GENERAL_ASPECTS_OF_ENERGY_MANAGEMENT_AND_ENERGY_AUDIT
(CHAPTER 5:MATERIAL AND ENERGY BALANCE)
Introduction
A material balance in its most broad definition is the application of the law of conservation of mass, which states matter is neither created nor destroyed. Matter may flow through a control volume and may be reacted to form another species, however, no matter is ever lost or gained. The same is true for nergy. As with material balances, we can apply the law that energy is neither created nor destroyed, it is simply converted into another form of energy. The law of conservation of mass and energy leads to what is called a mass (material) and energy balance.
Material balances, as they pass through processing operations, can describe material quantities. If there is no accumulation, what goes into a process must come out. This is true for batch operation. It is equally true for continuous operation over any chosen time interval.
Material balances are fundamental to the control of processing, particularly in the control of yields of the products. Material balance can be determined from conceptual stage to final production stage. Initially, material balance is estimated during the planning stage of a new process or equipment. This estimate is improved while carrying out pilot scale tests related to the new process. This material balance is verified during commissioning stage and finally used as a control measure during actual production stage.
Energy balances are means for industry to examine ways of reducing energy consumption in processing because of increasing energy cost. 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.
Components of Material and Energy Balance
Typical components of material and energy balance for a process or unit operation is shown in
Figure 5.1. It may be noted that recycle stream is shown along with input side.
Basic Principles of Material and Energy Balance If the unit operation, whatever its nature is seen as a whole it may be represented diagrammatically as a box, as shown in Figure. 5.2. The mass and energy going into the box must balance with the mass and energy coming out. The law of conservation of mass leads to what is called a mass or a material balance.
Mass In = Mass Out + Mass Stored
Raw Materials = Products + Wastes +Stored Materials.
If there are no chemical changes occurring in the plant, the law of conservation of mass will apply
also to each component, so that for component A:
mA in entering materials = mA in the exit materials + mA stored in plant.
For example, in a plant that is producing sugar, if the total quantity of sugar (mA) going into the plant is not equalled by the total of the purified sugar (mAp) and the sugar in the waste liquors (mAw) and accumulated in the process (mAs), then there is something wrong. Sugar is either being burned (chemically changed) or else it is going unnoticed down the drain somewhere. In this case:
Where, MAU,i s the unknown loss and needs to be identified. So the material balance is now:
Raw Materials = Products + Waste Products + Stored Products + Losses
Where, Losses are the unidentified materials.
Just as mass is conserved, energy is conserved in process operations. The energy coming into a unit
operation can be balanced with the energy coming out and the energy stored.
Energy In = Energy Out + Energy Stored
Energy balances are often complicated because forms of energy can be inter-converted, for example
mechanical energy to heat energy, but overall the quantities must balance.
Classification of Processes
Process can be viewed overall or as a series of units. Each unit is a unit operation that can be represented by a box as shown in Figure 5.3. Raw materials and energy go into the box and desired products, byproducts, wastes and energy come out of the box. The mass in and out of a control box must be equal. The equipment within the box will make the required changes with little waste and energy as possible. There are different types of process.
Steady-state process is one where none of the process variables change with time
Unsteady-state process is one where the process variables change with time.
B) Based on how the process was built to operate
A continuous process is one that has the feed streams and product streams moving into and out of the process all the time. Examples are oil refinery, distillation process etc.
batch process is one where the feed streams are fed to the process to get it started. The feed material is then processed through various process steps and finished products are taken out at specific times. Steps:
¢ Feed is charged into vessel
¢ Process is started
¢ No mass is added or removed from vessel (process parameters are usually monitored and controlled)
¢ Atsome conditions or after fixed time, products are removed.
Material Balance
Levels of Material Balance
The material balances can be developed at various levels:
Overall Material balance: This involves input and output steams for complete plant Section wise Material balances: This involves M&E balances to be made for each section/department/cost centre. This would help to prioritise focus areas for efficiency improvement
Equipment-wise Material balances: Material balances for key equipment would help assess performance of equipment, which would in turn help identify energy and material losses. The choice among the types of material balance depends on the reasons for making the balance. The major factor is the cost of the materials and so costly materials are more likely considered than cheaper ones and products more than waste materials
Material Balance Procedure
First step is to identify materials in, materials out and material stored. Next step is to consider whether
materials in each category have to be treated as whole (gross material balance) or whether individual
constituents in the material should be treated separately. For example, we can do material balance for
dry solids alone as opposed to total material. This means separating the material into two constituents,
non-water and water.
