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Energy. Kinetic Energy (E k ). Nomenclature: “ ˙ ” above means “rate” “ ˆ ” above means “per mole or per unit mass”. Potential Energy ( E p ). Energy due to position in a potential field (gravity / magnetic) or confirmation relative to equilibrium (eg, spring). Examples.

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Energy

Energy


Kinetic energy e k

Kinetic Energy (Ek)

Nomenclature: “ ˙ ” above means “rate” “ ˆ ” above means “per mole or per unit mass”


Potential energy e p

Potential Energy (Ep)

Energy due to position in a potential field (gravity / magnetic) or confirmation relative to equilibrium (eg, spring)


Examples

Examples

  • Water flows into a process unit through a 2-cm inner diameter(ID) pipe at a rate of 2 m3/hr. Calculate Ek for this stream (J/s).

  • Crude oil is pumped at a rate of 15.0 kg/s from a well 220 m deep to a storage tank 20 m above ground level. Calculate the rate at which potential energy increases (J/s).


Energy transfer

Energy Transfer

  • Exchange of fluid across system boundary

  • Heat (Q) across system boundary due to DT from Thigh to T low

    (positive Q from surroundings to system)

  • Work (W) across system boundary due to force, torque, voltage, etc.

    (positive W from surroundings to system)


Dimensions and units

Dimensions and Units

Energy [=] force * distance [=] M/L / t2 *L = M*L2/ t2

Energy [=] N*m = J, dyne * cm = erg, ft lbf


Extra practice problems

Extra Practice Problems

Handout Problem Set: IV-1 – IV-8


Physical properties

Physical Properties

• Extensive Properties = ƒ(amount of material)

examples: mass, volume, Ek, Ep

  • Intensive Properties ≠ ƒ(amount of material)

    examples: T, p,r

  • Specific Property: an intensive property obtained by dividing an extensive property by an amount of material

    example: specific volume is volume per unit mass

    example: Ek / m = Êk

    example: Ĥ  = Û + p V, where  Ĥ is enthalpy

    Û accounts for thermal effects on molecular motion, accounts for both thermal and pressure effects.

    Ĥ accounts for both thermal and pressure effects


State variables vs path functions

State Variables vs Path Functions

Consider a property l that is a function of x and y.

l(x,y) is a point function if & only if the finite Dl (or differential dl) is independent of path.

• that is, Dl = lf – li ... change in l is obtained by difference

• in this case, l is a point function or state variable

If l is not a point function or state variable - it is a path function or path variable

Consider the figure below showing a person starting at initial Point 1 and later at final Point 2

Consider path A and path B position does not depend on the path taken; that is,

position is apoint function


Energy

• position change is calculated by Dl = lf – li

• for a round trip ∫dl = lf – li = 0

• the work required to move from 1 to 2 depends on the path (A or B); that is, work is apath function

• for a round trip ∫dl = lf – li ≠ 0and depends on path

• all intensive properties are state variables (point functions)

• two important energy balance path functions are Q & W

2

Y

B

A

1

X


Internal energy u

Internal Energy (U)

Û = Û(T,V) ... Specific Internal Energy ... molecular motion

• Û is a point function because it only depends on the amount of energy at time of determination

• DÛ can be calculated using DÛ = ∫Cv(T)dT where Cv is the heat capacity at constant volume

• For ideal gas: this equation is accurate because Û is not a function of V

• For solids & liquids: this equation is a good approximation because Û is not a strong function of V

• For real gases: this equation is valid only if V is constant

• Note: only differences in Û can be calculated


Enthalpy

Enthalpy (Ĥ)

 Ĥ= Û + pV , where p is absolute pressure

• Ĥ is a point function = combination of two point functions - Ĥ =  Ĥ (T,p)

• Let Cp .... heat capacity at constant pressure

• DĤ  can be calculated using the following - DĤ = ∫Cp(T)dT

• for ideal gas: this equation is accurate because H is not a function of p

• for real gases: this equation is valid only if p is constant

• for solids & liquids: add D(V p) to RHS of equation

• Note: only differences in  Ĥ can be calculated, must be a reference state where Ĥ = 0


Heat capacities

Heat Capacities

Heat capacity definition - amount of heat /energy required to raise a unit of mass

by one degree

• Cp & Cv [=] J /mole °C [=] Btu / lbm °F ... functions of Temperature

• ideal gas: Cp = Cv + R, where R = Gas Constant.

