principles of food and bioprocess engineering (fs 231) solutions … · 2016-01-05 · principles...

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

pgiven by: c = 0.1 (1424) + 0.05 (1549) + 0.05 (1675) + 0.8 (4180) = 3647.6 J/kg K

2. The specific heat of the mixed stream is the weighted mean of the specific heats of the

pindividual streams and is given by: c = (10/15) (3800) + (5/15) (4180) = 3926.7 J/kg K

p(w) at 25 /C 13. Energy content of water at 25 /C = 5 [c (25)] = Q

s 2Energy content of saturated steam of 60% quality at 100 /C = 5 (H ) = Q

s v at 100 /C c at 100 /Cwith H = 0.6(H ) + 0.4(H )

2 1 v at 100 /C c at 100 /C p(w) at 25 /CTotal energy required = Q - Q = 5[0.6(H ) + 0.4 (H ) - c (25)]

p(w) at 40 /C 14. Energy content of water at 10 /C = m [c (10)] = Q

s 2Energy content of saturated steam of 80% quality at 100 /C = m (H ) = Q

s v at 100 /C c at 100 /Cwith H = 0.8(H ) + 0.2(H )

2 1 v at 100 /C c at 100 /C p(w) at 10 /CTotal energy required = Q - Q = m[0.8(H ) + 0.2 (H ) - c (10)]

p(w) at T /C 15. Energy content of 3 kg of water at T /C = 3 [c ]T = Q

s 2Energy content of saturated steam of 90% quality at 100 /C = 3 (H ) = Q

v at 100 /C c at 100 /C= 3 [0.9(H ) + 0.1(H ) ]

2 1 v at 100 /C c at 100 /C p(w) at T /C Total energy required = Q - Q = 3 [0.9(H ) + 0.1(H ) - c (T)]

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

s v cH = x (H ) + (1 - x) (H )Here, x = 0.9

v cFrom steam tables, we note that H = 2720.5 kJ/kg and H = 546.31 kJ/kg

sThus, H = 2503.08 kJ/kg

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

p(ice) 18. Energy required to convert ice at -10 /C to ice at 0 /C = m [c x 10] = Q

fus 2Energy required to convert ice at 0 /C to water at 0 /C = m (8 ) = Q

p(water) 3Energy required to convert water at 0 /C to water at 100 /C = m [c x 100] = Q

s c 4Energy required to convert water at 100 /C to steam at 100 /C = m (H - H ) = Q

s v at 100 /C c at 100 /C[H = 0.7(H ) + 0.3(H ) ]

1 2 3 4Total energy required = Q + Q + Q + Q

p9. a. Energy required = m c )T = 3 (4180) (353-323) = 376.2 kJ

fus pb. Energy required = m 8 + m c )T = 7 (333.2) + 7 (4.180) (353-273) = 4673.2 kJ

s v c10. Substitute H = 2650 kJ/kg, H = 2727.3 kJ/kg, H = 567.69 kJ/kg

This yields, x = 0.96

11. The energy gained by the product is given by:Q = = 2 (3800) (40) = 304,000 J/s = 304 kJ/sThe energy supplied by steam is given by:Q =

cHere, 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;

s304 = 0.15 (H - 567.69)

sSolving, we get: H = 2594.36 kJ/kg

s v cNow, H = x (H ) + (1 - x) (H )

s v cSubstituting the values of H , H (= 2727.3 KJ from steam tables), and H into the aboveequation, we get: x = 0.938

12. The system diagram is as follows:

Performing a mass balance, we get: 2 + 0.25 = mSolving, we get: m = 2.25 kg

p(w) at 10 /C s p(w) at 70 /CPerforming an energy balance, we get: 2 {c }10 + 0.25 (H ) = m {c }70From tables containing properties of water, we find that:

p(w) at 10 /C p(w) at 70 /Cc = 4195 J/kg K and c = 4187 J/kg K

sSolving, the energy balance equation, we get: H = 2302.2 kJ/kg

v cFrom steam tables, we note that H = 2706.3 kJ/kg and H = 503.71 kJ/kg

s v cNow, H = x (H ) + (1 - x) (H )

s v cSubstituting the values of H , H , and H , in the above equation, we get:2302.2 = x (2706.3) + (1 - x) 503.71Solving, we get: x = 0.82 Thus, steam quality = 82%

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 riseto a total of 6 equations. The specific heats of each stream is determined as a weighted meanof the specific heats of the individual components in the following manner:

p p p(c of product) = (solids fraction)(c of solid) + (water fraction)(c of water)

sThe enthalpy of steam is given by: H = 0.75 (2727.3) + 0.25 (567.69) = 2187.4 kJ/kg

1 2Overall mass balance in 1 sub-system: 6 + m = 2 + mst

1 2Solids balance in first sub-system: 0.3 (6) = 0.4 (2) + x (m )

p(1) 1 s p(2) 2 p(3)Energy balance in first sub-system: 6 {c 10} + m (H ) = 2 {c 30} + m {c T}

