Processteknikens grunder (”PTG”) Introduction to Process Engineering v.2014

3. conversion, balances, efficiency, equilibrium (Introduction to )

Ron Zevenhoven Åbo Akademi University Thermal and flow engineering / Värme- och strömningsteknik tel. 3223 ; [email protected]

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 1/124

3.1: Energy

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 2/124 What is energy? /1

. ”Energy is any quantity that changes the state of a closed system when crossing the system boundary” (SEHB06)

. ”Energy is the capacity to do ” (A83)

. ”Energy is the capacity to do work or produce heat” (ZZ03)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 3/124

What is energy? /2

. Energy cannot be produced or destroyed (First Law of Thermodynamics*) but it can be converted from one form into another, and vice versa.

. Energy can be degraded to lower quality, as a consequence of producing and converting heat (Second Law of Thermodynamics*)

* sv: Termodynamikens Första och Andra Huvudsatsen

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 4/124 Types of energy /1 . In a closed system, energy is . These are all relative to a present (”stored”) as reference frame or reference state:

potential energy* Ep and . potential energy relative to a kinetic energy** Ek position z = 0 in a gravity field (mechanical energy) and . kinetic energy relative to a non- internal energy*** U moving reference frame with (thermal energy) velocity v = 0 . For a mass m, with vertical . internal energy relative to zero position z in a gravity field and temperature T = 0, or surroundings velocity v defined as temperature T = T°.

Ep = m· g· z 2 2 Ek = ½· m· v = ½· m· v U = m·u (* sv: lägesenergi; * * sv: rörelseenergi; Picture: *** sv: inre energi) SEHB06

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 5/124

Types of energy /2

Unit: Joule, J ; 1 J = 1 N·m= 1 kg·m2/s2 Energy form Energy per unit mass Comment Kinetic energy ½v·v = ½v2 velocity v Potential energy g·z gravity g, height z Internal energy u Pressure energy p/ρ pressure p, density ρ Enthalpy u + p/ρ Surface energy (σ·A) /(ρ·V) surface tension σ surface A, volume V ......

Most important energy forms for (chemical) process engineering

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 6/124 7/124 Potential energy A visualisation of potential energy, showing two mechanical equilibrium points A and C, point C being the most stable, and point B being the most unstable

Picture: M62

8/124 Potential energy, kinetic energy

2 2 Conservation of mechanical energy: ½mv1 + mgz1 = ½mv2 + mgz2

2 2 For a closed system with constant mass: ½v1 + gz1 = ½v2 + gz2

Picture: KJ05 Picture: CB98

Transformation kinetic energy ↔ potential energy (without friction !) Internal energy, U /1 . Internal energy U is a consequence of the motion of constituent molecules of a substance and their interactions (T06) . Unit : J . Specific internal energy u = U/m (J/kg)

Picture: SEHB06

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 9/124

Internal energy, U /2 . Internal energy of a material or substance is related to temperature via the specific heat, also referred to as heat capacity, (sv: värmekapacitet) c (unit: J kg-1K-1 or J mol-1K-1) introduced by Gadolin at ÅA, 1784

. Defining u = 0 at reference temperature T° (for example T 0°C or 25°C) gives u = T° ∫ c dT ≈ c· (T – T°) Alternatively, a reference value u = u° can be defined for T T = T° = 0°C or 25°C, then u = u° + T° ∫ c dT As c = c(T), an approxation for c for the temperature

range T1 →T2 is c(Taverage) for Taverage = ½(T1+T2)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 10/124 11/124 Internal energy, U /3 . For gases, a distinction is made between specific Specific heat data for several materials heat at constant volume, cv and specific heat at constant pressure, cp (see below section 3.4)

The specific heat ratio

cp/cv = γ = κ

. Solids and (ideal) liquids are (practically) incompressible: cp= cv = c

12/124 Example: specific heat solids, liquids /1

. An aluminum ball with mass mb= 80 g and temperature Tb = 200°C is dropped into a volume Vw = 6 liters of water that has a temperature Tw = 22°C. The bath is well insulated.