• real gas: see Figure 23.4

• liquids and solids: Cv ~ Cp

  • Cp0 (J / gmol DK) → Cp0 = a + b(T) + c(T)2 + d(T)3 where a, b, c, d given in Appendix E

• for mixtures: Cp' = ∑ yi Cpi & Cv' = ∑ yi Cvi

• to obtain DH  for any process use DH  = ∫Cp(T) dT


Example

Example

Determine the enthalpy change for CO2 gas cooled from 300°C to 40°C, Compare the calculated value to the value obtained using Figure 23.4 at the average temperature of 170 °C.

a = 36.11 X 10-3

b = 4.233 X 10-5

c = -2.887 X10-8

d = 7.464X10-12


Enthalpy tables

Enthalpy Tables

Since only DĤ  can be calculated, define reference condition (T, p, state of aggregation) at this reference condition set   Ĥ = 0 then tabulated value is H  from this condition. This is, establishing a relative scale just like Celsius and Fahrenheit did for T

Different texts & different handbooks use different reference conditions so the tabulated values differ, but DĤ  is the same regardless of source of data but you can not mix data from sources with different reference points.


Steam tables

Steam Tables


P h diagram

P-H Diagram


Example1

Example

Steam at 150 psia (absolute) with 350°F of superheat is fed to a turbine at a rate of 2000 lbm/hr. The turbine operation is adiabatic, and the effluent is saturated vapor at 15 psia. Calculate the rate of enthalpy change in the turbine (kilowatts).


Extra practice problems1

Extra Practice Problems

Problem Handout Set: IV-15 – IV-49


Phase change operations

Phase Change Operations

When changing phases, bonds are broken in the first phase and reformed in the second phase

• if breaking old bonds requires more energy than is released by forming new bonds, then energy must be transferred into the system

• energy transfer in the form of heat

• for a closed system: DU + DEk + DEp = Q + W

• adding heat to system causes internal energy to increase but does not cause the temperature to increase

• use DĤ  rather than DÛ

• since DĤ  = DÛ + D(pV ) balance is now Q = DĤ  – D(pV)


Phase change operations1

Phase Change Operations

• if the phase change is liquid ↔ vapor / gas.

• assumed the gas is ideal, then Q = DĤ  – RDT

• if phase change is liquid to solid at constant T & p

• then Q = DĤ  ... heat causes a change in specific enthalpy

Consider three important phase changes

• for each change, DĤ  = strong ƒ(T) & weak ƒ(p)

• Specific enthalpy change of VAPORIZATION: DĤvap

• liquid →vaporat constant T & p

• Specific enthalpy change of CONDENSATION: DĤcond

• vapor → liquidat constant T & p

• DĤcond = – DĤ vap ...... usually tabulate as DĤvap


Phase change operations2

Phase Change Operations

• Specific enthalpy change of MELTING: DĤm

• solid → liquid at constant T & p

• Specific enthalpy change of FREEZING: DĤfreeze

• liquid →solid at constant T & p

• DĤfreeze = – DĤm

• Specific enthalpy change of SUBLIMATION: DHs

• solid → vapor at constant T & p

• each of these three quantities is the amount of energy required to change the phase of one unit mass or mole at constant T & p

• Also known as latent heat of .... from the Latin for “hidden”

• heat is added but T does not increase

• each of the DĤ change = ƒ(T)

• we usually refer to the normal boiling point or normal melting / freezing point .which occurs at p = 1 atm


Phase change data

Phase Change Data


Phase change data1

Phase Change Data


Phase change data2

Phase Change Data


State properties and hypothetical paths

State Properties and Hypothetical Paths

Even when you cannot find tabulated enthalpies for a substance, you will often find data that allows you to calculate enthalpy changes associated with the following:

• change in pressure at constant T

• change in T at constant p

• change in phase at a specific T & p (usually 1 atm)

• to determine DH  for a process, use a hypothetical process path for which you can calculate DH  for each step.


Example2

Example

If you are given DĤm (T1,1 atm) and you wish to calculate

DĤm ( T2 , P atm), make use of hypothetical process path:

(solid, T2, p atm)

(liquid, T2,p atm)

DĤm (T2)

∫ CpdT

∫ CpdT

(solid, T1,1 atm)

DĤm (T1)

(liquid, T1,1 atm)


10 minute problem

10 Minute Problem

Calculate DĤ  (J / gmol) for the process in which ice at –5°C and 1 atm is converted to steam at 300°C and 5 atm.