2 3Overall mass balance in 2 sub-system: m = 3 + mnd

1 2 3Solids balance in second sub-system: x (m ) = 0.3 (3) + 0.05 (m )

2 p(3) 4 p(w) 4 p(w)Energy balance in second sub-system: m {c T} + m {c 10} = m {c 30} +

p(4) 3 p(5)3 {c 20} + m {c 25}


pc of solids = 3600 J/kg K

pc of water = 4180 J/kg K

Mass balance equation in the first sub-system: 10 = 2 + m

pTotal solids balance equation in the first sub-system: 10 (0.6) = 2 (0.2) + m (x )

pSolving, we get: m = 8 kg/s x = 0.7

pc = 0.6 (3600) + 0.4 (4180) = 3832 J/kg K

pc ' = 0.2 (3600) + 0.8 (4180) = 4064 J/kg K

p p pc '' = x (3600) + (1 - x ) (4180) = 3774 J/kg K

v cFrom steam tables, H = 2720.5 kJ/kg, H = 546.32 kJ/kg at 270.1 KPa

s v cH = 0.9 (H ) + 0.1 (H ) = 2503.08 kJ/kg

Energy balance equation in the first sub-system:

p s s c p p10 (c ) (20) + m (H - H ) = 2 (c ') (40) + m (c '') (70)

Energy balance equation in the second sub-system:

cw pm (4180) (15) = m (c '') (40)

s cwSolving, we get: m = 0.85 kg/s m = 19.3 kg/s


1 2Overall mass balance in first sub-system: 10 +25 + m = m

2Solids balance in first sub-system: 0.3 (10) + 0.2 (25) = x (m )

p(1) p(2)Energy balance in first sub-system: 10 {c (283 - 273)} + 25 {c (313 - 273)} +

1 s 2 p(3)m (H ) = m {c (343 - 273)}

2 p(3) cw p(5)Energy balance in second sub-system: m {c (343 - 273)} + m {c (278 - 273)} =

2 p(4) cw p(6)m {c (303 - 273)} + m {c (298 - 273)}

p (1) p (s) p (w) at 10 /Cc = 0.3 c + 0.7 c

p (2) p (s) p (w) at 40 /Cc = 0.2 c + 0.8 c

p (3) p (s) p (w) at 70 /Cc = x c + (1 - x) c

p (4) p (s) p (w) at 30 /Cc = x c + (1 - x) c

p (5) p (w) at 5 /Cc = c

p (6) p (w) at 25 /Cc = c

s v at 270.1 kPa c at 270.1 kPaH = 0.7 (H ) + 0.3 (H )

Thus we have 4 basic equations and 4 basic unknowns OR 7 overall equations (includingspecific heat and enthalpy equations) and 7 overall unknowns.

p16. a. Energy required = m c )T = 3 (4180) (353-323) = 376.2 kJ

fus pb. Energy required = m 8 + m c )T = 7 (333.2) + 7 (4.180) (353-273) = 4673.2 kJ

v c at 316.3 kPa17. Energy released = (H - H ) = 2733.9 - 589.13 = 2144.77 kJ/kg

p (ice) 1 fus p (water) 218. Energy required = m c )T + m 8 + m c )T = 5 (2.050)(273 - 243) + 5 (333.2) +5 (4.180) (363 - 273) = (307.5 + 1666.0 + 1881.0) kJ = 3854.5 kJ

1 2 3 4 1 2 3 4 5 6 719. We have 12 unknowns in the system: m , m , m , m , x , x , x , x , x , x , x , T

First sub-system

1 2Overall mass balance: 10 + m = 4 + m

1 2Starch balance: 0.2 (10) = 4 (x ) + 0.4 (m )

2Vitamin balance (1): x = 0

3Vitamin balance (2): x = 0

p(1) 1 s p(2) 2 p(3)Energy balance: 10 {c }(283 - 273) + m H = 4{c }(293 - 273) + m {c }(T - 273)Second sub-system

2 4Overall mass balance: m + 6 = m

2 7Starch balance: 0.4 (m ) = 16 (x )

5Vitamin balance: 0.3 (6) = 16 (x )

2 p(3) 3 p(w) p(4)Energy balance: m {c } (T - 273) + {m c } (283 - 313) + 6 {c } (323 - 273) =

4 p(5)m {c } (298 - 273)

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 2x + x + 0.1 = 1.0 (The sum of the fractions of each component leaving the first sub-system at the top has to equal 1.0)

3 4x + x + 0.4 = 1.0 (The sum of the fractions of each component entering the second sub-system has to equal 1.0)

5 6 7x + x + x = 1.0 (The sum of the fractions of each component leaving the second sub-system has to equal 1.0)