Typical steps are as follows:
a) Define basis & units: Choose a basis of calculations on quantity (mass for batch process) or flow
rate (mass per hour for continuous process) of one of the process streams. Convenient Units are then
chosen as mass can be expressed in various ways: weight/weight (w/w), weight/volume (w/v), molar
concentration (M), mole fraction.
¢ The weight/weight concentration is the weight of the solute divided by the total weight of the solution and is the fractional form of the percentage composition by weight.
¢ The weight volume concentration is the weight of solute in the total volume of the solution. With gases, concentrations are primarily measured in weight concentrations per unit volume, or as partial pressures.
¢ The molar concentration is the number of molecular weights of the solute expressed in kg in 1 m^3 of the solution.
¢ The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all species present in the solution.
b) Draw a flowchart: Establish a boundary so that the flow streams in and out can be identified. The
identification and drawing up a unit operation/process is prerequisite for energy and material balance.
Flow charts are schematic representation of the production process, involving various input resources,
conversion steps and output and recycle streams. The process flow may be constructed stepwise 1.e.
by identifying the inputs / output / wastes at each stage of the process, as shown in the Figure 5.4.
¢ Inputs of the process could include raw materials, water, steam, energy (electricity, etc);
¢ Process Steps should be sequentially drawn from raw material to finished product. Intermediates and any other by-product should also be represented. The operating process parameters such as temperature, pressure, % concentration, etc. should be represented.
¢ The flow rate of various streams should also be represented in appropriate units like m /h or kg/h. In case of batch process the total cycle time should be included.
¢ Wastes / by products could include solids, water, chemicals, energy etc. For each process steps (unit operation) as well as for an entire plant, energy and mass balance diagram should be drawn.
¢ Output of the process is the final product produced in the plan
c) Write Material Balance Equations: The following examples are illustration for writing the material balance equations
Example 5.1
A solution which contains 10% solids is mixed with 25% solid solution. A single output which is 20%
solid is removed. If the 10% solution enters at 5.0 kg/s, what are the other rates? (Assume no
accumulation)
Solution
Basis:
Solution A (INPUT) = 5 kg/s
Solution B (INPUT) = x kg/s
Solution C (OUTPUT) = y kg/s
Therefore A+ B=C
5+X+Y............EQ-1
Since solution A contains 10% solids, solution B contains 25% solids and solution C contains 20%
solids the equation can be written as
Solids 0.1 *5 +0.25 * X=0.2 * Yo, EQ -2
Liquids 0.9 *5 +0.75*X = 0.8 * Yo. EQ -3
Substituting value of Y from EQ -1 in EQ-2
0.5+0.25 X =0.2 * (5 +X)
0.05 X = 0.5
x = 10 kg/s,
Substituting X in EQ-1, Y= 15 kg/s
Example 5.2
solution which is 80% oil, 15% usable by-products and 5% impurities, enters a refinery. One output
is 92% oil and 6% usable by-products. The other output is 60% oil and flows at the rate of 1000 lit/hr
(assume no accumulation, percents by volume)
— What is the flow rate of input?
— What is the percent composition of the 1000 lit/hr output?
— What percent of the original impurities are in the 1000 lit/hr output?
Solution
Basis:
Input Stream A = X lit/hr
Output Stream B = Y lit/hr
Output Stream C = 1000 lit/hr
Material balance equations
1) Total: X = Y + 1000.0ec. .e cece. eee eee EQ-1
2) Oil: 0.8 * X = 0.92 * Y + 0.6 * 1000...............084. EQ-2
3) UBP: 0.15 * X = 0.06 * Y + v * 1000.............0... EQ-3
4) IMP: 0.05 * X = 0.02 * Y + w *1000...............04. EQ-4
5) OUTPUT impurities & UBP: v+w=0.4........... EQ-5
Solving Equations | and 2; substituting value of X in EQ-2
0.8 (Y + 1000) = 0.92 * Y + 600
0.8 Y + 800 =0.92 Y + 600, Y = 1666 lit/hr
Substituting Y in EQ-1
X = 1666 + 1000 = 2666 lit/hr = Flow rate of input stream
Substituting value of X and Y in EQ-3, v = 30%
Substituting value of v in EQ-5, w = 10%
Thus composition of 1000 lit/hr output stream is 60% oil, 30% usable by products and 10% impurities
Impurities in input stream = 0.05 * 2666 = 133.3 lit/hr
Impurities in 1000 lit/hr stream (10%) =100 lit/hr
Therefore impurities in 1000 lit/hr stream as % input stream = 100/133.3 = 75%
Example 5.3: Constituent balance
Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to
contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk
contained 4.5% fat, calculate its composition assuming that fat only was removed to make the skim
milk and that there are no losses in processing.
