HW1

15)

a)

d(M1)/d(t)=10

d(M2)/d(t)=0

d(M3)/d(t)=0

d(S1)/d(t)=2-10*S1/M1

d(S2)/d(t)=10*S1/M1-10*S2/M2

d(S3)/d(t)=10*S2/M2-10*S3/M3

t(0)=0

M1(0)=1000

M2(0)=1000

M3(0)=1000

S1(0)=200

S2(0)=200

S3(0)=200

t(f)=60

b)

t M1 M2 M3 S1 S2 S3 SaltPC1 SaltPC2 SaltPC3 60 1600 1000 1000 222.5 181.7907 196.3368 13.90625 18.17907 19.63368

POLYMATH Report No TitleOrdinary Differential Equations 09-Sep-2014

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value

1 M1 1000. 1000. 1600. 1600.

2 M2 1000. 1000. 1000. 1000.

3 M3 1000. 1000. 1000. 1000.

4 S1 200. 200. 222.5 222.5

5 S2 200. 181.7907 200. 181.7907

6 S3 200. 196.3368 200. 196.3368

7 t 0 0 60. 60.

Differential equations 1 d(M1)/d(t) = 10

2 d(M2)/d(t) = 0

3 d(M3)/d(t) = 0

4 d(S1)/d(t) = 2-10*S1/M1

5 d(S2)/d(t) = 10*S1/M1-10*S2/M2

6 d(S3)/d(t) = 10*S2/M2-10*S3/M3

General Total number of equations 6Number of differential equations 6Number of explicit equations 0Elapsed time 0.000 secSolution method RKF_45Step size guess. h 0.000001Truncation error tolerance. eps 0.000001

Data file: c:\users\knghcmdell\google drive\fa\chee 400\1-15.pol

16)

a)

# Example 1 - Heat Exchange in a Series of Tanks

d(T1)/d(t) = (W * Cp * (T0 - T1) + UA * (Tsteam - T1)) / (M * Cp) # Temperature in the first tank (deg. C)

d(T2)/d(t) = (W * Cp * (T1 - T2) + UA * (Tsteam - T2)) / (M * Cp) # Temperature in the second tank (deg. C)

d(T3)/d(t) = (W * Cp * (T2 - T3) + UA * (Tsteam - T3)) / (M * Cp) # Temperature in the third tank (deg. C)

# The explicit equations

W = 100 # Feed flow rate (kg/min)

Cp = 2.0 # Heat capacity (kJ/kg -deg. C)

T0 = 20 # Feed temperature (deg C)

UA = 10. # Area*heat transfer coefficient (kJ/min *deg C)

Tsteam = 250 # Temperature of steam (deg. C)

M = 1000 # Total mass in a tank (kg)

# Initial values of the differential variables

T1(0) = 20

T2(0) = 20

T3(0) = 20

# Initial/final values of the independent differentiation variable

t(0) = 0

t(f) = 200

b)

d(T1)/d(t)=UA*(Tstream-T1)/(M*Cp)



UA=10

Tstream=250

M=1000

Cp=2.0

t(0)=0

T1(0)=30.95238

T2(0)=41.38322

T3(0)=51.31735

t(f)=180

c)

# Example 1 - Heat Exchange in a Series of Tanks

d(T1)/d(t) = (W * Cp * (T0 - T1) + UA * (Tsteam - T1)) / (M * Cp) # Temperature in the first tank (deg. C)

d(T2)/d(t) = (W * Cp * (T1 - T2) + UA * (Tsteam - T2)) / (M * Cp) # Temperature in the second tank (deg. C)

d(T3)/d(t) = (W * Cp * (T2 - T3) + UA * (Tsteam - T3)) / (M * Cp) # Temperature in the third tank (deg. C)

# The explicit equations

W = 100 # Feed flow rate (kg/min)

Cp = 2.0 # Heat capacity (kJ/kg -deg. C)

T0 = 20 # Feed temperature (deg C)

UA = 10. # Area*heat transfer coefficient (kJ/min *deg C)

Tsteam = 250 # Temperature of steam (deg. C)

M = 1000 # Total mass in a tank (kg)

# Initial values of the differential variables

T1(0) = 160.9419

T2(0) = 165.1827

T3(0) = 169.2217

# Initial/final values of the independent differentiation variable

t(0) = 0

t(f) = 200

Result:

t T1 T2 T3 88.33253 30.96457 41.50247 51.90634 91.53253 30.96109 41.47122 51.76597

Since 1.01*T3 (steady state)=51.8305, t is between 88.33 and 91.53 minutes.