Thus, we have 12 equations and 12 unknowns and can hence solve the set of equations oncewe have the following equations for the specific heats and enthalpy:

p(1) p(starch) p(water)c = 0.2 {c } + 0.8 {c }

p(2) 1 p(starch) 2 p(starch) p(water)c = x {c } + x {c } + 0.1 {c }

p(3) p(starch) 3 p(vit) 4 p(water)c = 0.4 {c } + x {c } + x {c }

p(4) p(vit) p(water)c = 0.3 {c } + 0.7 {c }

p(5) 5 p(vit) 6 p(water) 7 p(starch)c = x {c } + x {c } + x {c }

s v at 415.4 kPa c at 415.4 kPaH = 0.8 (H ) + 0.2 (H )

1 2 320. The system diagram is shown below. We have 4 unknowns in the system: m , m , m , T

First sub-system

2Overall mass balance: 4 + 2 = m

1 s c 2 p (1)Energy balance: 4(3500)(313 - 273) + 2(4000)(283 - 273) + m (H - H ) = m {c }(T -273)

Second sub-system


2 p (1) p (2) 3 p (3)Energy balance: m {c } (T - 273) + 3 {c } (278 - 273) = m {c } (303 - 273)

Thus, we have 4 equations and 4 unknowns and can hence solve the system of equations oncewe have the following expressions for specific heat and enthalpy:

p (1)c = (4/6) (3500) + (2/6) (4000) = 3666.67 J/kg K

p (2) p water at 5 /Cc = c = 4206 J/kg K

p (3) 2 2 p (1) 2 p water at 30 /Cc = {m /(m + 3)}{c } + {3/(m + 3)}c

s v at 145 /C c at 145 /CH = 0.8 (H ) + 0.2 (H )

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

1� H = 2748.6 kJ/kg

Enthalpy of saturated steam (quality = 90%) is given by:

2H = 0.9 (2683.8) + 0.1 (440.15) = 2459.4 kJ/kg

1 2Thus, energy released = H - H = 289.2 kJ/kg

b. Enthalpy of superheated steam at 150 /C and 75 kPa = 2778.2 kJ/kgEnthalpy of saturated vapor at 150 /C = 2746.5 kJ/kgThus, energy released = 2778.2 - 2746.5 = 31.7 kJ/kg


pc = 0.4 (3000) + 0.6 (4180) = 3708 J/kg K

pc ' = 0.6 (3000) + 0.4 (4180) = 3472 J/kg K

pc '' = x (3000) + (1 - x) (4180) = {4180 - 1180 x} J/kg K

v vH = 2733.9 kJ/kg at 316.3 kPa H = 2740.3 kJ/kg at 415.4 kPa

c cH = 589.13 kJ/kg at 316.3 kPa H = 610.63 kJ/kg at 415.4 kPa

Interpolating,

v cThis yields, H = 2736.08 kJ/kg, H = 596.44 kJ/kg

sThus, H = 0.8 (2736.08) + 0.2 (596.44) = 2308.15 kJ/kg

1 2Overall mass balance equation: 10 + m = m + 5

2Total solids balance equation: 0.4 (10) = x (m ) + 0.6 (5)

p 1 s 2 p pEnergy balance equation: 10 (c ) (10) + m (H ) = m (c '') (60) + 5 (c ') (30)

1 2Solving these equations yields: m = 0.65 kg/s, m = 5.65 kg/s, x = 0.18


sOverall mass balance equation: 10 + m = m

s s pEnergy balance equation: 10 (3000) (20) + m (H ) = m (c ') (70)

p s sSpecific heat of product: c ' = {10(3000) + m (4180)}/{10 + m )

c vFrom steam tables, H = 567.69 kJ/kg, and H = 2727.3 kJ/kg at 313 kPa

s v cEnthalpy of steam: H = 0.8 (H ) + 0.2 (H ) = 2295.38 kJ/kg

sSolving, we get: m = 10.75 kg/s m = 0.75 kg/s

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

1 2 3 4Other 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.

First sub-system

1 3Overall mass balance: 5 = m + m

1 3Solids balance: 0.3 (5) = 0.2 (m ) + 0.4 (m )

2 s c 1Energy balance: 5 (4000) (283 - 273) + m (H - H ) = m (4200) (313 - 273) +

3 p(1)m c (T - 273)

Second sub-system

3Overall mass balance: m + 4 = 7

3Solids balance: 0.4 (m ) + 0.25 (4) = x (7)

3 p(1) 4 p(2)Energy balance: m c (T - 273) + m c (283 - 273) + 4 (3800) (288 - 273) =

p(4) 4 p(3)7 c (293 - 273) + m c (308 - 273)

p(2) p(3)We note that c and c are the specific heats of water at 10 /C and 35 /C respectively andcan 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(1) 3 2 3c = (5/m )(4000) + (m /m )(4200)

p(4) 3 p(1)c = (m /7)(c ) + (4/7)(3800)

s v at 313 kPa c at 313 kPaH = 0.9 (H ) + 0.1 (H )

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

1 2 s 1 2 p sunknowns in the system (m , m , m , x , x , T). We do not include c and H as unknowns.However, if we decide to include them as unknowns, we then include 2 more equations -- 1

p sequation to compute c and the other to compute H so that we have a total of eight equationsand eight unknowns.

