8–1
Energy balance
Md Khairul Islam (Ph.D)
Applied Chemistry & Chemical Engineering University of Rajshahi
Energy
The classical description of energy is the ability of a system to perform work, but as energy exists in so many forms, it is hard to find one comprehensive definition. It is the property of an object that can be transferred from one object to another or converted to different forms but cannot be created or destroyed.
Forms of Energy
The total energy of a system has three components-
Potential energy (mgh); energy due to the motion of the system as a whole relative to some frame reference.
Kinetic energy (1/2 mv2): energy due to the position of the system in a potential field.
Internal energy (U): energy due to the motion of molecules relative to the center of mass of the system.
For closed system, no mass transferred across, energy may be transferred in the following two ways-
Thermal energy – heat (Q) supplied to or removed from a process
Work energy – e.g. work done by a pump (W) to transport fluids m – mass (kg)
g – gravitational constant, 9.81 ms-2 v – velocity, ms-1
Sources of Energy
Renewable sources of energy are available plentiful in nature and are sustainable. These resources of energy can be naturally replenished and are safe for the environment.
Examples: Solar energy, geothermal energy, wind energy, biomass, hydropower and tidal energy.
A non-renewable resource is a natural resource that is found underneath the earth. These type of energy resources do not replenish at the same speed at which it is used. They take millions of years to replenish. The main examples of non-renewable resources are coal, oil and natural gas. Examples: Natural gas, coal, petroleum, nuclear energy and hydrocarbon gas liquids.
Advantages of Renewable Energy
These types of energy sources are environmentally friendly.
They do not release toxic gases like carbon dioxide.
These energy sources do not run out and replenish naturally.
They are safer for our health as they don’t generate toxic residues harmful to people.
They also freely exist in nature.
Disadvantages of Renewable Energy
Solar energy is not sufficiently stored when there is a cloudy day. Only on sunny days, a greater amount of solar energy can be stored.
In the case of windmills, the amount of electricity obtained from one windmill is very less hence this makes the entire setup costs of windmills very high.
Maintenance costs of windmills are very high.
Renewable sources of energy are not abundant at every place on earth.
Obtaining renewable energy can be random because it depends on natural phenomena.
Advantages and Disadvantages of Non-renewable Energy
Advantages
Coal is relatively cheap and hence used up mostly in a thermal power plant.
Produces lots of energy.
Disadvantages
It is non-renewable
It produces pollution.
Renewable Non-renewable The resources that can be
renewed once they are consumed are called renewable sources of energy.
The resources that cannot be
renewed once they are consumed are called non-renewable sources of energy.
These resources do not cause any environmental pollution.
These resources cause environmental pollution.
Renewable resources are inexhaustible.
Non- Renewable resources are exhaustible.
Renewable resources are not affected by human activities.
Non- Renewable resources are affected by human activities.
Examples of Renewable resources- Air, water and solar energy.
Examples of Non-renewable resources- natural gas, coal and nuclear energy.
Differences in Energy sources
Types of Natural Sources of Energy
Conventional Non-conventional These resources are
exhaustible.
These resources are inexhaustible.
These resources cause
pollution as they emit smoke and ash.
These resources are usually pollution-free.
These resources are very expensive to be maintained, stored and transmitted.
These resources are less
expensive for local use and can easily be maintained.
Examples- coal, natural gas, petroleum, and water power.
Examples- solar, biomass, wind, biogas, and tidal, geothermal.
Energy Balance
An energy balance is a consideration of the energy input, output, and consumption or generation in a process or stage. Energy balance is based on the conservation of energy; first law of thermodynamics i.e. accumulation of energy in a system = energy input – energy output.
Energy balances are used in the examination of the various stages of a process, over the whole process and even extending over the total production system from the raw material to the finished product.
Energy Balance on closed system
Final system energy – Initial system energy
= Net energy transferred to the system.