Super Heated Steam T= 300 °C

P = 5 atm

Saturated Water

T= 100 °C

P = 1 atm

DĤ Steam Tables

DĤ  = ∫ Cp dT

Ice

T= 0 °C

P = 1 atm

Subcooled Water

T= 0 °C

P = 1 atm

DĤfusion

Data:

Cp(Ice) = 23.7 J/gmol DK

Cp (water) = 75.4 J/gmol DK

DĤ  = ∫ Cp dT

Ice

T= -5 °C

P = 1 atm


Closed systems

Closed Systems

Integral Energy Balance

Summation of energy changes between two points in time (initial and final)

Accumulation = final – initital

final = Uf + Ekf +Epf

initial = Ui + Eki + Epi

transfer = Q + W

First Law of Thermodynamics for Closed System:

DU + DEk + DEp = Q + W

  • If T surroundings = T system, Q = 0

  • If well insulated or adiabatic, Q = 0

  • If no change in T, phase, or composition, DU = 0

  • W is done against a resistance

  • W = 0 if no moving parts (shaft work)


Internal energy u1

Internal Energy (U)

Û = Û(T,V) ... Specific Internal Energy ... molecular motion

• Û is a point function because it only depends on the amount of energy at time of determination

• DÛ can be calculated using DÛ = ∫Cv(T)dT where Cv is the heat capacity at constant volume

• For ideal gas: this equation is accurate because Û is not a function of V

• For solids & liquids: this equation is a good approximation because Û is not a strong function of V

• For real gases: this equation is valid only if V is constant

• Note: only differences in Û can be calculated


Example3

Example

A gas is contained in a cylinder fitted with a movable piston. The initial gas temperature is 25°C. The cylinder is placed in boiling water with the piston held in a fixed position. Heat in the amount of 2 kcal is

absorbed by the gas, which equilibrates at 100°C (and a higher pressure). The piston is then released, and the gas does 100 J of work in moving the piston to its new equilibrium position. The final gas temperature is 100°C.

Write an energy balance equation for each of the two stages of this process, and in each case solve for the unknown energy term in the equation. In solving this problem, consider the gas in the cylinder to

be the system, neglect the change in potential energy of the gas as the piston moves vertically, and assume the gas behaves ideally. Express all energies in joules.


Extra practice problems2

Extra Practice Problems

Handout Problem Set IV-9 - IV-14


Open continuous flow systems

Open (Continuous Flow) Systems


Balances on open systems

Balances On Open Systems

• Continuous Process (heating water flowing in pipe)

– continuous flowing streams; thus, use flow rates

– Differential balance = analysis at an instant in time

• Consider W as the sum of two types of work: Ws + Wflow

• Ws = Shaft Work or Mechanical Work

• Due to moving parts within system (turbine, propeller)

• Due to movement of system boundary (piston)


Energy

• Wflow = Flow Work = pV Work due to fluid acting on adjacent fluid: in – out

• Consider pipe full of flowing fluid due to Dp ... p1 > p2

• Volume of system = length (L) l area (A)

• force new fluid volume (dashed box) into the system

L

• likewise Wout = poutVout

• thus, Wf = piVi – poVo

• for multiple inlets and outlets Wf = ∑ piVi – ∑ poVo

• since

• multiple inlets and outlets

A


Energy

• Wtotal = Wshaft + Wflow = Wshaft + ∑ pi mi Vi – ∑ pomoVo

Do energy balance on the system

• Accumulation = In – Out + Generated – Consumed

• at steady state balance yields: In = Out

• isolating “in”/“positive” on LHS ... ∑Ei + Q + W = ∑Eo

• rearranging gives: Q + W = ∑Eo – ∑Ei

• substituting for W gives: Q + Wshaft + ∑pimiVi – ∑pomoVo =

∑m(Ûo + Êko + Êpo ) – ∑m (Ûi + Êki + Êpi )

• rearranging gives Q + Wshaft = ∑mo (Ûo+ poVo + Êko + Êpo ) –∑mi (Ûi+ piVi + Êki + Êpi)

• recall   Ĥ = Û + pV ;

  • therefore, Q + Wshaft = ∑mo ( Ĥo + Êko + Êpo ) - ∑mi ( Ĥi + Êki + Êpi)


First law of thermodynamics

First Law of Thermodynamics

Open Systems

DH + DEk + DEp = Q + Wshaft

• if Tsurround = Tsystem, Q = 0

• if well insulated or adiabatic, Q = 0

• Wshaft = zero if there are no moving parts


Enthalpy1

Enthalpy (Ĥ)