First sub-system

1OMB: 5 + 2 = m

1 1Solids balance: 0.3 (5) + 0.4 (2) = x (m )

s s c 1 pEnergy balance: 5(3500)(293 - 273) + m (H - H ) + 2(3800)(298 - 273) = m c (T - 273)

Second sub-system

1 2OMB: m + 10 = 5 + m

1 1 2 2Solids balance: x (m ) = x (5) + 0.35 (m )

1 p 2Energy balance: m c (T - 273) + 10(4180)(278 - 273) = 5(4000)(303 - 273) + m (3600)(318- 273)

pAlso, c = (5/7)(3500) + (2/7)(3800)

s v at 313 kPa c at 313 kPaH = 0.8 (H ) + 0.2 (H )

26. The system diagram is as shown below:

First sub-system

p(1) 1 s c p(1)Energy balance equation: 3 c (283 - 273) + m (H - H ) = 3 c (353 - 273)

Second sub-system

2 3Overall mass balance equation: 3 + m = m

3Solids balance equation: 0.7 (3) = x (m )

p(1) 2 p(w) 3 p(2)Energy balance equation: 3 c (353 - 273) + m c (278 - 273) = m c (293 - 273)

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

p(1) p(s) p(w)c = 0.7 c + 0.3 c

p(2) 3 p(1) 2 3 p(w) p(s) p(w)c = (3/m ) c + (m /m ) c with c = 4000 J/kg K, c = 4180 J/kg K

s v at 313 kPa c at 313 kPa v cH = 0.8 (H ) + 0.2 (H ) with H = 2727.3 kJ/kg, H = 567.69 kJ/kg

p(1) sThis yields: c = 4054 J/kg K, H = 2295.38 kJ/kg

p(2)Substituting the expression for c in the energy balance equation for the second sub-system,we get:

2 p(1) 2 p(w)3 (4054) (80) + m (4180) (5) = [3 c + m c ] (20)

2 2Simplifying, we get: 972960 + 20900 m = [3 (4054) + m (4180)] (20)

2Solving, we get: m = 11.64 kg/s

3Thus, m = 14.64 kg/s, x = 0.14

p(2)c = 4154 J/kg K

1m = 0.49 kg/s

27. The system diagram for this problem is as follows:

1 2Overall mass balance: 5 + m = m

2Solids balance: 0.3 (5) = 0.27 (m )

p(1) 1 v 2 p(2)Energy balance: 5 c (283 - 273) + m (H ) = m c (348 - 273)

Using the idea of weighted means, we write the expressions for specific heats as follows:

p(1) p(s) p(w) at 10 /Cc = 0.3 c + 0.7 c = 0.3 (3600) + 0.7 (4195) = 4016.5 J/kg K


2From the solids balance equation, we get: m = 5.56 kg/s

1Thus, from the overall mass balance equation, we get: m = 0.56 kg/s

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

v5 (4016.5) (10) + 0.56 (H ) = 5.56 (4030.7) (75)

vThis yields: H = 2642.8 kJ/kg

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


First sub-system

1 2Overall mass balance: 10 + 3 + m = m

Second sub-system

2Overall mass balance: m + 4 = 21

2 p(2) p(3)Energy balance: m c (353 - 273) + 5 c (278 - 273) + 4 (3000) (313 - 273) =

p(4) p(5)5 c (298 - 273) + 21 c (T - 273)

2From 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

1m = 4 kg/s

Using the idea of weighted means, we write the following expressions for specific heat:

p(2) p(1)c = [(10/17)(4000) + (3/17)(c ) + (4/17)(3600)]

p(5) 2 p(2)c = [(m /21)(c ) + (4/21)(3000)]

p(1)We note that: c = specific heat of water at 80 /C = 4194 J/kg K

p(3)c = specific heat of water at 5 /C = 4206 J/kg K

p(4)c = specific heat of water at 25 /C = 4178 J/kg K

p(2) p(5)Thus, we get: c = 3940 J/kg K, c = 3761 J/kg K

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



p(w) at 20 /C 1 s 2 p(w) at 70 /CEnergy balance: 2 c (20) + m (H ) = m c (70)

s v (at 232. 1kPa) c (at 232.1 kPa)Also, H = 0.9 {H } + 0.1 {H } = 0.9 (2713.5) + 0.1 (524.99)= 2494.65 kJ/kg = 2494650 J/kg

p(w) at 20 /C p(w) at 70 /Cc = 4182 J/kg K and c = 4187 J/kg K

1 1Solving, we get: (2) (4182) (20) + m (2494650) = (2 + m ) (4187) (70)