Final system energy = Ui + Eki + Epi Initial system energy =Uf + Ekf + Epf Net energy transfer = Q – W
(Uf – Ui) + (Ekf – Eki) + (Epf – Epi) = Q – W
∆U + ∆Ek + ∆Ep = Q – W
Energy Balance on open system
In open system, mass is transferred across the system boundaries while the process is taking place.
Therefore work must be done on open system to push mass in and work is done on the surrounding by mass that emerges from the systems.
Both work terms must be include in the energy balance for open system
The net work done by an open system
W = Ws + Wfl and Wfl = W out – W in = PoutVout - PinVin Ws = shaft work
rate of work done by the process fluid on a moving part within the system such as a pump rotor.
Wfl = flow work
rate of work done by the fluid at the system outlet minus rate of work done by the fluid at the system inlet.
Steps to follow in Energy Balance Calculations
Prepare a block diagram for the unit.
Write down the available information such as the flow rates of the incoming and outgoing streams, their enthalpies, heat and work input, heat losses, etc.
Try to identify a tie substance.
Select a suitable basis for calculation.
If no chemical reaction takes place within the system, consider whether there is a phase change such as vaporization and condensation, fusion and solidification, sublimation, transition from one crystal structure to another, crystallization etc. Consider enthalpy changes accompanying such a change, if any.
Steps to follow in Energy Balance Calculations
If chemical reactions occur, consider enthalpy changes accompanying the reactions.
Tabulate all energy input items in one column and all output items in another. For convenience, all entries in each column are arranged to be of positive sign. Thus, any work done on the system will appear as an input item while work done by the system will appear as an output item. Heat absorbed is an input item, and heat evolved and/or losses are output items. The enthalpies of all entering streams are entered as input items, and those of the outgoing streams as output items. Heats of reaction, if negative, are input items, while they are considered to be output items if positive.
Energy Balance Calculations
Problem 1: 500 kilograms per hour of steam drive a turbine.
The steam enters the turbine at 44 atm and 450°C at a linear velocity of 60 m/s and leaves at a point 5 m below the turbine inlet atmospheric pressure and a velocity of 360 m/s.
The turbine delivers shaft work at a rate of 700 kW, and the heat loss from the turbine is estimated to be 104 kcal/h.
Calculate the specific enthalpy change associated with the process.
Energy Balance Calculations
Solution:
Energy Balance Calculations
We know, ∆H = Q + Ws + ∆Ek + ∆Ep
Energy Balance Calculations
Energy Balance Calculations
Problem 2: Saturated steam at 1 atm is discharged from a turbine at a rate of 1000 kg/h. Superheated steam at 300°C and 1atm 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 1atm. The mixing unit operates adiabatically. Calculate the amount of steam at 300°C produced, and the required volumetric flow rate of the 400°C steam.
Energy Balance Calculations
Solution:
Energy Balance Calculations
Energy Balance Calculations
Energy Balance Calculations
Problem 3: A steam flowing at a rate of 500 mol per minute containing 35% nitrogen and 65% hydrogen is to be heated from 25°C and 200°C . Calculate the heat that must be transferred . Cp of nitrogen = 5.4 + 1.2 × 10-3 T ; Cp of hydrogen = 7.8+2.6 × 10-3 T, where Cp is in cal mol-1 K-1 and T in K .
Solution:
Solution:
Total heat transferred = 176.66+489.13
= 665.79 kcal
Problem-4: An evaporator is to be fed with 1000 kg/h of a solution containing 1% solute by weight at a temperature 40°C. It is to be concentrated to a solution of 2% solute by weight in the evaporator operating at a pressure of 100 kPa in the vapour space. The heating surface is supplied with standard steam at 136 kPa (ts = 108.4°C). calculate the weight of the vapour produced and the weight of the steam required.
If the overall heat transfer coefficient of the evaporator is 1400 W/m2K calculate the necessary heating surface.
