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Principles of Food and Bioprocess Engineering (FS 231)

Solutions to Example Problems on Mass and Energy Balances

1 The specific heat of the product is the weighted mean of the individual specific heats and is

6 The energy content of saturated steam at 270.1 kPa and of 90% quality is given by:

Energy released during condensation = H - H = 2503.08 - 546.31 = 1956.77 kJ/kg

7 Energy required to convert ice at -20 /C to ice at 0 /C = 5 (2050 x 20) J = 205 kJ

Energy required to convert ice at 0 /C to water at 0 /C = 5 (333.2) kJ = 1666 kJ

Energy required to convert water at 0 /C to water at 100 /C = 5 (4180 x 100) kJ = 2090 kJEnergy required to convert water at 100 /C to vapor at 100 /C = 5 (2257.06) kJ = 11285.3 kJTotal energy required = 205 + 1666 + 2090 + 11285.3 = 15246.3 kJ

Here, H = 567.69 kJ/kg (from steam tables at 313 kPa)

Equating the energy gained by the product to the energy supplied by steam, we get;

12 The system diagram is as follows:

Performing a mass balance, we get: 2 + 0.25 = m

Solving, we get: m = 2.25 kg

p(w) at 10 /C s p(w) at 70 /C

From tables containing properties of water, we find that:

13 The system diagram is shown below:

The above system has 6 unknowns Thus, we need 6 equations to solve for the unknowns Wehave 2 mass balance equations and 1 energy balance equation for each sub-system, giving rise

to a total of 6 equations The specific heats of each stream is determined as a weighted mean

of the specific heats of the individual components in the following manner:

We thus have 9 equations, but 12 unknowns In order to be able to solve the set of equations,

we need to have 3 more equations Closer examination reveals that:

1 2

sub-system at the top has to equal 1.0)

3 4

sub-system has to equal 1.0)

5 6 7

sub-system has to equal 1.0)Thus, we have 12 equations and 12 unknowns and can hence solve the set of equations once

we have the following equations for the specific heats and enthalpy:

Overall mass balance: m + 3 = m

Thus, we have 4 equations and 4 unknowns and can hence solve the system of equations once

we have the following expressions for specific heat and enthalpy:

21 a Enthalpy of saturated steam (quality = 100%) is given by:

1

ˆ H = 2748.6 kJ/kgEnthalpy of saturated steam (quality = 90%) is given by:

2

H = 0.9 (2683.8) + 0.1 (440.15) = 2459.4 kJ/kg

1 2

Thus, energy released = H - H = 289.2 kJ/kg

b Enthalpy of superheated steam at 150 /C and 75 kPa = 2778.2 kJ/kg

Enthalpy of saturated vapor at 150 /C = 2746.5 kJ/kg

Thus, energy released = 2778.2 - 2746.5 = 31.7 kJ/kg

22 The system diagram is shown below:

Solving these equations yields: m = 0.65 kg/s, m = 5.65 kg/s, x = 0.18

23 The system diagram is shown below:

Other than the specific heats and enthalpies, we have 6 unknowns (m , m , m , m , T, x) Thus,

we should be able to write 6 equations and also expressions to determine all the specific heatsand enthalpies

can be obtained from tables containing properties of water.

Using the idea of weighted means, we write down the following expressions for specific heatsand enthalpy:

25 The system diagram for this problem (along with the unknowns) is as follows:

In the above system, there are two components in each of the subsystems Hence we have twomass balance equations in each of the sub-systems and one energy balance for each sub-system, yielding a total of six equations As we can see from the figure, there are six

Using the idea of weighted means, we compute specific heat and enthalpy as follows:

27 The system diagram for this problem is as follows:

Thus, from the overall mass balance equation, we get: m = 0.56 kg/s

Substituting these values in the energy balance equation, we get:

Looking at the steam tables we see that this value of H corresponds to a pressure of ~ 47 kPa

28 The system diagram is as shown below:

From the overall mass balance equation in the second sub-system we get: m = 17 kg/sSubstituting this in the overall mass balance equation in the 1 sub-system, we get:st 1

