C) F Faraday constant (Na .e = 96485 C/kmol) hRP (lower) enthalpy of combustion of hydrogen (119,953 kJ/kg) I current i current density Na Avogadro’s number (6.022 x 1023 molecules/kmol) m mass flow rate P power V voltage Vc individual cell voltage x excess air ratio
fg∆ change in Gibbs free energy of formation (molar) η stack efficiency ηc individual cell efficiency µf fuel utilization coefficient Introduction: A fuel cell is a power source that uses a fuel to provide electric current without combustion. Many different fuel cells have been developed and used in a variety of applications. All fuel cells function much like a battery, having an anode and a cathode that conduct current. Gaseous fuel enters the anode side while oxygen or air is introduced on the cathode side. On the anode side, electrons are separated from the fuel. The electrons flow from the anode through an external load to the cathode side. Of the fuel cells that have been researched and manufactured, the most widely used and promising include alkaline, phosphoric acid and PEM fuel cells.
Dr. A. Pourmovahed, Kettering University 99 Introduction to Energy Systems
Alkaline fuel cells (Figure 1) are among the simplest fuel cells used in an application. They use pure hydrogen as fuel and oxygen to output DC current, water and heat. The United States space program used alkaline fuel cells in the Apollo space missions. The electricity was harnessed to power devices on the spacecraft and the astronauts collected the water for use. In an alkaline fuel cell, pure hydrogen (H2) is input on the anode side of the fuel cell. In the presence of a platinum catalyst, it splits into two hydrogen ions (H+) and two electrons. The electrons are passed through an external load to the cathode. Pure oxygen gas (O2) is input to the cathode side where it splits into separate oxygen atoms. In the electrolyte, potassium hydroxide solution (KOH), the oxygen reacts with the electrons and electrolyte solution to create hydroxyl ions (OH-). The hydroxyl ions then pass across the electrolyte to the anode where they react with the hydrogen ions in the anode to produce water (H2O). Output water can be gathered and stored. Alkaline fuel cells are good energy producers but cannot be exposed to air. Carbon dioxide is poisonous to the alkaline electrolyte.
2 H2O + -
_OH _OH
H O
H O
Electrolyte Anode Cathode
2 H2 O2
Figure 1: Alkaline Fuel Cell
Phosphoric acid fuel cells were the first commercially available fuel cells. They have been used since the 1970’s to provide power for stationary buildings such as hotels and hospitals. In the mid 1990’s phosphoric acid fuel cells provided by International Fuel Cells Corporation were tested in buses. Phosphoric acid fuel cells (Figure 2) use compressed hydrogen gas and oxygen. The fuel cell uses a phosphoric acid electrolyte and produces water and electricity. The hydrogen gas reacts with a platinum catalyst on the anode, splitting electrons from the hydrogen molecule. Electrons are passed through an external load and returned to the cathode.
Dr. A. Pourmovahed, Kettering University 100 Introduction to Energy Systems
Hydrogen ions are passed through the phosphoric acid to the cathode. Oxygen taken from air then reacts with the hydrogen ions to create water. PEM fuel cells are becoming an industry standard for consumer applications. The defining characteristic is the electrolyte. Instead of phosphoric acid, PEM fuel cells use a polymer film as an electrolyte. With operating temperatures ranging from 30 to 100oC, modern PEM fuel cell stacks are used in a variety of applications such as vehicles, mobile applications and for low power “combined heat and power” (CHP) systems. Currently, the most sought after use is to power electric motors for transportation vehicles. The future dubbed as “The Hydrogen Economy” will incorporate PEM fuel cells into all aspects of everyday life. Future applications of electricity created by PEM fuel cells will include anything from cell phones, laptops and other portable devices to larger applications such as to power large buildings as well as residential homes independent of large power grids.
2 H2O-+
H4PO4 OH H OH3PO4
Electrolyte Anode Cathode
2 H2 O2
Figure 2: Phosphoric-Acid Fuel Cell
Having low operating temperatures, PEM fuel cell stacks can be used in mobile applications. The solid flexible electrolyte makes it possible to avoid cracks and leaks. Emissions from PEM fuel cells are very clean. Automotive companies can lower carbon dioxide output from vehicles by implementing PEM fuel cell stacks. Output from the fuel cell will only be water and excess hydrogen and air. Fuel cells produce a small amount of voltage individually. Stacks of cells are used and their output to an inverter can provide a usable constant voltage. PEM fuel cells are expensive to manufacture and platinum catalysts are not cheap. Fuel must be processed before hydrogen gas can be input to the system. If impure hydrogen is used, carbon monoxide will quickly contaminate the polymer membrane electrolyte. Additionally, since the electrolyte is a thin film, constant and equal pressure must be applied to each side of the membrane or it will balloon and burst.