Problem-5: A continuous fractionating column operating at a pressure of 100 kPa is to be used to separate 1000 kg/h of a solution of benzene and toluene, containing 0.40 mass fraction benzene, into an overhead product containing 0.97 mass fraction benzene and a bottom product containing 0.02 mass fraction benzene. A reflux ratio of 3.5 kg/h of reflux per 1 kg of product is to be used. The feed will be liquid at its boiling point, and the reflex will be returned to the column at 38°C.
Calculate the quantity of top and bottom products in kg/h.
Calculate the condenser duty, if all the vapour entering the condenser is condensed.
Calculate the rate of heat input to the reboiler, if the liquid leaving the reboiler is saturated liquid.
Problem-6: Ammonium sulphate is to be crystallized from a solution containing 48% (NH4)2SO4 by cooling it in a counter flow crystallizer from 85 to 35°C. During cooling the amount of water that evaporates at 5% of the mass of the feed solution. If the feed rate is 1000 kg/h, calculate
(i) the rate of formation of crystals
(ii) the cooling water rate, if it is heated from 18 to 29°C, and (iii)
[Solubility of (NH4)2SO4 at 35°C = 75 kg/00 kg of water Specific heat of 48% (NH4)2SO4 solution = 2.97 kJ/kgK Heat of crystallization of (NH4)2SO4 = 75.2 kJ/kg
Latent heat of vaporization of water at 35°C = 2414 kJ/kg
Problem 7: Flue gases leaving the stack of a boiler at 523K have the following molar composition:1.31% carbon dioxide, 13.04% water vapor, 2.17% oxygen, 73.48% nitrogen.
Calculate the heat loss for 1 kmol of gas mixture above 298 K using heat capacity data.
Also calculate the average heat capacity of dry flue gases.
Cp value for carbon dioxide = 19.02 +79.63 x 10-3 T -73.7 x 10-
6T2 kJ mol-¹K-1
Cp value for water vapor = 34.05-9.65 x 10-3T+32.99 x 10-6T2 kJ mol-¹K-1
Cp value for oxygen =29.88-11.38 x 10-3T +43.78 x 10-6T2 kJ mol-1K-1
Cp value for nitrogen= 29.41-3.00 x 10-3T+5.45 x 10-
6T2 kJ kmol-1K-1
Solution: For a gaseous mixture the heat changes can be calculated as given below:
ΔΗ= nCp(mix) dT Cp(mix) = ∑ ni Cp(i)
where ni is the mole fraction of the ith constituent and Cp(i) is the heat capacity of the of ith constituent.
If Cp is given as a + b*10 - 3 *T + c* 10- 6 * T 2 then
Cp(max) = ∑ niai +(∑ nibi )*10-3 T +(∑ nic i ) *10 - 6 * T-2 of the gaseous mixture is found by substituting the relevant values.
Values given as mole per cent are converted to mole fraction by dividing by 100.
Cp(mix)= (0.1131 x 19.02+0.1304 x 34.054-0.0217x
29.88+0.7348 x 29.41)+ (0.1131 x 79.63-0.1304 x 9.65-0.0217 x 11.38 -0.7348* 3.00) x 10-3 T+ ( - 0.1131 * 73.7 + 0.1304 * 32.99 + 0.0217 * 43.78 + 0.7348 * 5.45)*106T2
= 28.85+5.546*10-3 T+0.923*10-6 T2
ΔΗ= 28.85(523 - 298) + (55.46/2)(5.232 - 2.982) + (0.923/3)(5.233- 2.983)= 7039.37 kJ
ΔΗ =T1T2∫ (a + bT + cT²)dT
Composition of dry gas is worked out on a mole fraction basis leaving aside the water vapour content is given:
Cp(mix) for dry flue gas is calculated as above and the heat change is then computed. This value on being divided by the temperature change gives the average Cp value of the dry flue gas.