. Calculate the final temperature Teq of the ball and the water at equilibrium. Consider only internal energy. . Specific heat data:

cwater = 4.19 J/(g· K); cAl = 0.88 J/(g·K) 13/124 Example: specific heat solids, liquids /2

•Answer: only internal energy needs to be considered; the sphere cools

down from Tb to Teq while the water heats up from Tw to Teq. The energy for heating the water comes from cooling the sphere; the internal energy balance gives

mw·cw·(Teq-Tw) = mb·cAl·(Tb-Teq) *) 3 3 with mass mw = 0.006 m ·1000 kg/m = 6 kg

The unknown temperature Teq is then found after rewriting *)

Teq = (mb·cAl·Tb+mw·cw·Tw) / (mw·cw+mb·cAl ) = 22.26°C

Total energy of a system

. Total energy of a system

E = Ek + Ep + U (J)

= m· ek + m· ep + m· u (J)

with specific energies ek, ep and u (J/kg)

. . A mass stream m (kg/s) corresponds to an energy stream Ė . . Ė = Ėk + Ėp + U = m· (ek + ep + u) (J/s)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 14/124 3.2: Energy transfer, First Law of Thermodynamics, Work, Power, Heat

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 15/124

Energy transfer . For a closed system ENERGY can be transferred across the boundary by WORK or by HEAT transfer

. For an open system, MASS may cross the bounday and the ENERGY associated with that can enter the system, in addition to WORK or by HEAT transfer (as for a closed system)

. Generally, for solid matter (”rigid objects”), work changes kinetic energy and potential energy of a system, while heat changes internal energy. . For gases and liquids (”fluids”) matters are more complicated: friction in fluid flow gives viscous heat, compression of a gas at constant volume increases its temperature, etc. Picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 16/124 First Law of Thermodynamics Also known as the ”conservation of energy” principle

. ”The energy of an isolated system is constant” (A83) . Energy can be converted from one form into another, and vice versa for example heat Q can be converted into work W and vice versa - but there are limitations ! (see section 3.6)

Closed system ∆E = W + Q (or: dE = δW + δQ) . . Open system ∆E = W + Q + min· ein -mout· eout

. Energy cannot be produced and speaking about ”energy production” is wrong and silly

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 17/124

Work /1 (sv: arbete)

. ”Work is force acting over a distance” (ZZ03)

. ”Work is the transfer of energy across a system or control-volume boundary, exclusive of energy carried across the boundary by a flow, and not the result of a temperature gradient at the boundary or a difference in temperature between the system and the surroundings” (T06)

. ”Work is mechanical energy in transit” (BB10)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 18/124 Work /2 (sv: arbete)

. ”Work is mechanical energy in transit” (BB10)

. ”Work stimulates organised motion” (A83)

. Work is a transfer process for mechanical energy

. Note: work (a path function) is not a system property

. Sign of work W : W > 0 implies work done ON the system; W < 0 implies work done BY the system . In many cases it is defined otherwise (with opposite sign) in engineering calculations (see for example KJ05) !

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 19/124

20/124 Work /3

. Most important types of work: shaft work (sv: axelarbete), expansion / compression work, electrical work

. For a constant force F, moving a body from position s1 to s2 requires work

. W12 = F· (s2 –s1) (note vectors!) . For a non-constant force F = F(t) or F = F(s), the work

follows from integration over the path: W12 = path ∫ δw = s2 s1∫ F· ds

Pictures: KJ05 Work /4 work is a path function!

. Work is a path function (an inexact differential) and a differerential ”dw” makes no sense

. ”W1” or ”W2” at start / end does not mean anything !

