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Lecture 5 Energy Balance. Ch 61. ENERGY BALANCE. Concerned with energy changes and energy flow in a chemical process. Conservation of energy – first law of thermodynamics i.e. accumulation of energy in a system = energy input – energy output. Forms of energy. Potential energy ( mgh )

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energy balance
ENERGY BALANCE
  • Concerned with energy changes and energy flow in a chemical process.
  • Conservation of energy – first law of thermodynamics i.e.

accumulation of energy in a system = energy input – energy output

forms of energy
Forms of energy
  • Potential energy (mgh)
  • Kinetic energy (1/2 mv2)
  • Thermal energy – heat (Q) supplied to or removed from a process
  • Work energy – e.g. work done by a pump (W) to transport fluids
  • Internal energy (U) of molecules

m – mass (kg)

g – gravitational constant, 9.81 ms-2

v – velocity, ms-1

energy balance1
Energy balance

W

system

mass out

mass in

Hin

Hout

Q

Paul Ashall, 2008

iupac convention
IUPAC convention

- heat transferred to a system is +ve and heat transferred from a system is –ve

- work done on a system is+ve and work done by a system is -ve

steady state non steady state
STEADY STATE/NON-STEADY STATE
  • Non steady state - accumulation/depletion of energy in system
slide7
Uses
  • Heat required for a process
  • Rate of heat removal from a process
  • Heat transfer/design of heat exchangers
  • Process design to determine energy requirements of a process
  • Pump power requirements (mechanical energy balance)
  • Pattern of energy usage in operation
  • Process control
  • Process design & development
  • etc
slide8

1) No mass transfer (closed or batch)

∆E = Q + W

2) No accumulation of energy, no mass transfer (m1 = m2= m)

∆E = 0 Q = -W

slide10

Air is being compressed from 100 kPa at 255 K (H = 489 kJ/kg) to 1000kPa and 278 K (H = 509 kJ/kg) The exit velocity of the air from the compressor is 60 m/s. What is the power required (in kW) for the compressor if the load is 100 kg/hr of air?

sample problem
SAMPLE PROBLEM

Water is pumped from the bottom of a well 4.6 m deep at a rate of 760 L/hour into a vented storage tank to maintain a level of water in a tank 50 m above the ground. To prevent freezing in the winter a small heater puts 31,600 kJ/hr into the water during its transfer from the well to the storage tank. Heat is lost from the whole system at a constant rate of 26,400 kJ/hr. What is the temperature of the water as it enters the storage tank assuming that the well water is at 1.6oC? A 2-hp pump is used. About 55% of the rated horse power goes into the work of pumping and the rest is dissipated as heat to atmosphere.

enthalphy balance
ENTHALPHY BALANCE
  • p.e., k.e., W terms = 0
  • Q = H2 – H1 or Q = ΔH

, where H2 is the total enthalpy of output streams and H1is the total enthalpy of input streams, Q is the difference in total enthalpy i.e. the enthalpy (heat) transferred to or from the system

continued
continued
  • Q –ve (H1>H2), heat removed from system
  • Q +ve (H2>H1), heat supplied to system.
example steam boiler
Example – steam boiler

Two input streams: stream 1- 120 kg/min. water, 30 deg cent., H = 125.7 kJ/kg; stream 2 – 175 kg/min, 65 deg cent, H= 272 kJ/kg

One output stream: 295 kg/min. saturated steam(17 atm., 204 deg cent.), H = 2793.4 kJ/kg

continued1
continued

Ignore k.e. and p.e. terms relative to enthalpy changes for processes involving phase changes, chemical reactions, large temperature changes etc

Q = ΔH (enthalpy balance)

Basis for calculation 1 min.

