chemical process and principles ch. 14 answer key

7/22/2019 Chemical Process and Principles Ch. 14 Answer Key

1/11

14.2 14.3Basis: 100 kg dry coal/min

Coal Comp Dry wt% MW kmoles/min O2/element Moles O2/element

C 75.2 12.01115 6.261 1 6.261

H 5 1.00797 4.960 0.25 1.240

N 1.6 14.0067 0.114 0 0.000

S 3.5 32.064 0.109 1 0.109

O 7.5 15.9994 0.469 -0.5 -0.234

Ash 7.2 Total O2= 7.376 kmol/min

Moist 4.8 kg/100 kg dry co 0.267

HHV 30,780 kJ/kg dry coal

Dry coal Cp 1.046 kJ/kg-C

Ash Cp 0.921 kJ/kg-C

14.4Excess O2= 15%

n_O2= 8.482 kmol/min

n_N2= 31.909 kmol/min(b) At 25 C, 50% RH

P= 1 atm

760 mm Hg

1.01325 bar

P*_H2O= 23.756 mm Hg

50% of P*= 11.878 mm Hg MW y_i*MW_i

y_H2O= 0.0156 mole frac H2O 18 0.28 Check 0.015629

n_H2O= 0.641 kmol/min H2O Note: y_H2O=n_H2O/(n_H2O+n_O2+n+N2)

Tot moles= 41.03 kmol/min

y_O2= 0.207 32 6.61

y_N2= 0.778 28 21.77

Avg MW= 28.67Dew Point (from Table B.3) 13.85 C

Degrees Superheat= 11.15 C

( c) Air volumetric flow rates

from above,

Tot moles air= 41.03 kmol/min

919.7 standard m^3/min

1004.0 m^3/min


2/11

14.5from coal kmol/min Species mol/elem kmol/min

C 6.261 CO2 1 6.261

H 4.960 H2O 0.5 2.480

N 0.114 N2 0.5 0.057

S 0.109 SO2 1 0.109

Moist in coal H2O 0.267

Most in Air 0.641

O2 left 1.106

N2 31.909

Entering scrubbers kmol/min y_i MW_i kg/min per scrubber (kmol/min)

CO2 6.261 0.1462 44.00995 275.54 3.130

H2O 3.388 0.0791 18.01534 61.04 1.694

N2 31.966 0.7463 28.0134 895.48 15.983

SO2 0.109 0.0025 64.0628 6.99 0.055

O2 1.106 0.0258 31.9988 35.40 0.553

Total 42.831 1274.45

Each 21.41526085 637.22

Total Ash 7.2 kg/min

Fraction in Bottoms= 20% or 1.44 kg/min

Fraction in Flue Gas= 80% or 5.76 kg/min

% removed in precipitator= 99.90%

Rate of ash removal in precipitator= 5.75 kg/min

Fly ash still in flue gas after precipitator= 0.00576 kg/min

14.6Assuming that 90% of sulfur is removed in each scrubber

T= 53 C

P*= 107.2 mm Hg

y_H2O= 0.1411 mole frac H2O

% SO2 removed= 0.9000 MW kg/min

SO2 removed 0.0491 kmol/min in each reactor 64.0628 3.147 removed from gas

CO2 formed 0.0491 kmol/min in each reactor 44.00995 2.162 becomes part of gas

CaCO3 consumed 0.0491 kmol/min in each reactor 100.08935 4.916 removed from solid

CaSO3 hydrate formed 0.0491 kmol/min in each reactor 129.14987 6.344 becomes part of solid

H2O consumed 0.0246 kmol/min in each reactor 18.01534 0.442 becomes part of solid