. W = W12 depends on the path from position or state ”1” to position or state ”2” Two paths (top view !) for . It is preferable to use moving a crate against notation δw or đwinstead friction with the floor (but nonethless ”dw” is used in general.....) Picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 21/124

22/124 Compression/expansion of gas /1

. Assume a frictionless, rigid, perfect-fit piston . inside: a gas

. pressure outside = pex

pressure inside = pin

pex < pin → the piston moves outwards/upwards, doing work (w < 0 )

. δw= Fz· dz = (Fz/A)·(A·dz) Picture: A83 = - pex· dV . For ∆V = V –V : Note: work = force * distance final initial = (pressure*surface) * distance work W = - pex· ∆V = pressure * (surface*distance) = pressure * volume change 23/124 Compression/expansion of gas /2 . The work

W = - pex· ∆V can also be shown in a p,V plot or indicator diagram

. The work depends on the external pressure;

against a vacuum (pex=0) W = 0 To be continued ...... Picture: A83

24/124 Power . Power (sv: effekt, (kraft)) is the rate of doing work . notation P = W/∆t or W or δw/dt

also: work W = time∫ Power dt . Unit: Watt, W = J/s USCS units: 1 BTU/h = 0.29307 W 1 horsepower hp = 745.7 W . . W = path∫ F· ds means that W = F· ds/dt = F· v

. for example: liquid pumping power

if Tout = Tin, then . . . P = Win = (m/ρ)· (pout -pin) = V· ∆p with liquid density ρ (kg/m3) Picture: SEHB06 Shaft work, electrical work . Mixing a fluid . An electric circuit

W = time∫ FRωdt (J) W = time∫ ξIdt (J) Power = FRω (Watt) with electric current I (A), with impeller radius R (m), resistance R (Ω = V/A), impeller angular velocity voltage ξ = IR (V) ω = dθ/dt (1/s), Power = ξI = I²R (VA = Watt) tangential force F (N)

Pictures: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 25/124

Heat /1 (sv: värme) . ”Heat is energy transferred, without transfer of mass, across a boundary of a system (or across a control surface) because of a temperature difference between the system and the surroundings or a temperature gradient at the boundary” (T06)

. ”Heat involves the transfer of energy between objects due to a temperature difference” (ZZ03)

. ”Heat is thermal energy in transit” (BB10)

. ”Heat stimulates random motion (thermal motion)”(A83)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 26/124 Heat /2 (sv: värme) . ”Heat is thermal energy in transit” (BB10)

. Heat is a transfer process for thermal energy

. Note: heat (a path function, like work) is not a system property - but temperature is!

. Sign of heat Q : Q > 0 implies heat INTO the system; Q < 0 implies heat OUT OF the system

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 27/124

28/124 Heat /3 . Heat occurs at the boundary of a system a system cannot ”contain” heat. . Heat is an energy transfer process (similar to ”work”) and ”heat transfer” is the same as ”heat” ! (”After all, ”work transfer” is not used either !)

Picture: SEHB06 . Nomenclature for heat transfer and rate: heat (transfer) (sv: värmemängd) Q unit: J . heat transfer rate (sv: värmeström) Q unit: J/s = W . heat flux (sv: värmeströmtäthet) Q” unit: J/(s· m2)=W/m2

. A process without heat transfer is called adiabatic (which is not the same as isothermal !!) See also 29/124 chapter 5 Heat /4 . Heat transfer is the result of a temperature gradient and occurs through two physical mechanisms only: – Conduction (sv: värmeledning) (as a result of microscale motion) – Radiation (sv: strålning) (heat transfer by electromagnetic waves)

A third mechanism for heat transfer is convection (sv: konvektiv värmeöverföring), but here the heat energy is transported as a medium flow or the motion of bodies with different temperatures. At the boundary, the energy transfer is finally by conduction or radiation.

A metal spoon in a hot cup of tea Picture: M1888 Temperature at A > B > C > D > E

3.3: Properties of pure substances; Processes in p,V,T space

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 30/124 31/124 p,V,T surface for a pure component

For a substance that expands on melting

See also Source:http://newark.rutgers.edu/~huskey/images/PVT_phase_diagram.jpg chapter 4

32/124 Non-ideal, ”real” gases /1

. Very high pressures and temperatures near the critical

point (Tcrit, pcrit,Vcrit): a correction factor Z , to ideal gas law: pV = nZRT; with compressibility

Z = Vreal / Videal = pVreal / (nRT)

