- 1.0 Chapter Objectives
- 1.1 Classification of Transport Processes and Separation Processes (Unit Operations)
- 1.2 SI System of Basic Units Used in This Text and Other Systems
- 1.3 Methods of Expressing Temperatures and Compositions
- 1.4 Gas Laws and Vapor Pressure
- 1.5 Conservation of Mass and Material Balances
- 1.6 Energy and Heat Units
- 1.7 Conservation of Energy and Heat Balances
- 1.8 Numerical Methods for Integration
- 1.9 Chapter Summary
- Problems
- References
- Notation
This chapter is from the book
This chapter is from the book
This chapter is from the book
1.7 Conservation of Energy and Heat Balances
1.7A Conservation of Energy
In making material balances, we used the law of conservation of mass, which states that the mass entering is equal to the mass leaving plus the mass left in the process. In a similar manner, we can state the law of conservation of energy, which states that all energy entering a process is equal to that leaving plus that left in the process. In this section, elementary heat balances will be made.
Energy can appear in many forms. Some of the common forms are enthalpy, electrical energy, chemical energy (in terms of ΔH reaction), kinetic energy, potential energy, work, and heat inflow.
In many cases in process engineering, which often takes place at constant pressure, electrical energy, kinetic energy, potential energy, and work either are not present or can be neglected. Then, only the enthalpy of the materials (at constant pressure), the standard chemical-reaction energy (ΔH0) at 25°C, and the heat added or removed must be taken into account in the energy balance. This is generally called a heat balance.
1.7B Heat Balances
In making a heat balance at steady state, we use methods similar to those used in making a material balance. The energy or heat coming into a process in the inlet materials plus any net energy added to the process are equal to the energy leaving in the materials. Expressed mathematically,
)
where ΣHR is the sum of enthalpies of all materials entering the reaction process relative to the reference state for the standard heat of reaction at 298 K and 101.32 kPa. If the inlet temperature is above 298 K, this sum will be positive. 
= standard heat of the reaction at 298 K and 101.32 kPa. The reaction contributes heat to the process, so the negative of is taken to be positive input heat for an exothermic reaction. Also, q = net energy or heat added to the system. If heat leaves the system, this item will be negative. ΣHp = sum of enthalpies of all exiting materials referred to the standard reference state at 298 K (25°C).
Note that if the materials coming into a process are below 298 K, ΣHR will be negative. Care must be taken not to confuse the signs of the items in Eq. (1.7-1). If no chemical reaction occurs, then simple heating, cooling, or phase change is occurring. Use of Eq. (1.7-1) will be illustrated by several examples. For convenience, it is common practice to call the terms on the left-hand side of Eq. (1.7-1) input items, and those on the right, output items.
EXAMPLE 1.7-1. Heating of a Fermentation Medium
A liquid fermentation medium at 30°C is pumped at a rate of 2000 kg/h through a heater, where it is heated to 70°C under pressure. The waste heat water used to heat this medium enters at 95°C and leaves at 85°C. The average heat capacity of the fermentation medium is 4.06 kJ/kg · K, and that for water is 4.21 kJ/kg · K (Appendix A.2). The fermentation stream and the wastewater stream are separated by a metal surface through which heat is transferred and the streams do not physically mix with each other. Make a complete heat balance on the system. Calculate the water flow and the amount of heat added to the fermentation medium assuming no heat losses. The process flow is given in Fig. 1.7-1.
Solution: It is convenient to use the standard reference state of 298 K (25°C) as the datum to calculate the various enthalpies. From Eq. (1.7-1) the input items are as follows:
Input items. ΣHR of the enthalpies of the two streams relative to 298 K (25°C) (note that Δt = 30 – 25°C = 5°C = 5 K):
)
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Figure 1.7-1. Process flow diagram for Example 1.7-1.
Output items. ΣHP of the two streams relative to 298 K (25°C):
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Equating input to output in Eq. (1.7-1) and solving for W,
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The amount of heat added to the fermentation medium is simply the difference of the outlet and inlet liquid enthalpies:
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Note in this example that since the heat capacities were assumed constant, a simpler balance could have been written as follows:
Then, solving, W = 7720 kg/h. This simple balance works well when cp is constant. However, when cp varies with temperature and the material is a gas, cpm values are only available between 298 K (25°C) and T K, and the simple method cannot be used without obtaining new cpm values over different temperature ranges.
EXAMPLE 1.7-2 Heat and Material Balance in Combustion
The waste gas from a process of 1000 g mol/h of CO at 473 K is burned at 1 atm pressure in a furnace using air at 373 K. The combustion is complete and 90% excess air is used. The flue gas leaves the furnace at 1273 K. Calculate the heat removed in the furnace.
Solution: First, the process flow diagram is drawn in Fig. 1.7-2, and then a material balance is made:
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(from Appendix A.3)
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Figure 1.7-2. Process flow diagram for Example 1.7-2.
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For the heat balance relative to the standard state at 298 K, we follow Eq. (1.7-1).
Input items
H(CO) = 1.00(cpm)(473 – 298) = 1.00(29.38)(473 – 298) = 5142 kJ/h
(The cpm of CO of 29.38 kJ/kg mol · K between 298 and 473 K is obtained from Table 1.6-1.)
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(This will give a negative value here, indicating that heat was removed.)
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Output items
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Equating input to output and solving for q,
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Hence, heat is removed: –34 837 W.
Often, when chemical reactions occur in a process and the heat capacities vary with temperature, the solution in a heat balance can be trial and error if the final temperature is the unknown.
EXAMPLE 1.7-3. Oxidation of Lactose
In many biochemical processes, lactose is used as a nutrient, which is oxidized as follows:
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The heat of combustion 
in Appendix A.3 at 25°C is –5648.8 × 103 J/g mol. Calculate the heat of complete oxidation (combustion) at 37°C, which is the temperature of many biochemical reactions. The cpm of solid lactose is 1.20 J/g · K, and the molecular weight is 342.3 g mass/g mol.
Solution: This can be treated as an ordinary heat-balance problem. First, the process flow diagram is drawn in Fig. 1.7-3. Next, the datum temperature of 25°C is selected and the input and output enthalpies calculated. The temperature difference ΔT = (37 – 25)°C = (37 – 25) K.
Input items
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(The cpm of O2 was obtained from Table 1.6-1.)
Output items
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Figure 1.7-3. Process flow diagram for Example 1.7-3.
(The cpm of liquid water was obtained from Appendix A.2.)
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(The cpm of CO2 is obtained from Table 1.6-1.)
ΔH37°C: Setting input = output and solving,
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