Basis: 100 kg of skim milk.
This contains, therefore, 0.1 kg of fat. Let the fat which was removed from it to make skim milk
be x kg.
Total original fat =(x+0.1) kg
Total original mass = (100 + x) kg
and as it is known that the original fat content was 4.5% so
So the composition of the whole milk is then fat = 4.5%, water = 90.5/104.6 = 86.5 %, protein =
3.5/104.6 = 3.3 %, carbohydrate= 5.1/104.6 = 4.9% and ash = 0.8%
Example 5.4 Continuous Process balance
In a continuous centrifuging of milk, if 35,000 kg of whole milk containing 4% fat is to be separated
in a 6 hour period into skim milk with 0.45% fat and cream with 45 % fat, what is the flow rate of the
two output streams from the continuous centrifuge which accomplishes the separation?
In continuous processes, time also enters into consideration and the balances are related to unit time.
Thus in considering a continuous centrifuge separating whole milk into skim milk and cream, if the
material holdup in the centrifuge is constant both in mass and in composition, then the quantities of
the components entering and leaving in the different streams in unit time are constant and a mass
balance can be written on this basis. Such an analysis assumes that the process is in a steady state,
that is flows and quantities held up in vessels do not change with time.
Solution
Basis:
Total mass input per hour = 35000/6 =5833 kg
Total mass output for skim milk = Y
Total mass output for cream = z
Material Balance Equations:
1) Mass In = Mass Out: 5833 =Y+Z (ie. Z=5833-Y) ... EQ-1
2) Fat In = Fat Out: 0.04 * 5833 = 0.0045 * Y + 0.45 * Z ... EQ-2
Substituting Z from EQ-1 to EQ-2
0.04 * 5833 = 0.0045 Y + 0.45 * (5833 -Y)
Y = 5369 kg/hr
Substituting Y in EQ -1
Z = 464 kg/hr
Example 5.5 Concentrations
A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a
liquid of density 1323 kg/m’. Calculate the concentration of salt in this solution as a (a) weight fraction,
(b) weight/volume fraction, (c) mole fraction, (d) molal concentration.
Solution
(a) Weight fraction: 20 / (100 + 20) = 0.167
% weight / weight = 16.7%
(b) Weight/volume:
A density of 1323 kg/m3 means that 1 m3 of solution weighs 1323 kg, but 1323 kg of salt
solution contains (20 kg of salt x 1323 kg/m3) / (100 + 20) kg = 220.5 kg salt / m3
1 m} solution contains 220.5 kg salt.
Weight/volume fraction = 220.5 / 1000 = 0.2205
And so weight / volume = 22.1%
c) Moles of water = 100 / 18 = 5.56
Moles of salt = 20 / 58.5 = 0.34
Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058
d) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m3
Note that the mole fraction can be approximated by the (moles of salt/moles of water) as the number
of moles of water is dominant, that is the mole fraction is close to 0.34 / 5.56 = 0.061. As the solution
becomes more dilute, this approximation improves and generally for dilute solutions the mole fraction
of solute is a close approximation to the moles of solute / moles of solvent.
Example 5.6
In a textile mill, an evaporator concentrates a liquor containing solids of 6% by w/w (weight by weight) to produce an output containing 30% solids w/w. Calculate the evaporation of water per 100 kg of feed to evaporator?
Solution
Inlet solid contents = 6%
Outlet solid contents = 30%
Feed =100 kg
Solid content in kg in feed = 100 * 0.06 = 6 kg
Since mass in = mass out, the outlet solid content should be 6 kg
Output = (100/30)*6 = 20 kg
Quantity of water evaporated = [100 —20] = 80 kg
Example 5.7 Air Composition
If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:
(a) The mean molecular weight of air,
(b) The mole fraction of oxygen,
(c) The concentration of oxygen in mole/m? and kg/m’ if the total pressure is 1.5 atmospheres and
the temperature is 25 °C
Solution
(a) Taking the basis of 100 kg of air: it contains 77/28 moles of N, and 23/32 moles of O,
Total number of moles = 2.75 + 0.72 = 3.47 moles.