Leaving

Each scrubber kmol/min y_i MW_i kg/min

CO2 3.180 0.1385 44.00995 139.93

H2O 3.239 0.1411 18.01534 58.34

N2 15.983 0.6961 28.0134 447.74

SO2 0.005 0.0002 64.0628 0.35

O2 0.553 0.0241 31.9988 17.70

Total 22.960 663.99

(b) amount of slurry entering each scrubber

Slurry feed ratio= 15.2 kg slurry liquid/kg inlet gas

Solid/liq ratio in 0.1111 kg solids/kg liquid

Slurry l iquid feed rate= 9685.8 kg slurry l iquid/min

Solids entering 1076.2 kg solids in slurry/min

Total slurry feed rate= 10762.0 kg slurry/min

(c) estimate solid/liquid loading out i n slurry

First do a water balance around each scrubber

CaSO3-hydrate formed = 0.0491 kmol/min

Solubilities Coming In with Liquid

CaCO3 0.002 kg CaCO3/100 kg liq water 0.19370664 kg CaCO3 in liq/min 0.001935 kmol/min

CaSO3 0.003 kg CaSO3/100 kg liq water 0.29055996 kg CaSO3 in liq/min 0.00225 kmol/min

Water in = 9685.33 kg H2O liq/min

For Each Scrubber

Gas In Liq In Solid In Gas Out Liq Out Solid Out

Species kmol/min kmol/min kmol/min kmol/min kmol/min kmol/min

H2O 1.694 537.616 ? 3.239 536.047 ? Liq H2O from H2O balance, with Del CaSO3 fro

CaCO3 0.00194 ? 0.00193 ?

CaSO3 0.00225 ? 0.00224 ?

kg/min kg/min kg/min kg/min kg/min kg/min

Ash 0.00288 0.00288

Gas 637.22 663.99

Slurry 9685.8 1076.2 9657.549 ?

Total slurry out= 10735.3 kg/min from total mass balance


3/11


4/11

Total solids out= 1077.7 kg/min

Solid/liq ratio= 0.1116 kg solids/kg liquid

Wt% solids= 10.04%

Total IN= 11399.2 kg/min

Total OUT= 11399.2 kg/min

Diff= 0.0 kg/min

(d) Feed of fresh ground l imestone to blending tank for each scrubber

n_SO2 absorbed 0.0491 kmol/min in each scrubber

Excess factor 1.052 kmol CaCO3/kmol SO2Purity 92.10% kg CaCO3/kg limestone

MW CaCO3 100.08935 kg CaCO3/mole CaCO3

m_limestone to tank 5.62 kg limestone/min 5.166479

(e) Flow rates in the wet solids for each scrubber Dry amount

Inerts (in = out) 0.44 kg/min 0.44

CaCO3 0.0026 kmol/min 0.2557 kg/min 0.2555

CaSO3 0.0491 kmol CaSO3 6.344 kg CaSO3/min 6.344

Fly ash 0.00288 kg fly ash/min 0.00288

Total w/o water 7.0461 kg/min 7.0457

Wt% liquid 50.20%

Water out (guessed)= 7.102 kg/min

CaCO3_aq 0.000142 kg/min

CaSO3_aq 0.0002131 kg/min

Total liquid flow rate out of filter= 7.102 kg/min

Total mass flow out of filter= 14.15 kg/min (i.e., wet solids)

Iterate on guessed amount of water until the correct wt% liquid is achieved ratio-.502= -6.40E-10

Mass Fraction of wet solids (i.e., in filter cake on a wet basis)

Inerts (in = out) 0.0314

CaCO3 0.0181

CaSO3 0.4484

Fly ash 0.0002

Liquid 0.5020 This is supposed to be 50.2%!! (We were right!)

% of CaCO3 in liq 0.056%

% of CaSO3 in liq 0.0034%

(f)

Solid/Liquid ratio out of scrubber= 0.1116 (from 14.6c) 0.1115

m_solids_out= 7.05 kg/min (from 14.11e) 7.06

m_liquid_out= 7.102 kg/min (from 14.11e) 7.103

m_solids_out/(m_liquid_out+m_recycle_liq_out)=solid/liquid ratio

Assuming all solids are filtered, m_recycled filtrate= 56.04 kg rec filtrate/min So % going to filter= 0.65%

Recycled slurry liquid = liquid out of scrubber - liquid out of filter 9650.4 kg liq/min1070.7 kg solid/min

Volumetric flow rate calculation

Density= 0.988 kg/L

Volumetric liquid flow rate= 9767.7 L liq/min

Makeup water + water in flue gas into scrubber = water out of filter + water out with scrubbed gas + water in hydrate

Water out of filter = 7.102 kg/min for each filter

Water out with scrubbed gas = 58.34 kg/min

Water into scrubber = 30.52 kg/min

Water in hydrate= 0.44 kg/min

Makeup water = 35.367 kg/min or 35.367 L/min

(for each scrubber)


5/11

14.7I want to calculate the heat of formation of coal.

