Transcript SAMPLE PROBLEM - R. M. FABICON's BLOG

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
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Forms of energy
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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 balance
W
mass in
mass out
system
Hin
Hout
Q
Paul Ashall, 2008
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
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Non steady state - accumulation/depletion of
energy in system
Uses
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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 + W
2) No accumulation of energy, no mass transfer (m1 =
m2= m)
∆E = 0 Q = -W
3. No accumulation of energy but with mass flow
Q + 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 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
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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
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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
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
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Enthalpy values (H kJ/kg) at various P, T
Enthalpy changes
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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= ∑∆Hfo(products) - ∑∆Hfo(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/mol
H2O(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
streams
2)
How much material must be introduced into the
entering stream to provide for a specific amount
of heat transfer.
aA + bB → cC + dD
T1
A
reactants
T2
B
T3
C
products
T4
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.
Products
Components
Amount kmole
∆Hf kJ/mol
FeS2
-177.9
Fe2O3
-822.156
N2
0
O2
0
SO2
-296.90
nx∆H
Products
Components
FeS2
Fe2O3
Amount kmole
∆Hf kJ/mol
-177.9
-822.156
N2
0
O2
0
SO2
-296.90
nx∆H
Material Balance First!!!!
∆E = 0, W = 0, ∆E = 0, ∆PE = 0, ∆KE = 0
Therefore Q = ∆H
Products
Components
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
Reactants
Components
Amount kmole
∆Hf kJ/mol
FeS2
0.7803
Fe2O3
0
N2
21.983
0
O2
5.8437
0
SO2
0
-177.9
-822.156
-296.90
nx∆H
-126.007
SAMPLE PROBLEM
Q=
Latent heats (phase changes)
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Vapourisation (L to V)
Melting (S to L)
Sublimation (S to V)
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
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Bernoulli equation (F=0, W=0)
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
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
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
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‘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
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
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
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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
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

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
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ΔHoR –ve (exothermic reaction)
ΔHoR +ve (endothermic reaction)
Paul Ashall, 2008
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
Hreactants
Hproducts
system
Qr
+ve
-ve
Qp
Note: enthalpy values must be calculated with reference to a
temperature of 25 deg cent
ENERGY BALANCE TECHNIQUES
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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 TECHNIQUES
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Adiabatic temperature: Qp = 0
Paul Ashall, 2008
EXAMPLES
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Reactor
Crystalliser
Drier
Distillation
References
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The Properties of Gases and Liquids, R. Reid
Elementary Principles of Chemical Processes,
R.M.Felder and R.W.Rousseau
Paul Ashall, 2008
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