1Rearranging, we get: m [(2494650) - (4187)(70)] = (2)(4187)(70) - (2) (4182)(20)

1Solving, we get: m = 0.19 kg


First sub-system

s 1Overall mass balance equation: 2.5 + m = m

1 1Solids balance equation: 0.6 (2.5) = m x

p(1) s s 1 p(2)Energy balance equation: 2.5 c (293 - 273) + m (H ) = m c (343 - 273)

Second sub-system

1 2 3Overall mass balance equation: m + m = m

1 1 3 2Solids balance equation: m x = m x

1 p(2) 2 p(w) 3 p(3)Energy balance equation: m c (343 - 273) + m c (277 - 273) = m c (283 - 273)

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

p(1) p(s) p(w)c = 0.6 c + 0.4 c

p(2) 1 1 p(w) p(s) p(w)c = (x ) 3800 + (1-x ) c with c = 3800 J/kg K, c = 4180 J/kg K

p(3) 2 2 p(w)c = (x ) 3800 + (1-x ) c

s v at 313 kPa c at 313 kPa v cH = 0.7 (H ) + 0.3 (H ) with H = 2727.3 kJ/kg, H = 567.69 kJ/kg

p(1) sThis yields: c = 3952 J/kg K, H = 2079.42 kJ/kg

1 s 1 1 p(1) p(2)Substituting the values of m (= 2.5 + m ), m x (= 1.5), c (= 3952), and c (= 4180 -

1380x ) in the energy balance equation for the first sub-system, we get:

s s197600 + 2079420 m = 70 (2.5 + m ) 4180 - 70 (380) (1.5)

sSolving, we get: m = 0.276 kg/s

1 1Thus, m = 2.78 kg/s, x = 0.54

3 2 3 2 p(2) 1 p(3)Substituting the values of m (= 2.78 + m ), m x (= 1.5), c (= 4180 - 380 x ), and c (=

24180 - 380x ) in the energy balance equation for the second sub-system, we get:

2 2773496.1 + 16720 m = 41800 (2.78 + m ) - (3180) (1.5)


3 2Thus, m = 29.2 kg/s, x = 0.05


First sub-system

1 2Overall mass balance: 3 + 4 + m = m

p p p 2 p 1Energy balance equation: 3 (c ) (75) + 4 (c ) (75) + 3 (c ) (75) = m (c ) (T )

Second sub-system

2Overall mass balance: m + 5 + 5 = 20

2 p 1 p p p 2Energy balance equation: m (c ) (T ) + 5 (c ) (7) + 5 (c ) (45) = 20 (c ) (T )

1 2 1 2Solving, we get: m = 3 kg/s, m = 10 kg/s, T = 75 /C, T = 50.5 /C

32. The system diagram for this problem is as follows:


2Solids balance: 0.35 (3) = 0.3 (m )

p(1) 1 v 2 p(2)Energy balance: 3 c (278 - 273) + m (H ) = m c (368 - 273)

Using the idea of weighted means, we write the expressions for specific heats as follows:



From the overall mass balance and solids balance equation, we get:

2 1m = 3.5 kg/s, m = 0.5 kg/s

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

v3 (3993.9) (5) + 0.5 (H ) = 3.5 (4024.2) (95)

vThis yields: H = 2556.28 kJ/kg

vFrom the steam tables we see that this value of H corresponds to a temperature of ~ 30 /C

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

For the first sub-system:

1Overall mass balance equation: 10 + m + 2 = 3 + 13

1Total Sugar balance equation: 0.3 (10) + 0.6 (2) = 0.1 (3) + x (13)

For the second sub-system:

2Overall mass balance equation: 13 + 2 = m + 2 + 5

1 3 2Total Sugar balance equation: x (13) + 0.05 (2) = x (m ) + 0.6 (2)

2 2Total Starch balance equation: 0.05 (2) = 0.02 (5) + x (m )

1 2 1 2 3Solving, we get: m = 4 kg/s, m = 8 kg/s, x = 0.3, x = 0, x = 0.35


1 2 3We have 4 unknowns in the system: m , m , m , T

First sub-system

2Overall mass balance: 3 + 2.5 = m

1 s c 2 p (1)Energy balance: 3(3500)(303 - 273) + 2.5(4100)(288 - 273) + m (H - H ) = m {c }(T - 273)

Second sub-system


2 p (1) p (2) 3 p (3)Energy balance: m {c } (T - 273) + 2 {c } (283 - 273) = m {c } (308 - 273)

Thus, we have 4 equations and 4 unknowns and can hence solve the system of equations oncewe have the following expressions for specific heat and enthalpy:

p (1)c = (3/5.5) (3500) + (2.5/5.5) (4100) = 3772.7 J/kg K

p (2) p water at 10 /Cc = c = 4195 J/kg K

p (3) 2 2 p (1) 2 p water at 10 /Cc = {m /(m + 2)}{c } + {2/(m + 2)}c

s v at 130 /C c at 130 /CH = 0.75 (H ) + 0.25 (H )

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

1 2 3 4 1 1Other 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.