. According to the Law of Corresponding States, as all gases (at not too extreme pressures) obey the same equation of state, Z can be related to

reduced pressure pR = p / pcrit

reduced temperature TR = T / Tcrit

reduced volume VR = V / Vcrit Non-ideal, ”real” gases /2

. Compressibility factor Z as function of reduced state variables

(PR, TR)

Picture: T06

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34/124 Non-ideal, ”real” gases /3 . First improvement: Van der Waals equation of state

an (p  )(V  nb)  nRT V  RT RT  with a  crit , b  crit , and Z   . pcrit pcrit 

. Alternatively, use virial coefficients to calculate Z, or, for example, Berthelot’s equation of state:

 9p  6  RT p V  nR'T 1 R 1  with R' crit R R R 128  2  P V   TR  crit crit 35/124 p,V and T,V diagrams

T = constant for ideal gas (isothermal)

. For n moles of an ideal gas, pV = nRT Picture: SEHB06

Example: a simple thermodynamic process

Pictures: T06

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 36/124 Polytropic processes of ideal gases /1 .A general expression for a process can be written as pVn = constant with polytropic exponent n

. n = 0: p = constant (isobar) n = 1 : pV = constant = nRT (isotherm) γ , n = γ = cp/cv:pV= constant (isentrope)* ** n = ∞: V = constant(isochore, isometric)

* isentropic = with constant entropy S : ** This requires that cp ∆S = ∆Q/T = 0, i.e. an adiabatic reversible and cv ≈ constant ! process (see section 3.6 for more detail) Picture: T06

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Polytropic processes of ideal gases /2

calculating work for ideal gases .Pressure-volume work W is calculated as W =n p  V  V2    W12 = - V1∫ pdV, and for any polytropic p  V  n n n n process pV = constant = C, i.e. p1V1 = p2V2 T  V     T  V  n  n n p T .This means p = C/V and for n ≠ 1:    p  T  V2 V2 dV C 1n 1n p1V1  p2V2 W12  W   pdV  - C  - (V2  V1 )     Vn 1 - n n 1 V1 V1

and for an (n = 1, pV = nmolRT): V2 V2 dV V W  W   pdV  - n RT  - n RT (lnV  lnV )  n RTln 2 12  mol  V mol 2 1 mol V V1 V1 1 Note: work done ON (=into) system = -∫pdV, work done BY (=out of) system = ∫pdV Example: calculating p,V work /1

out . Oxygen at 300 K expands slowly and isothermally from 100 kPa to 45 kPa.

The mass of the O2 is 0.052 kg; calculate the work Wout done by the gas.

Source and picture: KJ05

System energy increases by Q + W = Q - Wout

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 39/124

Example: calculating p,V work /2

. A well insulated piston-cylinder set-up contains 0.031 m3 air at 40°C and 102 kPa; calculate the work needed to compress this air slowly to 350 kPa. Consider the process adiabatic. For air,

γ = cp/cv = 1.4

Source and picture: KJ05

System energy increases by W = Win

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 40/124 3.4: Enthalpy ; Specific heat

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 41/124

42/124 Enthalpy /1

. Consider the addition of heat to a gas (assume an ideal gas for the moment) in two different situations (a) a rigid tank, i.e. a constant volume process (b) a cylinder-piston assembly, i.e. a constant pressure process

picture: KJ05 43/124 Enthalpy /2 (a) constant volume The energy balance (no potential energy or kinetic energy effects): ∆U = Q + W, and since no work is done: ∆U = Q This implies that all heat is used to increase the temperature of the gas at constant volume.