 Ĥ= Û + pV , where p is absolute pressure

• Ĥ is a point function = combination of two point functions - Ĥ =  Ĥ (T,p)

• Let Cp .... heat capacity at constant pressure

• DĤ  can be calculated using the following - DĤ = ∫Cp(T)dT

• for ideal gas: this equation is accurate because H is not a function of p

• for real gases: this equation is valid only if p is constant

• for solids & liquids: add D(V p) to RHS of equation

• Note: only differences in  Ĥ can be calculated, must be a reference state where Ĥ = 0


Examples1

Examples

500 kg of water vapor per hour drive a turbine. The steam enters at 44 atm and 450°C at a velocity of 60 m/s. It leaves 5 m lower at atmospheric pressure and 360 m/s. The turbine delivers W = 700 kW and there is a heat loss of 10,000 kcal/hr. Calculate the change in specific enthalpy in kJ/kg.

Two streams of liquid water (one 120 kg/min @ 30°C and the other 175 kg/min @ 65°C) are mixed to form the feed to a boiler operating at 17 bars absolute. The exiting steam emerges from the boiler through a 6-cm ID pipe. Calculate the required heat input to the boiler in kilojoules/minute if the emerging steam is saturated at the boiler pressure. Neglect the kinetic energies of the liquid inlet streams.


10 minute problem1

10 Minute Problem

Saturated steam at 1 atm is discharged from a turbine at a rate of 1000 kg/hr. Superheated steam at 300°C and 1 atm is needed as a feed to a heat exchanger; to produce it, the turbine discharge stream is mixed with superheated steam available from a second source at 400°C and 1 atm. The mixing unit operates adiabatically. Calculate the rate of steam (kg H2O(v)/hr) at 300°C produced, and the required volumetric flow rate (m3 H2O(v)/hr) of the 400°C steam


Extra practice problems3

Extra Practice Problems

Problem Set Handout: IV-50 – IV-104


Energy balances and chemical reactions

Energy Balances and Chemical Reactions


Enthalpy heat of reaction

Enthalpy (Heat) of Reaction

Define: DHrxn (T,p) = ∑Hproducts – ∑Hreactants

• true if and only if (i) feed is in stoichiometric quantities

(ii) the reaction is complete ... review

(iii) feed and products are at the same T,p

At low to moderate pressure, DHrxn ≠ƒ(p)

• DHrxn < 0 .... means product has less enthalpy than reactants

(referred to as exothermic reaction at constant T)

• DHrxn > 0 .... product has more enthalpy than reactants

(referred to as endothermic reaction at constant T)


Enthalpy heat of reaction1

Enthalpy (Heat) of Reaction

Consider the following reaction:

a0R0 + a1R1 + a2R2 + .... → b1P1 + b2P2 + b3P3 + ...

Divide by a0 so that one coefficient is unity yields:

R0 + c1R1 + c2R2 + .... → g1P1 + g2P2 + g3P3 +

Specific enthalpy of reaction (in terms of R0 reacted)

• DHrxn (kJ or kcal/ mole R0) = ∑(gjHP,j) – ∑ (cHR,i +  HR,o)

• specific enthalpy of rxn in terms of Rior Pj

DH’rxn (kJ or kcal/ mole Rior Pj ) = DH rxn (kJ or kcal/ mole R0)

ci or gi


Examples2

Examples

Example

The reaction 2A + 3B → 4C is carried out in a continuous reactor. Heat flowing out of the reactor is 50 kJ.

Open System Energy Balance: DH + DEk + DEp = Q + Wshaft Q = – 50 kJ

DHrxn (kJ/mole A reacted) = –50/2 = –25

DH’rxn (kJ/mole B reacted) = –50/3

DH’’rxn (kJ/mole C produced) = –50/4

Example:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O .... DH = –5756 kJ

DHrxn (kJ/mole C4H10 reacted) = –2878

DH’rxn (kJ/mole O2 reacted) = –442.8

DH’’rxn (kJ/mole CO2 produced) = –719.5

DH’’’rxn  (kJ/mole H2O produced) = –575.6


Standard heats of reaction

Standard Heats of Reaction

Rather than tabulate DHrxn at every possible T,p combination we use :

Standard Heat of Reaction(DH°rxn  )  @ 25°C & 1 atm

• superscript ° indicates “standard T and p”

If a reaction is at some other pressure:

• assume at low to moderate pressure, DHrxn  ≠ƒ(p)