First sub-system

1 2 3Overall mass balance: 4 + m + m = m

1 1 3Solids balance: 0.25 (4) + 0.2 (m ) = x (m )

p(1) 2 s 1 p(2) 3 p(3) 1Energy balance: 4 (c ) (15) + m (H ) + m (c ) (40) = m (c ) (T )

Second sub-system

3Overall mass balance: m = 4 + 10

1 3Solids balance: x (m ) = 0.35 (4) + 0.15 (10)

3 p(3) 1 4 p(6) p(4)Energy balance: m c (T ) + m c (5) = 4 (c ) (15) +

p(5) 4 p(7)10 c (20) + m c (25)

p(6) p(7)We note that c and c are the specific heats of water at 5 /C and 25 /C respectively and canbe obtained from tables containing properties of water.

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

p(1) p(solids) p(water)c = 0.25 [c ] + 0.75 [c ]


p(3) 1 p(solids) 1 p(water)c = x [c ] + (1 - x ) [c ]



s v at 150 /C c at 150 /CH = 0.8 (H ) + 0.2 (H )

s s36. The energy content of steam is given by m H

v at 316.3 kPa c at 316.3 kPa The energy content of steam initially = 3 [0.5 (H ) + 0.5 (H )]= 3 [0.5 (2733.9) + 0.5 (589.13)] = 4984.55 kJ

v at 316.3 kPa c at 316.3 kPaThe energy content of steam finally = 3 [0.9 (H ) + 0.1 (H )]= 3 [0.9 (2733.9) + 0.1 (589.13)] = 7558.27 kJ

Thus, energy required for conversion = 7558.27 - 4984.55 = 2573.72 kJ

p 37. The energy content of water is given by m c T Thus, in this problem, the energy of water = 5 (4184) (65) = 1359.8 kJ

v at 120 /C c at 120 /CThe energy content of steam = 5 [0.75 (H ) + 0.25 (H )]= 5 [0.75 (2706.3) + 0.25 (503.71)] = 10778.26 kJ

Thus, energy required for conversion = 10778.26 - 1359.8 = 9418.46 kJ

v at 150 /C c at 150 /C38. Energy of steam initially = 4 [0.95 (H ) + 0.05 (H )]= 4 [0.95 (2746.5) + 0.05 (632.2)] = 10563.14 kJ

c at 150 /CEnergy of steam finally = 4 (H ) = 4 (632.2) = 2528.8 kJ

Thus, energy released = 10563.14 - 2528.8 = 8034.34 kJ


For the first sub-system

s 1Overall mass balance: 8 + m = m

1 1Solids balance: 0.3 (8) = m x

p(1) s s 1 p(2)Energy balance: 8 c (20) + m H = m c (80)

sWe thus have 3 equations and 3 unknowns as long as we can write expressions for H and theindividual specific heats (without introducing any new unknowns).

s v at 135 /C c at 135 /CH = 0.85 (H ) + 0.15 (H ) = 0.85 (2727.3) + 0.15 (567.69) = 2403.3585 kJ/kg= 2403358.5 J/kg

p(1)c = 0.3 (4000) + 0.7 (4180) = 4126 J/kg K

p(2) 1 1 1c = 4000 x + 4180 (1 - x ) = 4180 - 180 x

We note that each of the first 3 equations written have more than 1 unknown in it and hence, we

s 1cannot solve for the unknowns easily, one at a time. We thus eliminate the unknowns m and xas follows:

s 1For the first equation (overall mass balance equation), we get: m = m - 8

sSubstituting these and the values of the specific heats and H into the energy equation, we get:

8 (4126) (20) + (m1 - 8) (2403358.5) = m1 (4180 - 180 x1) (80)

1 1From the second equation (solids balance equation), we get: x = 2.4/mSubstituting this into the simplified energy balance equation above yields:

1 1 18 (4126) (20) + (m - 8) (2403358.5) = m [4180 - 180 (2.4/m )] (80)

Simplifying, we get:

1 1 1660160 + 2403358.5 m - 19226868 = 80 m (4180 - 432/m )

1This reduces to: m [2403358.5 - 80(4180)] = 19226868 - 660160 - 80 (432)