Cp(mix) = (0.13*19.02+0.025*29.88+0.845*29.41) + (0.13 * 79.63 - 0.025 * 11.38 - 0.845 * 3) 10-3 * T
+ (0.13 * 73.7 + 43.78 * 0.025 + 5.45 * 0.845) * 10-6 * T2
=28.07 +7.53 x 10 T-3.88 x 10-6T2 kmol-1K-1
ΔΗ= 28.07*(523-298)+(75.3/2)*( 5.232 -
2.982)+(3.88/3)*(5.233 - 2.983)= 6.860.35kJ Average value of Cp(mix)
= 6890.35/225 = 30.49 kjmol-1K-1 ΔΗ =T1T2∫ (a + bT + cT²)dT
Problem 8: Two streams of water are mixed to form the feed to a boiler. Process data are as follows:
Feed stream 1- 120 kg/min at 30°C Feed stream 2- 175 kg/min at 65°C Boiler pressure 17 bar (absolute),
The exiting steam emerges from the boiler through a 6-cm ID pipe. Calculate the required heat input to the boiler in kilojoules per minute if the emerging steam is saturated at the boiler pressure. Neglect the kinetic energies of the liquid inlet streams.
175 kg H20 (l)/min 650 C, Ĥ =271.9kj/kg 120 kg H20 (l)/min 300 C, Ĥ = 125.7kj/kg
295 kg H20 (v)/min
17 bar, saturated (2040 C) Ĥ = 2793kj/kg
6-cm ID pipe HEAT
Qͦ (kj/min)
For this open system process, Qͦ͘ - Wͦ = ΔHͦ +Δ Eͦk + Δ Eͦp
Wͦs = 0 (No moving parts)
ΔEͦp = 0 (generally assumed unless displacement through large heights are involved)
Qͦ= ΔHͦ +Δ Eͦk Evaluate ΔHͦ
From equation ,
ΔHͦ = Σ mͦI ĤI (outlet) - Σ mͦI ĤI (Inlet)
295 kg 2793 kj min kg
120 kg 125.7 kj
min kg
=
120 kg 125.7 kj min kg
=
= 7.61 X 105 Kj/min
Evaluate Δ Eͦk
From Table B.6, the specific volume of saturated steam at 17 bar is 0.1166 m³/kg, and the cross- sectional area of the 6-cm ID pipe is
A = π R2
= 3.1416 (3.00)
2 cm2 1 m2 104 cm2
= 2.83 X 10-3 m3 The steam velocity is
u(m/s) = Vͦ(m3 /s) / A (m2)
295 kg 1 min 0.1166 m3
min 60s kg 2.83X 10-3m2
=
= 202 m/s
Then, since the kinetic energies of the inlet streams are assumed negligible,
ΔEͦk ≈ (Eͦk )outlet stream = mͦu2 /2
295 kg/min (202)2 m2 1 N 1 kj
2 s2 1 kg m/s2 103 N.m = 6.02 X 103kJ/min Finally,
Qͦ= ΔHͦ +Δ Eͦk
= [ 7.61 X 105 + 6.02 X 103 ] kj/min
= 7.67 X 105 Kj/min
Problem 9: In burning limestone to produce lime, the reaction is CaCO3 CaO + CO2 ∆H = 19278 kcal kmol-1
If 25% of the heat of combustion of coal goes up the stack and 5%
is lost by radiation, how many kilograms of coal of heating value 7780 kcalkg-1 must be used to supply the heat of reaction of 1 ton of limestone, assuming it to be 100% calcium carbonate.
Solution :
Basis: One tonne of lime stone
Heat required for 100 kg calcium carbonate = 19278 kcal
Heat required for 1000 kg of calcium carbonate = 192780 kcal Given, 25% of the heating value of coal is lost to the flue gas and 5% is lost by radiation.
Effective heat that can be recovered from 1 kg of coal
= 7780 x (70 ⁄100)
= 5446 kcal
Coal required to supply the required heat
= 192 780 ⁄ 5446 = 35.4kg
Problem 10: In the production of metallic manganese, 12 kg of Mn3O4 were heated with 4 kg of coke. The resulting product was found to contain 5.2 kg of manganese 2.8 kg MnO, unconverted charge and carbon monoxide gas. Calculate the standard heat of the reaction of this process. ΔHf° of Mn3O4, MnO, CO and coke are -331.4, -92.0, 2.64, and 2.6 kcal mol-1 respectively.