So mean molecular weight of air = 100 / 3.47 = 28.8
Mean molecular weight of air = 28.8
b) The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21
Mole fraction of oxygen = 0.21
(c) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm/
mole K and at a temperature of 25 °C = 25 + 273 = 298 K, and where V= 1 m°
pV =nRT
and so, 1.5 x 1 =n x 0.08206 x 298
n=0.061 mole/m3
weight of air =n * mean molecular weight
= 0.061 x 28.8=1.76 kg/m
and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m3
Concentration of oxygen = 0.4kg/m3
or 0.4/32 = 0.013 mole /m3
When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by
first calculating the number of moles of gas using the gas laws, treating the volume as the volume of
the liquid, and then calculating the number of moles of liquid directly.
Example 5.8: Gas composition
In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3
volumes of gas to one volume of water at 0 °C and atmospheric pressure. Calculate (a) the mass
fraction and (b) the mole fraction of the CO, in the drink, ignoring all the components other than CO,
and water.
Solution
Example 5.9: Dust balance
A bag filter is used to remove the dust from the inlet gas stream to meet the emission standards in
cement, fertilizer and other chemical industries.
Inlet gas to a bag filter is 1,69,920 m*/hr and the dust loading is 4577 mg/m3.
Outlet gas from the bag filter is 1,85,040 m°?/hr and the dust loading is 57 mg/m3
What is the maximum quantity of ash that will have to be removed per hour from the bag filter hopper
based on test results?
Solution.
1. Calculation in/ out dust quantity
Inlet dust qty =169920 (m*/hr) x 4577 (mg/m3) x 1/1000000 (kg/mg)= 777.7 kg/hr
Outlet dust qty = 185040 (m*/hr) x 57 (mg/m3) x 1/1000000 (kg/mg) = 10.6 kg/hr
2. Calculate ash qty (to be removed from the hopper)
Hopper ash = Inlet dust qty—Outlet dust qty = 777.7 kg/hr—10.6 kg/hr = 767.1 kg/hr
Energy Balance
Energy is the capacity to do work or to transfer heat. The law of conservation of energy states that energy can neither be created nor destroyed. The total energy in the materials entering the processing plant, plus the energy added in the plant must equal the total energy leaving the plant. This is a more complex concept than the conservation of mass, as energy can take various forms such as kinetic energy. Potential energy, heat energy, chemical energy, electrical energy and so on. During processing, some of these forms of energy can be converted from one to another; say for instance mechanical energy in a fluid can be converted through friction into heat energy. It is the sum total of all these norms of energy that is conserved.
For example, in the pasteurizing process for milk, the milk is pumped through a heat exchanger and is first heated and then cooled. The energy affecting the product is the heat energy in the milk. Heat energy is added to the milk by the pump and by the hot water passing through the heat exchanger. Cooling water then removes part of the heat energy and some of the heat energy is also lost to the surroundings. The heat energy leaving in the milk must equal the heat energy in the milk entering the pasteurizer plus or minus any heat added or taken away in the plant.
Heat energy leaving in milk = initial heat energy + heat energy added by pump + heat energy added in heating section - heat energy taken out in cooling section - heat energy lost to surroundings.
The law of conservation of energy can also apply to part of a process. For example, considering only the heating section of the heat exchanger in the pasteurizer, the heat lost by the hot water must be equal to the sum of the heat gained by the milk and the heat lost from the heat exchanger to its surroundings.
Conservation of Energy
In a system, if the storage does not change, the ingoing and outgoing energy must be equal (Figure 5.5 a).
If the storage changes, this must be reflected in the energy balance and the energy input to a system might not balance the energy that goes out. For instance as in Figure 5.5 b, the input is 75 units of energy but only 60 units go out. Since the first law requires that the energy be conserved, system had to gain 15 units of energy. In the Figure 5.5 c, the input is short by 15 units of energy so it can be inferred that the system must have lost 15 units.
The sinks are depositories of leakage or rejected energy. It is usually low-grade heat, as in, radiation
losses from boilers or heat carried away by cooling water. The outputs represent useful work.
The Figure 5.6 below illustrates power cycle schematic. In this input heat energy (Qn) resulting from combustion of fuel is transferred to water in a steam generator (boiler). The fluid feed water is pumped using input energy Win The steam is used to drive the turbine and perform useful work Wout- and the steam is condensed in condenser giving its heat energy Qout , The working fluid is feed water which changes its state from water to steam and back to condensate.
Applying the law of conservation of energy, if a system undergoes a process by heat and work transfer, then the net heat supplied, Q plus the net work input, W. is equal to the change of intrinsic (internal) energy, DU of the working fluid, 1.e.