1Thus, m (2068958.5) = 18532148


sSubstituting this into the overall mass balance equation, we get: m = 0.96 kg/s

1From the solids balance equation, we get: x = 0.27

For the second sub-system

1 p(2) p(w) 1 p(2) p(w) at TEnergy balance: m c (80) + 30 c at 5 /C (5) = m c (60) + 30 c (T)

p(w) at TIn the above equation, the unknowns are c and T

p(w) at TWe cannot solve for 2 unknowns from this equation. So we approximate the value of c as4180 J/kg K

p(2) 1Also, c = 4180 - 180x = 4180 - 180 (0.27) = 4131.4 J/kg KThus, we get:8.96 (4131.4)(80) + 30 (4206) (5) = 8.96 (4131.4) (60) + 30 (4180) (T)

Solving, we get T = 10.9 /C

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

For the first sub-system

1Overall mass balance: 5 = 3 + m

1 1Solids balance: 0.3 (5) = 0.1 (m ) + x (3)

p(1) s s c 1 p(2) p(3) 1Energy balance: 5 c 10 + m (H - H ) = m c 35 + 3 c T

For the second sub-system

2Overall mass balance: 3 = 1 + m

1 2 2Solids balance: x (3) = 0.05 (1) + x (m )

p(3) 1 p(w) at 15 /C p(4) 2 p(5) p(w) at 40 /CEnergy balance: 3 c T + 4 [c ] (15) + 1 c (30) + m c 25 + 4 [c ] (40)

1 s 2 1 2 1We thus have 6 equations and 6 unknowns (m , m , m , x , x , T ). We can solve for these as longas 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:

s v at 140 /C c at 140 /C v cH = 0.8 (H ) + 0.2 (H ) with H and H values being obtained from steam tables






41. The energy supplied by the heater is given by:

pQ = m (c ) (DT)

pThus, DT = Q / {m (c )}

pSince Q is the same for both liquids, the one with lower value of m (c ) will have a higher DT

pFor the first liquid, m (c ) = 3 (2000) = 6000 J/K

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

pSince 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 2The 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.

First sub-system

1Overall mass balance: 4 + 2.5 = m

s s 1 1 s cEnergy balance: 4 (3800) (35) + m (H ) + 2.5 (4100) (15) = m (4000) (T ) + m (H )

s s v at 130 /C c at 130 /CThe expression for H in the above equation is: H = 0.75 (H ) + 0.25 (H )

v cThe vales of H and H can be determined from steam tables.

Second sub-system

1 2Overall mass balance: m + 3 = 1 + m

1 1 2Energy balance: m (4000) (T ) + 3 (3750) (20) = 1 (3000) (15) + m (3900) (35)

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


s 1 cw 2 1 2We have 6 unknowns in this problem: m , m , m , m , x , xWe 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 6unknowns and can solve for them.

First sub-system

1 1Overall mass balance: 4 = 1 + m Thus, m = 3 kg/s

1 1 1Solids balance: 0.25 (4) = x (m ) Thus, x = 0.33

p(1) s s 1 p(2) s cEnergy balance: 4 {c }20 + m (H ) = 1 (4200) (70) + m {c }(60) + m (H )In the above equation,

p(1){c } = 0.25 (3600) + 0.75 (4200) = 4050 J/kg K

p(2) 1 1{c } = x (3600) + (1 - x ) (4200) = 4002 J/kg K

s v(at 130 /C) c(at 130 /C)H = 0.85 {H } + 0.15 {H } = 0.85 (2720.5) + 0.15 (546.31) = 2394.371 kJ/kg= 2394371 J/kg

sSubstituting these values in the energy equation, we get: m = 0.37 kg/s

Second sub-system

1 2 2Overall mass balance: m + 2 = m Thus, m = 5 kg/s

1 1 2 2 2Solids balance: x (m ) + 0.3 (2) = x (m ) Thus, x = 0.32

1 p(2) cw p(3) 2 p(4) cwEnergy balance: m {c }(60) + m (4200) (10) + 2 {c } (40) = m {c }(30) + m(4200) (35)

In the above equation,

p(3){c } = 0.3 (3600) + 0.7 (4200) = 4020 J/kg K

p(4) 2 2{c } = x (3600) + (1 - x ) (4200) = 4008 J/kg K

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


1 1 s cwThe unknowns are: m , x , m , m . Thus, we need 4 equations to solve for these unknowns.

1OMB for 1 sub-system: 2 = 0.3 + mst

1 1TSB for 1 sub-system: 2 (x ) = 0.6 (m )st

p(1) s s 1 p(2) s cEB for 1 sub-system: 2 {c }(20) + m H = 0.3 (4180) (70) + m {c }(70) + m (H )st

1 p(2) cw cw 1 p(2)EB for 2 subs-system: m {c }(70) + m (4180) (5) = m (4180) (20) + m {c }(25)nd

p(1) 1 1 p(2)c = x (3000) + (1 - x ) 4180 c = 0.6 (3000) + (0.4) 4180

s v(at 110 /C) c(at 110 /C)H = (0.8){H } + (0.2){H }

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

s v at 130 /C c at 130 /CH = 0.6 (H ) + 0.4 (H ) = 0.6 (2720.5) + 0.4 (546.31) = 1850.82 kJ/kg

