lecture 5 energy balance
DESCRIPTION
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 ) - PowerPoint PPT PresentationTRANSCRIPT
LECTURE 5ENERGY 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) 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
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Energy balance
systemmass in
Hin
mass out
Hout
W
Q
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
Non steady state - accumulation/depletion of energy in system
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
1) No mass transfer (closed or batch)∆E = Q + W2) No accumulation of energy, no mass transfer (m1 = m2= m)
∆E = 0 Q = -W
3. No accumulation of energy but with mass flowQ + W = ∆ [(H + K + P)m]
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 PROBLEMWater 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 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 Q –ve (H1>H2), heat removed from
system Q +ve (H2>H1), heat supplied to
system.
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
continuedIgnore 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 stateQ = Hout – HinQ = [295 x 2793.4] – [(120 x 125.7) + (175 x 272)] Q = + 7.67 x 105 kJ/min
STEAM TABLES Enthalpy values (H kJ/kg) at various P, T
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
Heat of Reaction:∆Hrxn= ∑∆Hf
o(products) - ∑∆Hf
o(reactants)
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/molH2O(g): -214.826 kJ/mol
∆H = -904.696 kJ/mol
ENERGY BALANCE THAT ACCOUNT FOR CHEMICAL REACTIONS:
Most common:1) What is the temperature of the
incoming or exit streams2) How much material must be introduced
into the entering stream to provide for a specific amount of heat transfer.
aA + bB → cC + dD
T4
T2
T3T1
reactants productsA
B
C
D
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.
ProductsComponents Amount
kmole∆Hf kJ/mol nx∆H
FeS2 -177.9
Fe2O3 -822.156
N2 0
O2 0
SO2 -296.90
ProductsComponents Amount
kmole∆Hf kJ/mol nx∆H
FeS2 -177.9
Fe2O3 -822.156
N2 0
O2 0
SO2 -296.90
Material Balance First!!!!
∆E = 0, W = 0, ∆E = 0, ∆PE = 0, ∆KE = 0
Therefore Q = ∆H
ProductsComponents Amount
kmole∆Hf kJ/mol nx∆H
FeS2 0.0242 -177.9 -4.305
Fe2O3 0.342 -822.156 -281.177
N2 21.998 0 0
O2 3.938 0 0
SO2 1.368 -296.90 -406.159
ReactantsComponents Amount
kmole∆Hf kJ/mol nx∆H
FeS2 0.7803 -177.9 -126.007
Fe2O3 0 -822.156
N2 21.983 0
O2 5.8437 0
SO2 0 -296.90
SAMPLE PROBLEM
Q =
Latent heats (phase changes) Vapourisation (L to V) Melting (S to L) Sublimation (S to V)
Mechanical energy balance Consider mechanical energy terms only Application to flow of liquidsΔP + Δ v2 + g Δh +F = Wρ 2where W is work done on system by a pump and F
is frictional energy loss in system (J/kg)ΔP = P2 – P1; Δ v2 = v2
2 –v12; Δh = h2 –h1
Bernoulli equation (F=0, W=0)
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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.
continuedΔP/ρ + Δv2/2 + gΔh +F = W
ΔP = P2 – P1 (Pa)Δv2 = v2
2 – 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
continuedVolumetric flow is 20/(1000.60) m3/s= 0.000333 m3/sv1 = 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 – 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
continuede.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
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-1 or 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 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 phaseA 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.]
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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, ΔHo
fB, ΔHofC, ΔHo
fD (heats of formation)
ΔHoR = c ΔHo
fC + d ΔHofD - a ΔHo
fA - bΔHofB
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Heat (enthalpy) of reaction ΔHo
R –ve (exothermic reaction) ΔHo
R +ve (endothermic reaction)
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ENTHALPHY BALANCE - REACTOR
Qp = Hproducts – Hreactants + Qr
Qp – heat transferred to or from processQr – reaction heat (ζ ΔHo
R), where ζ is extent of reaction and is equal to [moles component,i, out – moles component i, in]/ νi
system
Qr
HreactantsHproducts
Qp
+ve
-ve
Note: enthalpy values must be calculated with reference to a temperature of 25 deg cent
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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
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ENERGY BALANCE TECHNIQUES Adiabatic temperature: Qp = 0
EXAMPLES Reactor Crystalliser Drier Distillation
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References The Properties of Gases and Liquids, R.
Reid Elementary Principles of Chemical
Processes, R.M.Felder and R.W.Rousseau
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