DelH_comb= 30,780 kJ/kg dry coal (HHV) (really should be negative because it is exothermic)

Basis: 100 kg dry coal DelH_form

Coal Comp Dry wt% MW kmoles/min Species mol/elem kmol/min kJ/mol ni*delH_fi

C 75.2 12.01115 6.261 CO2 1 6.261 -393.5 -2463644

H 5 1.00797 4.960 H2O 0.5 2.480 -285.84 -708950

N 1.6 14.0067 0.114 N2 0.5 0.057 0 0

S 3.5 32.064 0.109 SO2 1 0.109 -296.6 -32375.9

O 7.5 15.9994 0.469 O2 0.5 0.234 0 0Ash 7.2 Sum= -3204970 kJ/100 kg dry coal

-32049.7 kJ/kg dry coal

Moist 4.8 kg/kg dry coal 0.267

HHV -30,780 kJ/kg dry coal

Dry coal Cp 1.046 kJ/kg-C DelH_f of coal = sum(ni*delH_fi),prod-DelH_c = -1,270 kJ/kg dry coal

Ash Cp 0.921 kJ/kg-C (this should be close to zero)

Heat Capacity Table(collected from different probl ems and examples)

Cp coefficients

a b c d T Form DelH_f DelH_vap T_boil ( C)

CO2 3.61E-02 4.23E-05 -2.89E-08 7.46E-12 C 1 -393.5

H2O(g) 3.35E-02 6.88E-06 7.60E-09 -3.59E-12 C 1 -241.83

N2 2.90E-02 2.20E-06 5.72E-09 -2.87E-12 C 1 0

O2 2.91E-02 1.16E-05 -6.08E-09 1.31E-12 C 1 0

H2O(l) 7.54E-02 C 1 -285.84

SO2 3.89E-02 3.90E-05 -3.10E-08 8.61E-12 C 1 -296.6

Dry Coal (per kg) 1.046 -1,270Ash (per kg) 0.921 0

Now we can construct an input and output table like we have been doing in the homework.

In to furnace

T( C) n or m (mol or kg) H_hat H=n*H_hat kJ/100 kg dry coal

Coal (dry) 25 100.0 kg -1,270 kJ/kg -1.27E+05

O2 315 8482.1 mol 8.95E+00 kJ/mol 7.59E+04

N2 315 31908.8 mol 8.57E+00 kJ/mol 2.73E+05

H2O(g) 315 641.3 mol -2.32E+02 kJ/mol -1.49E+05

H2O(l) 25 266.7 mol -2.86E+02 kJ/mol -7.62E+04

Ash 25 7.2 kg 0.00E+00 kJ/kg 0.00E+00

Sum -2.39E+03 kJ/100 kg dry coal

Out of furnace

T( C) n or m (mol or kg) H_hat H kJ/100 kg dry coal

Coal (dry) 25 0.0 kg 0 kJ/kg 0.00E+00

O2 330 1106.4 mol 9.43E+00 kJ/mol 1.04E+04

N2 330 31966.0 mol 9.02E+00 kJ/mol 2.88E+05

H2O(g) 330 3388.2 mol -2.31E+02 kJ/mol -7.83E+05H2O(l) 330 0.0 mol 0.00E+00 kJ/mol 0.00E+00