With specific heat at constant volume, cv (J kg-1K-1), gas mass m and temperature T:

∆U = m·cv·∆T, or ∆u = cv·∆T which can be written as cv = ∆u / ∆T or cv = du / dT IF u is a function of T only (like for ideal gas)!

picture: KJ05

44/124 Enthalpy /2 (b) constant pressure The energy balance (no potential energy or kinetic energy effects): ∆U = Q + W, and some work is done to increase the volume as to keep pressure constant: W = - V∫ p dV, with p = external pressure = internal pressure (at equilibrium: assume a slow or ”quasi- equilbrium” process !) As p is constant, W = - p·∆V, and thus ∆U = Q - p·∆V, or Q = ∆U + p·∆V = (at constant pressure) Q = ∆ (U + p· V) Part of the heat Q is used to increase volume V !

picture: KJ05 Enthalpy takes into 45/124 consideration that a Enthalpy /3 constant pressure process may give a volume (b) constant pressure (continued) change which implies The quantity U + p· V is referred to as exchange of work with the enthalpy, H (unit: J) environment. → the energy balance gives Q = ∆H = m· ∆h with specific enthalphy h With specific heat at constant pressure, -1 -1 cp (J kg K ), gas mass m and temperature T:

Q = ∆H = m· cp·∆T, or ∆h = cp· ∆T which can be written as

cp = ∆h / ∆T or cp = dh / dT IF h is a function of T only ! Simular to u, h must be fixed to a reference value h = h° at T = T°

picture: KJ05 Like U, p, T and V, enthalpy is a thermodynamic state property

46/124 Specific heat of gases /1 . Specific heat is also referred to as heat capacity

. For an ideal gas u = u(T) and cv = du/dT and h = h(T) and cp = dh/dT

. Specific heat ratio cp/cv = γ = κ

. According to the so-called State Principle ”any two independent thermodynamic properties are sufficient to describe the state of a system containing a single pure substance” (a simplification of the so-called phase rule by J.W. Gibbs for multi-phase, multi-component systems)

. Thus, u= u(p,T) or u= u(v,T) and h= h(p,T) or h= h(v,T) 47/124 Specific heat of gases /2

. Thus, u= u(p,T) or u= u(v,T) and h= h(p,T) or h= h(v,T) . This leads to partial differential expressions:  u   u   u  du    dT    dV  c vdT    dV  T V  V T  V T = 0  h   h   h  for an dh    dT    dp  cpdT    dp ideal gas  T p  p T  p T

pictures: T06

48/124 Partial differentials

U = U(V,T): Partial differentials with respect to (a): V (b): T (c): both V and T

Pictures: A83 Other thermodynamic properties . Starting with V = V(T,p)  V   V  dV    dT    dp  T p  p T dV 1  V  1  V  or :    dT    dp  βdT  κdp V V  T p V  p T defining isothermal compressibility κ and thermal expansion coefficient β (for ideal gas: κ = 1/p and β = 1/T) . Another important parameter is the Joule-Thomson

coefficient µJT, that quantifies how the temperature of a substance will increase or decrease when pressure is changed at constant enthalphy:  T   h  μ JT    and    c p  μ JT  p  h  p T

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50/124 Specific heat and enthalpy /1 .For ideal gases, the expression pV = nRT leads to a simple and-Heat.html relation between cp and cv : H = U + pV differentiated over temperature gives

dH/dT = dU/dT + d(pV)/dT -1/Temperature-

→ m· cp = m· cv + nR (J/K) for a gas with molar mass M, n=m/M, which gives

cp = cv + (n/M)·R eryday/Real-Life-Chemistry-Vol -1 -1 with unit J· kg K for cp and cv , BECAUSE OF WATER'S HIGH SPECIFIC HEAT CAPACITY, CITIES LOCATED NEXT or TO LARGE BODIES OF WATER TEND TO STAY WARMER IN THE WINTER cp = cv + R AND COOLER IN THE SUMMER. -1 -1 with units J· mol K for cv and cp

Note that both sets of units are used ! http://www.scienceclarified.com/ev 51/124 Specific heat and enthalpy /2 . Constant pressure processes are more common than constant volume processes → h more useful than u for engineering calculations (also if pressure ≠ constant)

. The use of cp, typically a polynomial cp(T), is essential for enthalpy calculations, with reference* state (Tref,pref) T h(T,p)  h(T)  h(Tref )   cpdT for ideal gas Tref

T Often Tref is more general : h(T,p)  h(T ,p )  c dT  c µ (p  p ) taken as = ref ref  p p JT ref T 0 ref Tsurroundings = T , or 0°C, or 25°C T check / be h(T,p)  h(T)  h(Tref )   cdT for liquid or solid careful / be Tref consiistent