Dr. A. Pourmovahed, Kettering University 101 Introduction to Energy Systems
The basis for operation of a PEM Fuel Cell is hydrogen and oxygen creating a voltage at two poles by reacting to form water. The basic fuel cell is composed of a membrane and electrode assembly known as an MEA, flow field plates, current collectors, and endplates (Figures 3 and 4). Heating or cooling plates may also be contained in the fuel cell. Hydrogen and oxygen react on opposite sides of the MEA to produce water, electricity, and heat. In order to accomplish this, the hydrogen and oxygen half reactions are separated by a solid polymer electrolyte (known as either a proton exchange membrane or a polymer electrolyte membrane). This electrolyte separates the two reactant gases while providing an ionic connection through which the protons may pass. The electrolyte has a catalytic electrode layer on each side to speed the reaction rate and provide electrical contact for the conduction of electrons. The reactant gases flow through channels cut into the flow field plates. While the gas flows parallel to the MEA the gas can diffuse the catalyst layer of the MEA to react. The product water is wicked out to the flow field, where it is swept away by the excess gas flow. The electrical potential is transferred to the current collectors, which can be connected to a load to perform electrical work. Electrochemical reactions in a PEMFC are shown schematically in Figure 5.
Figure 3: Schematic diagram of a PEM fuel cell (Courtesy of TVN Systems, Inc.)
Dr. A. Pourmovahed, Kettering University 102 Introduction to Energy Systems
Figure 4: Schematic diagram of a single PEMFC assembly (Courtesy of TVN Systems, Inc.)
H2 Air
Humidifier(Water)
Diffusion
Layer
H+
H2 O2 H2 → 2H+ + 2e- O2 + 4H+ + 4e- → 2H2O H20
Membrane Catalyst
(Pt) H20, Air H20, H2
Figure 5: Electrochemical reactions in a PEMFC Definitions and Theoretical Analysis:
Dr. A. Pourmovahed, Kettering University 103 Introduction to Energy Systems
Current density is the amount of current supplied by the fuel cell divided by the total area of the fuel cell electrode. This is analogous to the power to weight ratio of a car engine. The current density of a fuel cell is defined as
t
IiA
= (1)
and is usually expressed in (A/cm2). Current density is used to compare one fuel cell stack to another. For example, in automotive applications, it is important to optimize this number such that the overall size of the fuel cell stack is reduced for any given output. Electrical power output of the fuel cell (usually expressed in Watts) is easily found from
P VI= (2) Specific Power and Power Density are defined as
( / )powerSpecific Power W kgmass
= (3)
3 ( / )powerPower Density W mvolume
= (4)
The reversible open circuit voltage (OCV) of a fuel cell using pure hydrogen and oxygen at standard pressure (0.1 MPa) is given by (Larminie and Dicks, 2003)
2fg
EF
−∆= (5)
where fg∆ is the Gibbs free energy released and F is the Faraday constant (96485
C/kmol). The values of fg∆ for the simple reaction OHOH 2221
2 →+ at various temperatures are shown in Table 1.
Dr. A. Pourmovahed, Kettering University 104 Introduction to Energy Systems
Table 1 fg∆ for the reaction OHOH 2221
2 →+ at various temperatures Phase of Water Produced Temperature, ºC
fg∆ (kJ/kmol) liquid 25 -237.2 liquid 80 -228.2 vapor 80 -226.1 vapor 100 -225.2 vapor 200 -220.4 vapor 400 -210.3 vapor 600 -199.6 vapor 800 -188.6 vapor 1000 -177.4 Source: Larminie, J. and Dicks, A. (2003), Fuel Cell Systems Explained, 2nd ed., Wiley. As current is drawn from a fuel cell, the cell potential (voltage) drops. This is shown in Figure 6. The voltage drop is caused by four factors:
1. Activation losses 2. Fuel crossover and internal currents 3. Ohmic losses 4. Mass transport or concentration losses
Detailed discussion of these factors can be found in Larminie, J. and Dicks, A. (2003).
Dr. A. Pourmovahed, Kettering University 105 Introduction to Energy Systems
.
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
.
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
.
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1.05
1.15
1.25
0.00 0.50 1.00 1.50 2.00 2.50Current Density (A/cm2)
Cel
l Pot
entia
l (V
) Open-circuit voltage is lower than thermodynamic voltage due to gas crossover and peroxide formation
Voltage Loss due to Activation Resistance
Voltage Loss due to OhmicResistance and Concentration
Polarization
Voltage Loss due to Mass Transport Limitation
Figure 6: Cell potential versus current density for a PEMFC (Courtesy of TVN
Systems, Inc.)
One way to define fuel cell stack efficiency is to compare the energy output from a fuel cell to the lower enthalpy of combustion of hydrogen.