Steady state

Q = Hout – Hin

Q = [295 x 2793.4] – [(120 x 125.7) + (175 x 272)]

Q = + 7.67 x 105 kJ/min

steam tables
STEAM TABLES
  • Enthalpy values (H kJ/kg) at various P, T
enthalpy changes
Enthalpy changes
  • Change of T at constant P
  • Change of P at constant T
  • Change of phase
  • Solution
  • Mixing
  • Chemical reaction
  • crystallisation
involving chemical reactions
INVOLVING CHEMICAL REACTIONS

Heat of Reaction:

∆Hrxn= ∑∆Hfo(products) - ∑∆Hfo(reactants)

sample problem1
SAMPLE PROBLEM:

Calculate the ∆Hrxn for the following reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Given the following ∆Hfo/mole at 25oC.

NH3(g): -46.191 kJ/mol

NO(g): +90.374 kJ/mol

H2O(g): -214.826 kJ/mol

energy balance that account for chemical reactions
ENERGY BALANCE THAT ACCOUNT FOR CHEMICAL REACTIONS:

Most common:

  • What is the temperature of the incoming or exit streams
  • How much material must be introduced into the entering stream to provide for a specific amount of heat transfer.
aa bb cc dd
aA + bB→ cC + dD

T3

T1

C

A

products

reactants

T2

D

B

T4

sample problem2
SAMPLE PROBLEM:

An iron pyrite ore containing 85.0% FeS2 and 15.0% inert materials (G) is roasted with an amount of air equal to 200% excess air according to the reaction

4FeS2 + 11O2→ 2Fe2O3 + 8 SO2

in order to produce SO2. All the inert materials plus the Fe2O3 end up in the solid waste product, whick analyzes at 4.0% FeS2. Determine the heat transfer per kg of ore to keep the product stream at 25oC if the entering streams are at 25oC.

latent heats phase changes
Latent heats (phase changes)
  • Vapourisation (L to V)
  • Melting (S to L)
  • Sublimation (S to V)
mechanical energy balance
Mechanical energy balance
  • Consider mechanical energy terms only
  • Application to flow of liquids

ΔP + Δ v2+g Δh +F = W

ρ 2

where W is work done on system by a pump and F is frictional energy loss in system (J/kg)

ΔP = P2 – P1; Δ v2 = v22 –v12; Δh = h2 –h1

  • Bernoulli equation (F=0, W=0)
example bernoulli eqtn
Example - Bernoulli eqtn.

Water flows between two points 1,2. The volumetric flow rate is 20 litres/min. Point 2 is 50 m higher than point 1. The pipe internal diameters are 0.5 cm at point 1 and 1 cm at point 2. The pressure at point 2 is 1 atm..

Calculate the pressure at point 2.

Paul Ashall, 2008

continued2
continued

ΔP/ρ + Δv2/2 + gΔh +F = W

ΔP = P2 – P1 (Pa)

Δv2 = v22 – v12

Δh = h2 - h1 (m)

F= frictional energy loss (mechanical energy loss to system) (J/kg)

W = work done on system by pump (J/kg)

ρ = 1000 kg/m3

continued3
continued

Volumetric flow is 20/(1000.60) m3/s

= 0.000333 m3/s

v1 = 0.000333/(π(0.0025)2) = 16.97 m/s

v2 = 0.000333/ (π(0.005)2) = 4.24 m/s

(101325 - P1)/1000 + [(4.24)2 – (16.97)2]/2 + 9.81.50 = 0

P1 = 456825 Pa (4.6 bar)

sensible heat enthalpy calculations
Sensible heat/enthalpy calculations
  • ‘Sensible’ heat – heat/enthalpy that must be transferred to raise or lower the temperature of a substance or mixture of substances.
  • Heat capacities/specific heats (solids, liquids, gases,vapours)
  • Heat capacity/specific heat at constant P, Cp(T) = dH/dT or ΔH = integral Cp(T)dT between limits T2 and T1
  • Use of mean heat capacities/specific heats over a temperature range
  • Use of simple empirical equations to describe the variation of Cp with T
continued4
continued

e.g. Cp = a + bT + cT2 + dT3

,where a, b, c, d are coefficients

ΔH = integralCpdT between limits T2, T1

ΔH = [aT + bT2 + cT3 + dT4]

2 3 4

Calculate values for T = T2, T1 and subtract

Note: T may be in deg cent or K - check units for Cp!

example
Example

Calculate the enthalpy required to heat a stream of nitrogen gas flowing at 100 mole/min., through a gas heater from 20 to 100 deg. cent.