52/124 Specific heat and enthalpy /3

. For cp(T) and cv(T) polynomials are used, with parameters found from tabelised data, for example

2 3 c p  a  bT  cT  dT  .... or d c  a  bT  cT2   .... p T

NOTE: BY DEFINITION: h = 0 for the Elements of the Periodic Table at 25°C, 1 atm, in their standard states. 53/124 Polynomial cp data for gases

Source: SEHB06

Enthalpy of water at 1 atm

h(kJ/kg) 2676 dh/dt = cp,steam

With h = 0 defined here ∆hvaporisation for liquid at 0°C

419 Note that ∆hmelt 0 cp = ∞ at dh/dt = cp,water phase -334 transitions dh/dt = cp,ice

0 100 T (°C) see also ÖS96 Fig. 2.3

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 54/124 55/124 (Enthalpy /4 Another approach: flow work)

. An alternative visualisation of enthalpy follows from the flow work concept: a small mass flow dm has a volume dV = dm/ρ (with density ρ) = A·dx, entering a system through area A. This is accompanied by an amount of flow work equal to δw = p·A·dx to the system, where p is the pressure inside the system.

. Thus, the amount of energy dEin entering the system equals 2 2 dEin = (gz + ½v + u)· dm + (p/ρ)· dm = (gz + ½v + u)· dm + p· dV . Similarly, for mass dm leaving the system 2 2 dEout = (gz + ½v + u)· dm + (p/ρ)· dm = (gz + ½v + u)· dm + p· dV . Internal energy + p.dV flow work → enthalpy h·dm, h = u + p/ρ

Picture: SEHB06

3.5: Energy balances; Conversion work → work, work → heat

General form Heat and/or work into/from system = Change of state (position, velocity, temperature, pressure, volume, enthalpy, ...) of system

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 56/124 The total energy balance

. The total energy of a system equals E = mgz + ½mv2 + U + pV = mgz + ½mv2 + H, or as energy concentration (and using m = ρ·V) e = gz + ½v2 + u + p/ρ = gz + ½v2 + h

. For a process that involves energy transfer to the system as heat and/or work: ∆E = Q + W ∆(mgz + ½mv2)+ ∆U + ∆(pV) = ∆(mgz + ½mv2) + ∆H = Q + W ∆e = q + w ∆(gz + ½v2)+ ∆u + ∆(p/ρ) = ∆(gz + ½v2)+ ∆h = q + w

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Example: potential energy gives heat

. A steel ball of 0.5 kg is dropped from a height of 60 m. When hitting the ground it becomes embedded; calculate the temperature rise of the ball at the impact. The specific heat of the steel is c -1 -1 = 235 J kg K Source and picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 58/124 Example: water pumped through a pipe

. Water is pumped at 1 kg/s through an isolated horizontal pipe with inner diameter 2 cm. The pressure drop ∆p over the pipe is 2 bar. The density ρ of the water is 1000 kg/m3; the specific heat is c = 4.2 kJ/(kgK). Calculate 1) the temperature rise ∆T of the water and 2) the pumping power needed.

. . .

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 59/124

The mechanical energy balance

. The mechanical energy balance disregards thermal effects, assuming an iso-thermal process without heat effects: ∆U = 0, Q = 0

→ the total energy balance reduces to

∆E = W ∆(mgz + ½mv2)+ ∆(pV) = W (J) ∆e = w ∆(gz + ½v2) + ∆(p/ρ) = w (J/kg)

A special case is Bernoulli’s equation if also work = 0 ∆(gz + ½v2) + ∆(p/ρ) = 0, or gz + ½v2 + p/ρ = constant

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 60/124 61/124 Example: Bernoulli Equation

. Question: Water forms a vertical jet when released from a pipe at 6 m/s vertical velocity. Calculate jet height h. Neglect fluid friction; assume inviscid flow.

