As illustrated in the following examples, the format described in Chapter 4 for material balances can be used as a foundation for energy balance calculations.
EXAMPLE 5.4 CONTINUOUS WATER HEATER
Water at 25C enters an open heating tank at a rate of 10 kg h21. Liquid water leaves the tank at 88C at a rate of 9 kg h21; 1 kg h21water vapour is lost from the system through evaporation.
At steady state, what is the rate of heat input to the system?
Solution 1. Assemble
(i) Select units for the problem.
kg, h, kJ,C
(ii) Draw the flow sheet showing all data and units.
The flow sheet is shown inFigure 5.4.
(iii) Define the system boundary by drawing on the flow sheet.
The system boundary is indicated inFigure 5.4.
2. Analyse
(i) State any assumptions.
process is operating at steady state system does not leak
system is homogeneous evaporation occurs at 88C vapour is saturated shaft work is negligible no heat losses
(ii) Select and state a basis.
The calculation is based on 10 kg water entering the system, or 1 hour.
(iii) Select a reference state.
The reference state for water is the same as that used in the steam tables: 0.01C and 0.6112 kPa.
(iv) Collect any extra data needed.
hðliquid water at 88Cị5368:5 kJ kg21ðTable D:1ị hðsaturated steam at 88Cị52656:9 kJ kg21ðTable D:1ị hðliquid water at 25Cị5104:8 kJ kg21ðTable D:1ị
Liquid water 10 kg h–1 25°C
Liquid water 9 kg h–1 88°C Water vapour
1 kg h–1 88°C
Heating tank
System boundary –Q
FIGURE 5.4 Flow sheet for continuous water heater.
(v) Determine which compounds are involved in reaction.
No reaction occurs.
(vi) Write down the appropriate mass balance equation.
The mass balance is already complete.
(vii) Write down the appropriate energy balance equation.
At steady state,Eq. (5.9)applies:
X
input streams
ðMhị2 X
output streams
ðMhị2Q1Ws50
3. Calculate
Identify the terms in the energy balance equation.
For this problemWs50. The energy balance equation becomes:
ðMhịliq in2ðMhịliq out2ðMhịvap out2Q50 Substituting the information available:
ð10 kgị ð104:8 kJ kg21ị2ð9 kgị ð368:5 kJ kg21ị2ð1 kgị ð2656:9 kJ kg21ị2Q50 Q5 24925:4 kJ
Qhas a negative value. Thus, according to the sign convention outlined inSection 5.2, heat must be supplied to the system from the surroundings.
4. Finalise
Answer the specific questions asked in the problem; check the number of significant figures;
state the answers clearly.
The rate of heat input is 4.933103kJ h21.
EXAMPLE 5.5 COOL ING IN DOWNSTREAM PROCESSING
In downstream processing of gluconic acid, concentrated fermentation broth containing 20%
(w/w) gluconic acid is cooled prior to crystallisation. The concentrated broth leaves an evapora- tor at a rate of 2000 kg h21 and must be cooled from 90C to 6C. Cooling is achieved by heat exchange with 2700 kg h21water initially at 2C. If the final temperature of the cooling water is 50C, what is the rate of heat loss from the gluconic acid solution to the surroundings? Assume the heat capacity of gluconic acid is 0.35 cal g21C21.
Solution 1. Assemble
(i) Units kg, h, kJ,C (ii) Flow sheet
The flow sheet is shown inFigure 5.5.
(iii) System boundary
The system boundary indicated inFigure 5.5separates the gluconic acid solution from the cooling water.
2. Analyse
(i) Assumptions steady state no leaks
other components of the fermentation broth can be considered water no shaft work
(ii) Basis
2000 kg feed, or 1 hour (iii) Reference state
H50 for gluconic acid at 90C H50 for water at its triple point (iv) Extra data
The heat capacity of gluconic acid is 0.35 cal g21C21; we will assume thisCpremains constant over the temperature range of interest. Converting units:
Cpðgluconic acidị50:35 cal
gC 4:187 J1 cal
1000 J1 kJ
1000 g1 kg
51:47 kJ kg21C21 hðliquid water at 90Cị5376:9 kJ kg21ðTable D:1ị
hðliquid water at 6Cị525:2 kJ kg21ðTable D:1ị hðliquid water at 2Cị58:4 kJ kg21ðTable D:1ị hðliquid water at 50Cị5209:3 kJ kg21ðTable D:1ị (v) Compounds involved in reaction
No reaction occurs.
Cooler
Q (to cooling water) Q (loss) Feed stream
2000 kg h–1
400 kg h–1 gluconic acid 1600 kg h–1 water 90°C
Product stream 2000 kg h–1
400 kg h–1 gluconic acid 1600 kg h–1 water 6°C
System boundary
FIGURE 5.5 Flow sheet for cooling gluconic acid solution.
(vi) Mass balance equation
The mass balance equation for total mass, gluconic acid, and water is:
mass in5mass out The mass flow rates are as shown inFigure 5.5.
(vii) Energy balance equation X
input streams
ðMhị2 X
output streams
ðMhị2Q1Ws50
3. Calculate
Ws50. There are two heat flows out of the system: one to the cooling water (Q) and one representing loss to the surroundings (Qloss). With symbols W5water and G5gluconic acid, the energy balance equation is:
ðMhịW in1ðMhịG in2ðMhịW out2ðMhịG out2Qloss2Q50 ðMhịW in5ð1600 kgị ð376:9 kJ kg21ị56:033105kJ
ðMhịG in50ðreference stateị
ðMhịW out5ð1600 kgị ð1:47 kJ kg21ị54:033104kJ
(Mh)G outat 6C is calculated as a sensible heat change from 90C usingEq. (5.12):
ðMhịG out5MCpðT22T1ị5ð400 kgị ð1:47 kJ kg21C21ị ð6290ịC
‘ðMhịG out5 24:943104kJ
The heat removed to the cooling water,Q, is equal to the enthalpy change of the cooling water between 2C and 50C:
Q5ð2700 kgị ð209:328:4ịkJ kg2155:423105kJ These results can now be substituted into the energy balance equation:
ð6:033105kJị1ð0 kJị2ð4:033104kJị2ð24:943104kJị2Qloss25:423105kJ50
‘Qloss57:013104kJ 4. Finalise
The rate of heat loss to the surroundings is 7.03104kJ h21.
It is important to recognise that the final answers to energy balance problems do not depend on the choice of reference states for the components. Although values ofhdepend on the reference states, as discussed inSection 5.3.1this dependence disappears when the energy balance equation is applied and the difference between the input and output enthalpies is determined. To prove this point, any of the examples in this chapter can be repeated using different reference conditions to obtain the same final answers.