CO2 330 6260.85 mol -3.81E+02 kJ/mol -2.38E+06

SO2 330 109.16 mol -2.83E+02 kJ/mol -3.09E+04

Ash 330 5.76 kg 2.81E+02 kJ/kg 1.62E+03

Ash 900 1.44 kg 8.06E+02 kJ/kg 1.16E+03

Sum -2.89E+06 kJ/100 kg dry coal

Qdot=Hout-Hin = -2.892E+06 kJ/100 kg dry coal

-2.892E+04 kJ/kg dry coal

If we have 100 kg dry coal/min, then

Qdot = -2.892E+06 kJ/min

-4.821E+04 kW

-4.821E+01 MW

For Steam Characteristics, H_hat

Water In Sat'd Liq 38 C 0.0662 bar 159.1 kJ/kg

Del H = VdP = 24093.38 J/kg 24.09 kJ/kg

Supercrit 38 C 24.1 MPa 183.19 kJ/kgSteam Out Supercrit 540 C 24.1 MPa

540 C 241 Bar 3313.8 kJ/kg

550 C 250 Bar 3337 kJ/kg

Note: This is just interpolation 500 C 250 Bar 3166 kJ/kg

540 C 250 Bar 3302.8 kJ/kg

550 C 221.2 Bar 3370 kJ/kg

500 C 221.2 Bar 3210 kJ/kg

540 C 221.2 Bar 3338 kJ/kg

Delta H for steam = 3130.61 kJ/kg

Steam Rate = 923.9 kg/min


6/11

14.8I did this problem by merely changing the % excess air in problem 14.3 and seeing what happened in problem 14.7

% Excess Air 5% 15% 25%

Steam (kg/min) 924.4 923.9 923.4

This did not change much at all!!!

The reason for the 15% excess air has more to do with the temperature (for NOx control) and enough O2 to burn out the carbon.

14.9 (Air Heater)

Q (kJ/min)

T_air_out 315 C

T_air_in 25 C

Cp coefficients

a b c d T Form DelH_f DelH_vap T_boil ( C)

O2 2.91E-02 1.16E-05 -6.08E-09 1.31E-12 C 1 0

N2 2.90E-02 2.20E-06 5.72E-09 -2.87E-12 C 1 0

H2O(g) 3.35E-02 6.88E-06 7.60E-09 -3.59E-12 C 1 -241.83

CO2 3.61E-02 4.23E-05 -2.89E-08 7.46E-12 C 1 -393.5

SO2 3.89E-02 3.90E-05 -3.10E-08 8.61E-12 C 1 -296.6

Ash (per kg) 0.921 0

Air kmol/min DelH_hat ni*H_hat

O2 8.482 8.95E+00 7.59E+04

N2 31.909 8.57E+00 2.73E+05 Using Answer key data:

H2O 0.641 1.01E+01 6.49E+03 10112.88 kJ/kmol

Sum= 3.56E+05 kJ/min 5.93E+03 kW

T_flue_in 330 C

T_flue_out 71.85 C (guessed at first)

Flue Gas mol/min DelH_hat ni*H_hat

O2 1106.4 8.04E+00 8.90E+03

N2 31966.0 7.66E+00 2.45E+05 Check Dew Point

H2O 3388.2 9.07E+00 3.07E+04 y_H2O= 0.079

CO2 6260.85 1.12E+01 7.01E+04 P_H2O= 0.079 atm

SO2 109.16 1.17E+01 1.28E+03 60.121 mm Hg

Ash (kg) 0.00576 2.38E+02 1.37E+00 T_dp= 41.6 C (from Table B.3)

Sum= 3.56E+05 kJ/min

Diff= 2.15E-07 kJ/min (this is the difference in Q's, which should be zero after the solver)

When Excess Air = 5%, T_flue_out= 74.50 C




7/11

Prob 14.10n_SO2_out= 0.00546 kmol/min from each scrubber

0.01092 kmol/min total

0.699 kg/min

6.99288E+11 ng/min

Coal feed= 100 kg dry coal/min

HHV= 30780 kJ/kg dry coal

Q fed to boiler= 3.08E+06 kJ/min

3.08E+09 J/min

S emission= 227.19 ng/J fed to boiler

EPA standard is met!