The energy of steam at 130 /C and quality of 85% is given by:

s v at 130 /C c at 130 /CH’ = 0.85 (H ) + 0.15 (H ) = 0.85 (2720.5) + 0.15 (546.31) = 2394.37 J/kg

Mass of steam = 3 lbs = 3 (0.45359) kg = 1.36 kg

Thus, amount of energy required to convert 3 lbs of steam at 130 /C from a quality of 60% toa quality of 85% is given by 1.36 (2394.37 - 1850.82) = 739.2 kJ

46. The system diagram (with all unknowns marked) is as follows:

1 2 s cw 1The 6 unknowns in the system are: m , m , m , m , x , T and we have the following 6 equations:


1OMB: 0.3 + 0.7 = m

1 1TSB: 0.25 (0.3) + x (0.7) = 0.25 (m )

p(1) s s p(1) s c 1 p(3) 1EB: 0.3 [c (5)] + m (H ) + 0.7 [c (40)] = m (H ) + m [c (T )]


1 2OMB: m = m + 0.4

1 2TSB: 0.25 (m ) = 0.35 (m ) + 0.35 (0.4)

1 p(3) 1 cw pw( cw pw 2 p(4) p(5)EB: m (c (T ) + m [c (5)] = m [c (20)] + m (c (25) + 0.4[c (20)]

In the above equations, specific heats and enthalpies are determined as follows:

p(1)c = 0.25 (3700) + 0.75 (4200)

p(2) 1 1c = x (3700) + (1 - x ) (4200)

p(3)c = 0.25 (3700) + 0.75 (4200)

p(4)c = 0.35 (3700) + 0.65 (4200)

p(5)c = 0.35 (3700) + 0.65 (4200)

s v(at 130 /C) (at 130 /C)H = 0.95 [H ] + 0.05 [Hc ]

v cH and H are obtained from steam tables at 130 /C


1 s cw 1 2The 6 unknowns in the system are: m , m , m , x , x , T and we have the following 6 equations:


s 1OMB: 6 + m = 0.5 + m

1TSB: 0.3 (6) = 0.1 (0.5) + x (m )

p(1) s s p(2) 1 p(3)EB: 6 [c (5)] + m (H ) = 0.5 [c (60)] + m [c (T)]


1OMB: m + 1.5 = 7.5

1 1 2TSB: x (m ) + 0.4 (1.5) = x (7.5)

1 p(3) p(4) cw pw cw pw p(5)EB: m [c (T)] + 1.5 [c (30)] + m [c (5)] = m [c (25)] + 7.5 [c (35)]

1From the OMB equation for sub-system 2, we get: m = 6 kg/s

sFrom the OMB equation for sub-system 1, we get: m = 0.5 kg/s

1From the TSB equation for the first sub-system, we get: x = 0.29

2From the TSB equation for the second sub-system, we get: x = 0.31

The specific heats and enthalpies are determined as follows:

p(1)c = 0.3 (3800) + 0.75 (4200) = 4080 J/kg K

p(2)c = 0.1 (3800) + 0.9 (4200) = 4160 J/kg K

p(3) 1 1c = x (3800) + (1 - x ) (4200) = 4084 J/kg K

p(4)c = 0.4 (3800) + 0.6 (4200) = 4040 J/kg K

p(5) 2 2c = x (3800) + (1 - x ) (4200) = 4076 J/kg K

s v(at 125 /C) c(at 125 /C)H = 0.9 [H ] + 0.1 [H ] = 0.9 (2713.5) + 0.1 (524.99) = 2494.65 x 10 J/kg3

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

cwFrom 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

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

pIn the above expression, c is the specific heat of waterStep 2: Converting water at 100 /C to steam at 100 /C and a quality of 60%

2 s c s cThe energy for this step is determined Q = m (H - H ) = 5 (H - H )

s v at 100 /C c at 100 /CIn the above expression, H = 0.6 (H ) + 0.4 (H )

c vH and H are determined from steam tables at a temperature of 100 /C

1 2 p s cThe total energy for the process is given by Q = Q + Q = 5 c (100 - 25) + 5 (H - H )


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,OMB: 0.5 + 1.2 + 2.8 = mTSB: 0.3 (0.5) + 0.35 (1.2) + 0.4 (2.8) = x (m)

For the second sub-system,

p s s p s cEB: m c (25) + m (H ) = m c (75) + m (H )

In the above equation,

pc = [0.5/(0.5+1.2+2.8)]3000 + [1.2/(0.5+1.2+2.8)]3500 + [2.8/(0.5+1.2+2.8)]4000

s v(at 125 /C) c(at 125 /C)H = (0.9)[H ] + (0.1)[H ]

principles of food and bioprocess engineering (fs 231) solutions … · 2016-01-05 · principles...

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