Substituting these values in the energy balance equation, we get: T = 341.6 K = 68.6 /C

29 The system diagram is shown below:

30 The system diagram is as shown below:

Using the idea of weighted means, we compute specific heat and enthalpy as follows:

31 The system diagram is shown below:

33 Since there are 2 components in both the first system and 3 components in the second system, we have a total of 5 equations.

sub-For the first sub-system:

1

1

For the second sub-system:

Thus, we have 4 equations and 4 unknowns and can hence solve the system of equations once

we have the following expressions for specific heat and enthalpy:

35 The complete system diagram along with the unknowns is as shown below:

1 2 3 4 1 1

Other than the specific heats and enthalpies, we have 6 unknowns (m , m , m , m , T , x ).Thus, we should be able to write 6 equations and also expressions to determine all the specificheats and enthalpies

be obtained from tables containing properties of water

Using the idea of weighted means, we write down the following expressions for specific heatsand enthalpy:

p

Thus, in this problem, the energy of water = 5 (4184) (65) = 1359.8 kJ

v at 120 /C c at 120 /C

= 5 [0.75 (2706.3) + 0.25 (503.71)] = 10778.26 kJThus, energy required for conversion = 10778.26 - 1359.8 = 9418.46 kJ

v at 150 /C c at 150 /C

= 4 [0.95 (2746.5) + 0.05 (632.2)] = 10563.14 kJ

c at 150 /C

Thus, energy released = 10563.14 - 2528.8 = 8034.34 kJ

39 The system diagram is as follows:

For the first sub-system

From the second equation (solids balance equation), we get: x = 2.4/m

Substituting this into the simplified energy balance equation above yields:

From the solids balance equation, we get: x = 0.27

For the second sub-system

40 The system diagram with all the unknowns marked is shown below:

For the first sub-system

We thus have 6 equations and 6 unknowns (m , m , m , x , x , T ) We can solve for these as long

as we can write down expressions for specific heats and enthalpies without introducing any newunknowns Accordingly, we write expressions for specific heats and enthalpies as follows:

For the second liquid, m (c ) = 4 (1500) = 6000 J/K

Since the m (c ) values are equal for both liquids, the rise in temperature will also be the same

42 The system diagram with all the unknowns marked is shown below:

s 1 1 2

The 4 unknowns in the above problem are m , m , T , and m We will be able to write 1 massbalance and 1 energy balance equation for each of the two sub-systems, for a total of 4 equationsane hence we can solve for 4 unknowns

Thus, we have 4 equations and 4 unknowns and can hence solve for the unknowns

43 The system diagram is shown below:

s 1 cw 2 1 2

We have 6 unknowns in this problem: m , m , m , m , x , x

We have 1 overall mass balance equation, 1 solids balance equation, and 1 energy balanceequation for each sub-system for a total of 6 equations Thus, we have 6 equations and 6

unknowns and can solve for them

Substituting these in the energy balance equation, we get: m = 4.2 kg/s

44 The system diagram is as follows:

45 The energy of steam at 130 /C and quality of 60% is given by:

47 The system diagram is as follows:

From the TSB equation for the second sub-system, we get: x = 0.31

The specific heats and enthalpies are determined as follows:

From the EB equation for the first sub-system, we get: T = 50.8 /C

From the EB equation for the second sub-system, we get: m = 4.25 kg/s

48 There are 2 steps to consider in this situation.

Step 1: Converting water at 25 /C to water at 100 /C

The energy required for this step is given by Q = m c (DT) = 5 c (100 - 25)

p

In the above expression, c is the specific heat of water

Step 2: Converting water at 100 /C to steam at 100 /C and a quality of 60%

The total energy for the process is given by Q = Q + Q = 5 c (100 - 25) + 5 (H - H )

49 The system diagram is as follows:

Note: In the first sub-system above, the stream that exits the system is at 25 /C since all theincoming streams are at 25 /C

For the first sub-system,