Source and picture: KJ05

The thermal energy balance

. The thermal energy balance considers mainly the change of internal energy U, in situations where potential and kinetic energy changes can be neglected. Often, work is neglected as well.

→ the total energy balance reduces to

∆U + ∆(pV) = ∆H = Q + W (J) ∆u + ∆p/ρ = ∆h = q + w (J/kg)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 62/124 63/124 Example: compression of a gas with heat transfer and shaft work /1 A gas is contained in a piston- cylinder system. The gas is compressed by doing 670 J work on it. Also, a propeller does 182 J work, and the internal energy is found to decrease by 201 J.

Question: How much heat was transferred during the process, and was the gas heated or cooled? Source and picture: KJ05

64/124 Example: compression of a gas with heat transfer and shaft work /2

Source and picture: KJ05 Example: electric heating of gas /1

. Air flows through a pipe at a rate of 0.0064 m³/s, entering at 25°C, 101 kPa. . The outside of the pipe is covered by an • A voltage of 120 V is electrically heated tape, supplied to the tape, which which is well insulated has an resistance of 30 Ω. from the surroundings • Calculate the exit by a thick insulation. temperature of the air, assuming constant values for

specific heat cp = 1.005 kJ/(kgK), density 1.18 kg/m3 Source and picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 65/124

Example: electric heating of gas /2

Heat production due to Ohmic losses = current × voltage. = voltage² /resistance = Q

Source and picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 66/124 Throttling valves . A throttling device is used to reduce the pressure of a flowing fluid, producing a large reduction in pressure over a short distance. . Mass and energy balances show – One inlet and one outlet Throttling devices – No work is done . Note that (partial) vaporisation – Steady state (usually) or condensation is possible, – No significant heat transfer giving a temperature effect! . → A constant enthalpy, ↔ Joule-Thomson coefficient ”isenthalpic” process,

hin = hout Source and picture: KJ05

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 67/124

Energy balances: special cases . In many applications a balance of only mechanical or thermal energy is needed . For thermal processes without work

– At constant volume: ∆U = Q = Qin –Qout

– At constant pressure: ∆H = Q = Qin –Qout . For continuous processes: .. . . . – Steady. state: Q = Qin –Qout = ∆U + ∆pV = ∆H . For non-continuous processes: . . . dE/dt = δQ/dt + δW/dt; dE/dt = δQ/dt + δW/dt

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 68/124 3.6: Reversible ↔ Irreversible processes; Entropy; Second Law of Thermodynamics Conversion heat → work; Thermal efficiency

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 69/124

Reversible/irreversible processes /1

. ”A system undergoes a . For example, charging a battery and reversible process if it can decharging it are reversible be returned to its initial processes, provided that no energy state with no change to the is dissipated as heat surroundings” (KJ05) . However, electric systems have Ohmic losses and if electric current . In thermodynamics ”a > 0, these losses are > 0. reversible change is one . Reversible only if the electric current that can be reversed by an equals ≈ 0. infinitesimal (infinitely small) modification of a variable” (A83) . Irreversible processes lead to dissipation (sv: förslösande) of energy as heat to the surroundings

Picture: ZZ03

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 70/124 Reversible/irreversible processes /2

.A reversible process is an .Close-to-reversible devices idealisation of a real, Compressors and turbines, pumps; irreversible process heat exchangers operating at very .Close-to-reversible: small temperature differerences Slow isothermal (or adiabatic) .Clearly irreversible devices compression ↔ slow isothermal Throttling valves; heat exchangers (or adiabatic) expansion operating at not-small temperature In reality, friction (sv: friktion, differerences rivning) introduces irreversibilities .Clearly irreversible: mixing Rule of thumb: irreversible of fluids; moving masses against processes are friction processes that look absurd when run backwards

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 71/124

72/124 Compression/expansion of gas /3 A reversible process • Assume the expansion process

considered earlier, but with pex changing and always slightly smaller

than pin: pex = pin - ∆ , with ∆ very small.