ΣQ + ΣW =ΔU
Applying this general principle to a thermodynamic cycle, when the system undergoes a complete cycle. 1.e. change in internal energy, ΔU = 0.
ΣQ +ΣW=0
Where,
ΣQ = The algebraic sum of the heat supplied to (+) or rejected from (-) the system.
ΣW = The algebraic sum of the work done by surroundings on the system (+) or by the system on
surroundings (-).
Applying the rule to the power plant gives:
Efficiency
The (thermodynamic) efficiency of a process is the ratio of useful output to input and is always less than 100%. In the energy balance shown in Figure 5.7, there is no internal storage, so the sum of the inputs must equal the outgoing energy. The efficiency of the process is 22.5% and the energy lost in the system is 100 - 22.5 = 77.5%.
How can energy be lost in a system?
Energy can manifest itself in many forms such as heat, kinetic energy, chemical energy, potential energy but because of inter-conversions, it is not always easy to isolate separate constituents of energy balances. Energy is “lost” really means it changes into a form that is not counted. Most often the “uncounted’ energy is work done against friction. This work changes other forms of energy into heat and wear. No energy is lost in the end, only the forms have changed.
This loss of usable energy is due to many causes
¢ In mechanical systems it is friction
¢ In electrical systems it is resistance
¢ In fluid systems it is turbulence, viscosity or mixing
Practically, approach to energy balance takes into account only “heat balances’ ignoring internal energies. When we are unfamiliar with the relative magnitudes of the various forms of energy (oil, gas, coal, steam, chilled water or electricity) entering and exiting a particular processing situation, it is best to put them all down and convert them to equivalent “Heat Energy”.
Heat Balances
The most common important energy form is heat energy and the conservation of this can be illustrated by considering operations such as heating and drying. In these, enthalpy (total heat) is conserved and as with the mass balances enthalpy balances can be written round the various items of equipment or process stages, or round the whole plant, and it is assumed that no appreciable heat is converted to other forms of energy such as work.
Enthalpy (H) is always referred to some reference level or datum, so that the quantities are relative to this datum. Working out energy balances is then just a matter of considering the various quantities of materials involved, their specific heats, and their changes in temperature or state i.e. latent heat.
Facility as an Energy System
In a production facility, the energy in form of coal, oil, gas and electricity enters the facility and is converted to more convenient forms of energy such as steam, compressed air, chilled water etc. The outgoing energy is usually in the form of heat and motion.
The energy usage in the overall plant can be split up into various forms such as (Figure 5.8):
¢ Electrical energy, which is usually purchased as HT and converted into LT supply for end use.
¢ Some plants generate their own electricity using DG sets or captive power plants.
¢ Fuels such as furnace oil, coal are purchased and then converted into steam or electricity.
¢ Boiler generates steam for heating and drying demand
¢ Cooling tower and cooling water supply system for cooling demand
¢ Air compressors and compressed air supply system for compressed air needs
All energy/utility system can be classified into three areas like generation, distribution and utilisation
for the system approach and energy analysis.
Example 5.10: Furnace
A furnace shell has to be cooled from 90°C to 55°C. The mass of the furnace shell is 2 tonnes; the
specific heat of furnace shell is 0.2 kcal/kg °C. Water is available at 28°C. The maximum allowed
increase in water temperature is 5°C. Calculate the quantity of water required to cool the furnace.
Neglect heat loss.
Solution
Energy Stream #1
Mass of furnace shell (m) = 2000 kg
Specific heat (Cp) = 0.2 kcal/kg °C
Initial furnace temperature (T1) = 90°C
Desired furnace shell temperature (T2) = 55°C
Total heat that has to be removed from the furnace = m x Cp x (T1 - T2) = 2000 x 0.2 x (90- 55)
Energy Stream #2
Quantity of water required =X kg
Specific heat of water = 1 kcal/kg °C
Inlet cooling water temperature (T3) = 28°C
Maximum cooling water outlet temperature (T,) = 33°C
Heat removed by water X x | x (33 — 28) = 5X kcal
For energy balance:
Energy Stream #1 = Energy Stream #2
or Quantity of water required (X) = 14000/5 = 2800 kg
Example 5.11: Dryer heat balance
A textile dryer is found to consume 4 m*/hr of natural gas with a calorific value of 800 kJ /mole. If the
throughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture,
estimate the overall thermal efficiency of the dryer taking into account the latent heat of evaporation
only.