2hH RP
VIm
η = (6)
This definition provides a method for comparing the power output of a fuel cell stack to that of an internal combustion engine utilizing the same fuel (hydrogen) with water vapor in the combustion products. The efficiency of an individual cell has been derived by Larminie, J. and Dicks, A. (2003, p. 35) as
Dr. A. Pourmovahed, Kettering University 106 Introduction to Energy Systems
1.48 1.25c
c fVor
η µ= (7)
Where 1.48 is the number to be used in conjunction with the “higher heating value” and 1.25 corresponds to the “lower heating value” of hydrogen. The fuel utilization coefficient, fµ , is usually taken to be 0.95.
Stoichiometric reaction is a complete reaction where the quantity of oxidizer is the precise amount needed to completely consume a quantity of fuel. Consider stoichiometric reaction of hydrogen with air inside the fuel cell stack,
H + 212
[O + 3.76 N 2 ] → H O + 1.88 N (8) 2 2 2
.( / )stoichA F = 0.5[32 3.76(28)]
2(1)+
= 34.32 (9)
2
( / ) Air
actual
H
mA F
m
•
•= (10)
Define excess air ratio (x) as
.
( / )( / )
actual
stoich
A Fx
A F= (11)
For example, if x = 2, we have 100% excess air or 200% theoretical air (double the amount of air required for complete reaction of hydrogen and oxygen). The actual reaction is
H + 2 2x [O + 3.76 N 2 ] → H O (liquid) + (1.88 x) N + 2 2 2
12
x − O (12) 2
The minimum (stoichiometric) reactant flow rate required for current generation at standard pressure and temperature may be determined as follows. The stoichiometric reaction of hydrogen and pure oxygen to form water can be written as
Dr. A. Pourmovahed, Kettering University 107 Introduction to Energy Systems
2
2 2
2 2 2
2 2 ( )1 2 2 ( )2
:1 ( )2
H H e a
O H e H O b
which combine to
H O H O a
+ −
+ −
→ +
+ + →
+ → + b
At standard temperature and pressure (STP), 1 kmol of any gas (such as hydrogen) occupies 22.41 liters and consists of 6.022 x 1023 molecules (Avogadro’s number, Na). The charge on one electron is e = 1.602 x 10-19 C. For each kmol of hydrogen that reacts, 2 electrons are released. The electric charge produced is 2 Na e = 2F = 192,970 C. Therefore, a hydrogen volumetric flow rate of 22.41 liters/s (at STP) corresponds to an electric current of 192,970 C/s (or Amps). To produce 1 Amp, the minimum volumetric flow rate of hydrogen at STP is 22410/192970 = 0.11613 ml/s = 6.97 ml/min ≈ 7 ml/min. For every oxygen molecule, two hydrogen molecules are needed. This will result in a 2:1 ratio for hydrogen/oxygen volumetric flow rates. If air is used instead of pure oxygen, since air is comprised of 21% oxygen and 79% nitrogen by volume, the volumetric flow rate of air would have to be ≅ 4.76 times greater than that of oxygen to react with the same amount of hydrogen. Table 2 shows the stoichiometric (minimum) reactant flow rates required for current generation at STP.
Table 2 Stoichiometric reactant flow rate required for current generation at standard temperature and pressure , STP (Courtesy of TVN Systems, Inc.)
Dr. A. Pourmovahed, Kettering University 108 Introduction to Energy Systems
Current H2 O2 Air Current H2 O2 Air (A) (ml/min) (ml/min) (ml/min) (A) (ml/min) (ml/min) (ml/min)0.5 3.5 1.7 8.3 10.5 73.1 36.6 174.1 1 7.0 3.5 16.6 11 76.6 38.3 182.4
1.5 10.4 5.2 24.9 11.5 80.1 40.0 190.7 2 13.9 7.0 33.2 12 83.6 41.8 199.0
2.5 17.4 8.7 41.5 12.5 87.1 43.5 207.3 3 20.9 10.4 49.7 13 90.5 45.3 215.6
3.5 24.4 12.2 58.0 13.5 94.0 47.0 223.9 4 27.9 13.9 66.3 14 97.5 48.8 232.2
4.5 31.3 15.7 74.6 14.5 101.0 50.5 240.4 5 34.8 17.4 82.9 15 104.5 52.2 248.7
5.5 38.3 19.2 91.2 15.5 108.0 54.0 257.0 6 41.8 20.9 99.5 16 111.4 55.7 265.3
6.5 45.3 22.6 107.8 16.5 114.9 57.5 273.6 7 48.8 24.4 116.1 17 118.4 59.2 281.9
7.5 52.2 26.1 124.4 17.5 121.9 60.9 290.2 8 55.7 27.9 132.7 18 125.4 62.7 298.5
8.5 59.2 29.6 141.0 18.5 128.8 64.4 306.8 9 62.7 31.3 149.2 19 132.3 66.2 315.1
9.5 66.2 33.1 157.5 19.5 135.8 67.9 323.4 10 69.6 34.8 165.8 20 139.3 69.6 331.7