• → for the work δw = - pexdV =

-(pin - ∆ )dV ≈ -pindV . For an ideal gas (pV = nRT) the changing see also pressure pin can be related to changing volume V section . For an isothermal reversible process, starting at 3.2 Vinitial = Vi and ending at Vfinal = Vf :

Vf Vf Vf nRT dV Vf W   pindV   dV  nRT   nRTln   V  V Vi Vi Vi Vi 73/124 Compression/expansion of gas /4 Reversible ↔ irreversible process

. The indicator diagram shows that for the (isothermal) expansion process :

Wreversible > Wirreversible Vf . W = - Vi∫ pdV in p,V plot . Here, the reversible path is isothermal irreversible work : W   pex (Vf  Vi ) V initial final reversible work : W   nRTln f Vi Picture: A83

Reversible/irreversible processes /3

Pictures: SEHB06 . Irreversible p,V work . Compression and expansion in a cylinder with friction.

1→2 Reversible adiabatic frictionless compression to volume Vmin

1→2A Compression with friction (p2A > p2 (and T2A > T2) to reach volume Vmin)

2A→1A Expansion with friction to Vmax (with T1A > T1) 1A→1 Heat transfer to surroundings, heat Q is area in p,V diagram The work done to compensate for this heat equals W = Q

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 74/124 Reversible/irreversible processes /4

Picture: SEHB06 . Irreversible heat transfer . Compression and expansion in a frictionless cylinder with heat transfer with a finite temperature difference. 1→2 Reversible adiabatic frictionless compression 1→2B Frictionless compression with heat transfer with surroundings 2B→1B Reversible adiabatic frictionless expansion 2B→1C Frictionless expansion with heat transfer with surroundings

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 75/124

76/124 Entropy /1 sv: entropi

Picture: SEHB06 Picture:A83

Processes that involve heat and/or work proceed only in one particular direction, even though the First Law of Thermodynamics allows for two (or more!) directions 77/124 Entropy /2 . A spontaneous process occurs (fast or slow) without outside intervention . Earlier it was thought that processes are spontaneous if exothermic (= producing heat, ∆H < 0), but, for example, ice melting at > 0°C is endothermic (= consuming heat, ∆H > 0) and spontaneous! . The driving force for a spontaneous process is an increase of the ENTROPY of the universe . Nature spontaneously proceeds towards the states that have the highest probability

http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter17/Text_Images/FG17_01.JPG

78/124 Entropy /3

. Entropy can be viewed as a measure of Ludwig Boltzmann (1844-1906) molecular randomness or disorder. . Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available (possible) to a system in a given state . With entropy S and possible

arrangements W with probability pi, Boltzmann derived http://ldc.upenn.edu/myl/llog/BoltzmannTomb.jpg S = - k· log W = - k Σ p · log p i i Entropy can also be . Boltzmann’s constant k (or kB) relates linked to the information temperature to energy: ”i” we have of a system, -23 for example kB = 1.38065×10 J/K (= R/NAvogadro !) ”i” = Smaximum -S 79/124 Entropy /4 . The entropy of the universe must increase:

∆Suniverse = ∆Ssystem + ∆Ssurroundings ∆Suniv > 0 : spontaneous process ∆Suniv < 0: spontaneous process in reverse direction ∆Suniv = 0: equilibrium

http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter17/Text_Images/FG17_03.JPG

80/124 Entropy /5

. ∆Suniv = ∆Ssys + ∆Ssurr

∆Ssurr is primarily determined by the flow of energy in or out the system as heat

. Exothermic processes: Q to surroundings, ∆Ssurr > 0

Endothermic processes: Q into system, ∆Ssurr < 0

. If ∆Ssurr or ∆Ssys < 0, temperature determines what will happen, for example: water → ice if T < 0°C (at 1 bar) . ∆S depends on temperature: surr Q = heat into the system quantity of heat from system to surroundings - Q ∆S   surr temperature T ∆H  (if no work is done, - Q  - ∆H)   system T 81/124 Entropy /6 The Definition

. For a reversible process path

 δQ δQ δQ δQ ∆S  S  S    ; and dS        T A T B T T

. For a reversible process

δQ δQ δQ  δQ  δQ  δQ  δQ   