Carbothermal Reduction 12Kg of
Mn3O4 4Kg
coke(carbon)
5.2kg Mn 2.8 kg MnO
CO gas
Unconverted Feed stream ΔHf° (Mn3O4) = -331.4 kcalmol-1
ΔHf° (MnO) = -92 kcalmol-1 ΔHf° (CaO) = 2.64 kcalmol-1 ΔHf° (coke) = 2.6 kcalmol-1
Solution:
Reaction that are taking place in the process:
Mn3O4 + 4C 3Mn + 4CO Mn3O4 + C 3MnO + CO
This is incomplete carbo-thermal reduction process. So we know that,
For incomplete reaction
Standard heat of reaction,
ΔHoR = (ΔHf° of actually formed product) – (ΔHf° of actually reacted reactant)
Actually Formed Product:
Given amount of formed product:
Mn = 5.2 Kg 1 kmol Mn
55 Kg = 0.0945 kmol
2.8 Kg 1 kmol MnO
71 Kg = 0.03943 kmol
MnO =
Actually reacted/consumed reactent:
Consumption of Mn3O4 : i. For Mn
formation - 3 kmol Mn is formed from 1 kmol Mn3O4
So, 0.0945 kmol Mn is formed from (0.0945/3) kmol Mn3O4
ii. For MnO
formation - Similarly, the consumed Mn3O4 = (0.03943/3) kmol
Total consumed or reacted Mn3O4 = (0.0945/3 + 0.03943/3) kmol
= 0.0446 kmol Consumption of coke carbon:
i. For Mn
formation - 3 kmol Mn is formed from 4 kmol of coke carbon So, 0.0945 kmol Mn is formed from (4 0.0945)/3) kmol of coke carbon
ii. For MnO
formation - Similarly, the consumed carbon(c) = (0.03943/3) kmol
Total consumed or reacted coke carbon
= ((4×0.0945)/3 + 0.0394/3) kmol
= 0.139 kmol coke carbon
1 kmol CO is formed from 1kmol carbon(coke) Hence, Amount of CO formed = 0.139 kmol
×
Heat of formation of reactant actually reacted:
ΔHf° of Mn3O4 = 0.0446 × (-331400) = -14780.4 kcal ΔHf° od coke = 0.139 × 2600 = 361.4 kcal ΔHf° of materials = -14419 kcal Heat of formation of products actually formed:
ΔHf° of MnO = 0.0394 × (-92000) = -3624.8 kcal ΔHf° od Mn = 0.0945 × 0 = 0 kcal
ΔHf° of CO = 0.139 × (26400) = -14419 kcal ΔHf° of products = -7294.4 kcal The standard heat of reaction for this reaction,
ΔHoR = (ΔHf° of actually formed product) – (ΔHf° of actually reacted reactant) = -7294.4 – (-14419) kcal
= 7124 .6 kcal
Problem-11: Calculate the values of ∆H°298 for the reaction in the transformation of glucose in an organism-
C6H12O6(s) = 2C2H5OH(l) + 2CO2(g)
C6H12O6(s) + 6O2(g) = 6CO2(g) + 6 H2O (l)
Which of these reaction supplies more energy to the organism?
The standard enthalpies of formation of C6H12O6(s) ,
C2H5OH(l), CO2(g), and H2O(l) are -1273.0, -277.6, -939.5 and - 285.8 kJ/mol, respectively.
Problem-12: Carbon monoxide is burnt with 100% excess air.
Determine the final temperature obtained by the gaseous products of combustion, if combustion is adiabatic and complete.
20% of the heat is lost to the surroundings. Mean heat capacity of oxygen, nitrogen, and carbon monoxide is 31.9, 32.15 and 51.8 Jmol-1K-1 respectively. Standard heat of the reaction is -283.2 kJmol-1 for carbon monoxide.