Solution
1) Initial moisture in wet cloth =60 x 0.55 = 33 kg moisture
2) Bone dry cloth = 60 x (1-0.55) =27 kg bone dry cloth
3) Final product moisture content 10% = 27/9 = 3 kg
4) So moisture removed /hr = 33 - 3 = 30 kg/hr
5) Latent heat of evaporation = 2257 kJ/kg
6) Heat used for drying cloth = 30 x 2257 = 6.8 x 10^4 kJ/hr
7) Assuming the natural gas to be at standard temperature and pressure at which | mole occupies
22.4 liters
8) Rate of flow of natural gas = 4 m^3/hr = (4 x 1000)/22.4 = 179 moles/hr
9) Heat available from combustion 179 x 800 =14.3 x 10^4 kJ/hr
10) Approximate thermal efficiency of dryer = heat needed I heat used = 6.8 x 10^4/14.3 x 10^4 = 48%
To evaluate this efficiency more completely it would be necessary to take into account the sensible heat of the dry cloth and the moisture, and the changes in temperature and humidity of the combustion air, which would be combined with the natural gas. However, as the latent heat of evaporation is the dominant term, the above calculation gives a quick estimate and shows how a simple energy balance can give useful information.
Example 5.12- Evaporation Rate
Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to paper machine is 40% and 95% respectively. Evaporated moisture temperature is 80 °C. To evaporate moisture, the steam is supplied at 35 kg/cm2. Latent heat of steam at 35 kg/cm2 is 513 kcal/kg.
24 hours/day operation a) Estimate the quantity of moisture to be evaporated b) Input steam quantity required for evaporation (per hour). Consider enthalpy of evaporated moisture as 632 kCal/kg. Assume
Solution
Production rate from a paper machine: 340 TPD or 14.16 TPH (tonnes per hour)
Inlet dryness to paper machine: 40%
Outlet dryness from paper machine: 95%
Estimation of moisture to be evaporated
Paper weight in final product: 14.16 x 0.95 = 13.45 TPH
Weight of moisture before dryer: [(100-40)/ 40] x 13.45 = 20.175 TPH
Weight of moisture after dryer: 14.16 x 0.05 = 0.707 TPH
Evaporated moisture quantity: 20.175 - 0.707 =19.468 TPH
Input steam quantity required for evaporation
Evaporated moisture temperature: 80 °C
Enthalpy of evaporated moisture: 632 kcal/kg
Heat available in moisture (sensible & latent): 632 x 19468 = 12303776 kcal/h
For evaporation minimum equivalent heat available should be supplied from steam
Latent Heat available in supply steam at 3.5 kg/cm2 = 513 kcal/kg
Quantity of steam required: 12303776/513= 23984 kg or 23.98 MT/hour
Energy Analysis and the Sankey Diagram
The basic data needed for an energy analysis is an energy balance of each process section. The objective is to define in detail the energy input, energy utilized, and the energy dissipated or wasted. This is best represented by a Sankey diagram. The Sankey diagram is very useful tool to represent an entire input and output energy flow in any energy equipment or system such as boiler generation, fired heaters, furnaces after carrying out energy balance calculation. Usually the flows are represented by arrows. The width of the arrows is proportional to the size of the actual flow. Better than numbers, tables or descriptions, this diagram represents visually various outputs (benefits) and losses so that energy managers can focus on finding improvements in a prioritized manner.
Sankey Diagram
The Figure 5.9 shows a Sankey diagram for an internal combustion engine. From the Figure, it is clear that exhaust flue gas losses are a key area for priority attention.Since the engines operate at high temperatures, the exhaust gases leave at high temperatures resulting in poor system efficiency. Hence a heat recovery device such as a waste heat boiler has to be necessarily part of the system. The lower the exhaust temperature, higher is the system efficiency. Further improvements can be thought of by utilising the heat in the cooling water circuit for a vapour absorption refrigeration system.
Solved Example:
An evaporator is to be fed with 10,000 kg/hr of a solution having 1 % solids. The feed is at 38°C. It is
to be concentrated to 2% solids. Steam is entering at a total enthalpy of 640 kcal/kg and the condensate
leaves at 100°C. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kcal/kg and that of the
vapour is 640 kcal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.
Heat Balance:
Heat input by steam + Heat in feed = Heat out in vapour + Heat out in thick liquor
[M x (640-100) + 38.1 x 10,000] = (32,00,000 + 5,04,000)
M x 540 = 33,23,000
Mass of steam required = 33,23,000/540
= 6153.7 kg/hr
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