By changing t(f), the time could be determined accurately at 89.956 minutes.

POLYMATH Report Example 1 - Heat Exchange in a Series of TanksOrdinary Differential Equations 09-Sep-2014


1 Cp 2. 2. 2. 2.

2 M 1000. 1000. 1000. 1000.

3 t 0 0 200. 200.

4 T0 20. 20. 20. 20.

5 T1 20. 20. 30.95238 30.95238

6 T2 20. 20. 41.38322 41.38322

7 T3 20. 20. 51.31735 51.31735

8 Tsteam 250. 250. 250. 250.

9 UA 10. 10. 10. 10.

10 W 100. 100. 100. 100.

Differential equations 1 d(T1)/d(t) = (W * Cp * (T0 - T1) + UA * (Tsteam - T1)) / (M * Cp)

Temperature in the first tank (deg. C)

2 d(T2)/d(t) = (W * Cp * (T1 - T2) + UA * (Tsteam - T2)) / (M * Cp)

Temperature in the second tank (deg. C)


Temperature in the third tank (deg. C)

Explicit equations 1 W = 100

Feed flow rate (kg/min)

2 Cp = 2.0

Heat capacity (kJ/kg -deg. C)

3 T0 = 20

Feed temperature (deg C)

4 UA = 10.

Area*heat transfer coefficient (kJ/min *deg C)

5 Tsteam = 250

Temperature of steam (deg. C)

6 M = 1000

Total mass in a tank (kg)


POLYMATH ReportOrdinary Differential Equations 09-Sep-2014


1 Cp 2. 2. 2. 2.

2 M 1000. 1000. 1000. 1000.

3 t 0 0 180. 180.

4 T1 30.95238 30.95238 160.9419 160.9419

5 T2 41.38322 41.38322 165.1827 165.1827

6 T3 51.31735 51.31735 169.2217 169.2217

7 Tstream 250. 250. 250. 250.

8 UA 10. 10. 10. 10.

Differential equations 1 d(T1)/d(t) = UA*(Tstream-T1)/(M*Cp)

2 d(T2)/d(t) = UA*(Tstream-T2)/(M*Cp)

3 d(T3)/d(t) = UA*(Tstream-T3)/(M*Cp)

Explicit equations 1 UA = 10

2 Tstream = 250

3 M = 1000

4 Cp = 2.0




1 Cp 2. 2. 2. 2.

2 M 1000. 1000. 1000. 1000.

3 t 0 0 200. 200.

4 T0 20. 20. 20. 20.

5 T1 160.9419 30.95238 160.9419 30.95238

6 T2 165.2217 41.38322 165.2217 41.38322

7 T3 169.2217 51.31737 169.2217 51.31737

8 Tsteam 250. 250. 250. 250.

9 UA 10. 10. 10. 10.

10 W 100. 100. 100. 100.

Differential equations 1 d(T1)/d(t) = (W * Cp * (T0 - T1) + UA * (Tsteam - T1)) / (M * Cp)





Temperature in the third tank (deg. C)

Explicit equations 1 W = 100


2 Cp = 2.0


3 T0 = 20


4 UA = 10.


5 Tsteam = 250


6 M = 1000

Total mass in a tank (kg)



Calculated values of DEQ variablesVariable Initial value Minimal value Maximal value Final value

1 Cp 2. 2. 2. 2.

2 M 1000. 1000. 1000. 1000.

3 t 0 0 89.956 89.956

4 T0 20. 20. 20. 20.

5 T1 160.9419 30.96266 160.9419 30.96266

6 T2 165.1827 41.48545 165.1827 41.48545

7 T3 169.2217 51.83049 169.2217 51.83049

8 Tsteam 250. 250. 250. 250.

9 UA 10. 10. 10. 10.

10 W 100. 100. 100. 100.

Differential equations1 d(T1)/d(t) = (W * Cp * (T0 - T1) + UA * (Tsteam - T1)) / (M * Cp)




3 d(T3)/d(t) = (W * Cp * (T2 - T3) + UA * (Tsteam - T3)) / (M * Cp)Temperature in the third tank (deg. C)

Explicit equations1 W = 100


2 Cp = 2.0


3 T0 = 20


4 UA = 10.


5 Tsteam = 250


6 M = 1000Total mass in a tank (kg)

GeneralTotal number of equations 9Number of differential equations 3Number of explicit equations 6Elapsed time 0.000 secSolution method RKF_45Step size guess. h 0.000001Truncation error tolerance. eps 0.000001

HW11-151-15a code1-15a1-151-15PCP

1-161-16code1-16a1-16b1-16c

1-16d

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HW1