(use mean Cp value 29.1J mol-1 K-1or Cp = 29 + 0.22 x 10-2T + 0.572 x 10-5T2 – 2.87 x 10-9 T3, where T is in deg cent)

heat capacity specific heat data
Heat capacity/specific heat data
  • Felder & Rousseau pp372/373 and Table B10
  • Perry’s Chemical Engineers Handbook
  • The properties of gases and liquids, R. Reid et al, 4th edition, McGraw Hill, 1987
  • Estimating thermochemical properties of liquids part 7- heat capacity, P. Gold & G.Ogle, Chem. Eng., 1969, p130
  • Coulson & Richardson Chem. Eng., Vol. 6, 3rd edition, ch. 8, pp321-324
  • ‘PhysProps’
example change of phase
Example – change of phase

A feed stream to a distillation unit contains an equimolar mixture of benzene and toluene at 10 deg cent.The vapour stream from the top of the column contains 68.4 mol % benzene at 50 deg cent. and the liquid stream from the bottom of the column contains 40 mol% benzene at 50 deg cent.

[Need Cp (benzene, liquid), Cp (toluene, liquid), Cp (benzene, vapour), Cp (toluene, vapour), latent heat of vapourisation benzene, latent heat of vapourisation toluene.]

energy balances on systems involving chemical reaction
Energy balances on systems involving chemical reaction
  • Standard heat of formation (ΔHof) – heat of reaction when product is formed from its elements in their standard states at 298 K, 1 atm. (kJ/mol)

aA + bB cC + dD

-a -b +c +d (stoichiometric coefficients, νi)

ΔHofA, ΔHofB, ΔHofC, ΔHofD (heats of formation)

ΔHoR = c ΔHofC + d ΔHofD - a ΔHofA - bΔHofB

Paul Ashall, 2008

heat enthalpy of reaction
Heat (enthalpy) of reaction
  • ΔHoR –ve (exothermic reaction)
  • ΔHoR +ve (endothermic reaction)

Paul Ashall, 2008

enthalphy balance reactor
ENTHALPHY BALANCE - REACTOR

Qp = Hproducts – Hreactants + Qr

Qp – heat transferred to or from process

Qr – reaction heat (ζ ΔHoR), where ζ is extent of reaction and is equal to [moles component,i, out – moles component i, in]/ νi

Paul Ashall, 2008

slide44

system

Hproducts

Hreactants

Qr

+ve

-ve

Qp

Note: enthalpy values must be calculated with reference to a

temperature of 25 deg cent

energy balance techniques
ENERGY BALANCE TECHNIQUES
  • Complete mass balance/molar balance
  • Calculate all enthalpy changes between process conditions and standard/reference conditions for all components at start (input) and finish (output).
  • Consider any additional enthalpy changes
  • Solve enthalpy balance equation

Paul Ashall, 2008

energy balance techniques1
ENERGY BALANCE TECHNIQUES
  • Adiabatic temperature: Qp = 0

Paul Ashall, 2008

examples
EXAMPLES
  • Reactor
  • Crystalliser
  • Drier
  • Distillation
references
References
  • The Properties of Gases and Liquids, R. Reid
  • Elementary Principles of Chemical Processes, R.M.Felder and R.W.Rousseau

Paul Ashall, 2008

sample problem4
SAMPLE PROBLEM:

Limestone (CaCO3) is converted into CaO in a continuous vertical kiln. Heat is supplied by combustion of natural gas (CH4) in direct contact with limestone using 50% excess air. Determine the kg of CaCO3 that can be processed per kg of natural gas. Assume that the following heat capacities apply.

Cp of CaCO3 = 234 J/mole –oC

Cp of CaO = 111 j/mole - oC