If a reaction is at some other temperature,

• use hypothetical path because H is a state variable

• DHrxn = ƒ(state of aggregation)

Example: A(g) + B(g) → C(g) + D(g) ...... DHrxn,1

A(g) + B(g) → C(g) + D(liq) ...... DHrxn,2

DHrxn,1 ≠ DHrxn,2they differ by the amount of heat required to evaporate D


Hess s law

Hess’s Law

If a series of reactions (1, 2, 3) can be rearranged/adjusted by a series of algebraic operations to yield the desired reaction equation, then the heat (enthalpy) of the desired reaction can be obtained by performing the same algebraic operations to DH°rxn,1, DH°rxn,2,DH°rxn,3


Example4

Example

2 C(s) + O2 (g) → 2 CO(g)

Determine DH°rxn ? .... can we do this experimentally ?

No cannot be carried out at 25°C & 1 atm

At higher temperatures: C + O2 → CO2 or a mix of CO2 & CO, but never pure CO

Hypothetical path with reactions for which we know or can experimentally measure DH°rxn,1

From tables of DH°rxn or experiments we can get:

1) C(s) + O2 (g) → CO2 (g) ..... DH°rxn,1  = –393.51 kJ/mole C

2) CO(g) + ½ O2→CO2 (g) ..... DH°rxn,2  = –282.99 kJ/mole CO


Energy

DH°rxn

C + ½ O2

CO

DH°rxn,1

-DH°rxn,2

CO2

  • Following the hypothetical path gives:

    • DH°rxn= DH°rxn,1 + (- DH°rxn,2 )= –393.51 + 282.99 = – 110.52

  • Same result by subtracting equation 2 from equation 1 to get 3

    • C + O2 – CO – ½ O2 → CO2 – CO2

  • • which gives C + ½ O2 → CO and DH°rxn= DH°rxn,1 + (- DH°rxn,2 )


  • Example5

    Example

    The standard heats of the following combustion reactions have been determined:

    (a.) C2H6 + 7/2 O2 → 2 CO2 + 3H2O : DH°rxn,a= –1559.8 kJ/mole

    (b.) C + O2 → CO2 : DH°rxn,b= –393.5 kJ/mole

    (c.) H2+ ½ O2 → H2O : DH°rxn,c= –285.8 kJ/mole

    Use Hess’s law and the given heats of reaction to determine the standard heat of the reaction:

    (d.) 2C + 3 H2 → C2H6 : DH°rxn,d= ?


    Enthalpy heat of formation

    Enthalpy (Heat) of Formation

    • Enthalpy (heat) of formation – enthalpy (heat) of a reaction in which one mole of a compound is formed from its atomic constituents as they appear in nature [for example, O2 rather than atomic O]

    Standard Heat of Formation ...... DH°f@ 25°C & 1 atm

    • must be in the state of aggregation that would exist at 25°C & 1 atm

    • for atomic species / elements DH°f = 0

    Hess's Law can be applied to DH°f just as we did for DH°rxn


    Table b 1

    Table B.1


    Example6

    Example

    Formation of benzene liquid state at 25°C & 1 atm

    6C(s) + 3H2(g) → C6H6(liq): DH°f = 48.66 kJ/mole

    • formation of benzene vapor, Table B.1 lists 82.93 kJ/mole

    • why is that value not equal to 48.66 + 30.77 (Table B.1)?

    Formation of ammonium nitrate NH4NO3 crystals (c)

    N2(g) + 2H2(g) + 3/2 O2(g)→NH4NO3(c): DH°f = –366.1 kJ/mole


    Standard enthalpy heat of combustion

    Standard Enthalpy (Heat) of Combustion

    Heat of Combustion - enthalpy (heat) of a reaction @ 25°C & 1 atm in which one mole of a compound undergoes complete combustion:

    • all C → CO2 (g)

    • all H → H2O(liq)

    • all S → SO2(g)

    • all N →N2(g)

    • all Cl → an infinitely dilute aqueous HCl solution

    • for O2(g) & combustion products listed above, DHC= 0

    {that is, you cannot oxidize combustion products}

    {thus, DÛ°c = 0 for combustion products and O2 (g)}

    Hess's Law can be applied to DH°C just as we did for DH°rxn

    Note the following:

    • DH°rxn = ∑(gjDH°f,j) – ∑ (ciDH°f,i ) --- products – reactants

    • DH°rxn = ∑(gjDH°c,j) – ∑ (ciDH°c,i ) --- products - reactants


    Table b 11

    Table B.1


    Heating value definitions

    Heating Value Definitions

    Q

    Products

    (vapor H2O)

    Air

    LHV

    Combustion

    Chamber

    Fuel

    Products

    (liquid H2O)

    HHV


    Examples3

    Examples

    • Use DH°fdata to determine the standard heat of reaction for the combustion of liquid n-pentane, assuming H2O(liq) is a combustion product:

      C5H12 (liq) + 8 O2 (g) → 5 CO2 (g) + 6 H2O(liq)

      2.Using DH°cdata, calculate DH°rxn(kcal/mole) for the reaction:

      2 CH4 (g) →C2H2 (g) + 3 H2 (g)


    Energy balances with chemical reactions

    Energy Balances with Chemical Reactions

    Q + Wshaft = DH + DEk + DEp

    • DH includes thermal effects as before plus reactions

    • often useful to define reference conditions

    Two methods to solve such problems:


    Enthalpy table t ref 298 k

    Enthalpy Table (TRef = 298 K)


    Method 1

    Method 1

    Convenient when you know or can calculate DH°rxn

    Reference Conditions:

    • reaction participants:  H = 0 for reactants & products at 25°C

    • this is the state for which DH°rxn is known.

    • inert species:  H = 0 for the inerts at any convenient T

    • use 25°C or inlet T or outlet T

    Where:

    • s = any reactant or product

    • ns = moles of “s” produced or consumed

    •gs & cs= stoichiometric coefficients

    cC+dD

    T2

    aA+bB

    T1

    DH°rxn

    aA+bB

    cC+dD


    Method 11

    Method 1

    If uA moles of A remains unreacted, treat it as uA moles of “product” A and take it from

    25°C to T2

    Inert / non-reactive components enter at T1 and leave at T2, thus, treated as we have in the past

    For multiple reactions, account for each reaction:


    Example 9 5 1

    Example 9.5-1

    The standard heat of reaction DH°rxn at 25°C and 1 atm for the oxidation of ammonia can be obtained from heat of formation data (Table F.1). One hundred g-moles NH3 /hr and 200 g-moles O2 /hr at 25°C are fed into a reactor in which the NH3 is completely consumed. The product stream emerges as a gas at 300°C.

    4NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g)

    Calculate the heat transferred to or from the reactor, assuming operation at 1 atm

    .


    10 minute problem2

    10 Minute Problem

    The dehydrogenation of ethanol to form acetaldehyde is carried out in an adiabatic reactor.

    C2H5OH (g) → CH3CHO(g) + H2(g) : DH°rxn(25°C) = +68.95 kJ/g-mole

    Ethanol vapor is fed to the reactor at 300°C, and a conversion of 30 % is obtained. Calculate the product temperature (°C) using the following heat capacities.

    Note: ethanol and acetaldehyde are gas throughout the process because you are operating at a pressure for which they do not condense.

    C2H5OH(g) : Cp = 0.110 [email protected]

    CH3CHO(g) : Cp = 0.080 [email protected]

    H2(g) : Cp = 0.029 [email protected]


    Method 2

    Method 2

    Easier than Method 1 for multiple reactions

    Reference Conditions:

    • reaction participants:  H = 0 for the elements that constitute reactants & products at 25°C

    • inert species:  H = 0 for the inerts at any convenient T

    • use 25°C or inlet T or outlet T

    Where:  H for a reactant or product is the sum of the enthalpy of formation at 25°C (DH°f) plus any specific sensible or latent heat required to go from 25°C to T1 or T2

    cC+dD

    T2

    aA+bB

    T1

    aA+bB 25°C DH°f

    cC+dD 25°C DH°f


    Example 9 5 2

    Example 9.5-2

    Methane is oxidized with air to produce formaldehyde in a continuous reactor.

    CH4(g) + O2 (g) → HCHO(g) + H2O(g) .....(1)

    A competing side reaction is the combustion of methane to form CO2.

    CH4 (g) + 2 O2 (g) → CO2 + 2 H2O(g) ...(2)

    100 moles/min of CH4 is fed to the reactor at 25°C along with enough air (at 100°C) to provide 100 moles O2 /min.

    The following product stream exits at 150°C: 60 moles CH4 /min, 30 moles HCHO/min, 10 moles CO2 /min, 50 moles H2O/min, 50 moles O2 /min, 376 moles N2 /min. The pressure of 1 atm is low enough to assume ideal gas. At what rate (kJ/min) must heat be withdrawn?


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