14.11Power= 500 Mwe

Efficiency= 39%

Heat Required= 1282.05 MW

for the 100 kg/min of dry coal case,

Q= -2.892E+06 kJ/min

-48207.4 kW

-48.2 MW

(a) Coal feed rate

Coal feed rate= 2659.5 kg/min 26.595 Scaling factor

1.596E+05 kg dry coal/hr

1.672E+05 kg moist coal/hr

(b) Air feed rate

Air Feed Rate= 41.03 kmol/min for 100 kg dry coal/min (from Prob 14.4)

1091.2 kmol/min for 500 Mwe

6.547E+04 kmol/hr

2.446E+04 standard m^3/min2.670E+04 m^3/min

(c) Flue gas flow rates (before precipitator

for 100 kg coal/minfor 500 MWe

(kmol/min) kmol/min kg/min kmol/hr kg/hr

O2 1.106 29.42 941.5 1.765E+03 5.649E+04

N2 31.966 850.12 23814.7 5.101E+04 1.429E+06

H2O 3.388 90.11 1623.3 5.406E+03 9.740E+04

CO2 6.261 166.50 7327.8 9.990E+03 4.397E+05

SO2 0.109 2.90 186.0 1.742E+02 1.116E+04

Ash (kg) 5.76 153.2 9.191E+03

Sum= 6.834E+04 2.043E+06

(d) Steam generation

Steam for 100 kg/min of d ry coal= 923.9 kg steam/min

Steam for 500 Mwe case= 2.457E+04 kg steam/min

1.474E+06 kg steam/hr

14.12For each scrubber, For gases only

100 kg/min case 500 MWe case 100 kg/min case 500 MWe casekg/min kg/min kg/hr kmol/min kmol/min kmol/hr stand m^3/min st m^3/hr T( C) m^3/min m^3/hr

Feed rate of slurry to each scrubber 1.076E+04 2.862E+05 1.717E+07

Flue gas feed to each scrubber (w/ash) 6.372E+02 1.695E+04 1.017E+06 21.42 5.695E+02 3.417E+04 1.277E+04 7.660E+05 72.9 1.62E+04 9.70E+05

Slurry out from each scrubber 10735.6 2.855E+05 1.713E+07

Flue gas leaving each scrubber (no ash 663.99 1.766E+04 1.060E+06 22.96 6.106E+02 3.664E+04 1.369E+04 8.212E+05 53 1.63E+04 9.81E+05

Wet solids from filter 14.15 3.762E+02 2.257E+04

Total to blender (solids + liquid) 10721.5 2.851E+05 1.711E+07

Filtrate (liquid) 56.02 1.490E+03 8.939E+04

Limestone feed rate to filter 5.62 1.493E+02 8.961E+03

Fresh water to blending tank 35.4 9.406E+02 5.643E+04

5.643E+04 L/hr of makeup water


8/11

14.13Heating the flue gas decreases the density and permits the gas to rise up the stack better.

Also, slight cooling of the saturated flue gas would cause condensation as it rises up the stack.

14.14 (Stack Heater)(a) Bypassing the scrubber with a portion of the gas will increase the sulfu r dioxide emissions

and the particulate emissions.

(b) Burning methane with 10% excess air

T_out= 80 C

T_in= 53 C

n_CH4= 78.87 kmol/hr (guessed at first and used in solver) 1.314 kmol/min

Excess O2 10%

n_O2 fed 1.74E+02 kmol/hr

n_N2 fed 6.53E+02 kmol/hr

n_H2O fed 1.31E+01 kmol/hr

Must do enthalpy balance (adiabatic, i.e., no heat lost from heater)

Cp coefficients

a b c d T Form DelH_f

O2 2.91E-02 1.16E-05 -6.08E-09 1.31E-12 C 1 0

N2 2.90E-02 2.20E-06 5.72E-09 -2.87E-12 C 1 0H2O(g) 3.35E-02 6.88E-06 7.60E-09 -3.59E-12 C 1 -241.83

CO2 3.61E-02 4.23E-05 -2.89E-08 7.46E-12 C 1 -393.5

SO2 3.89E-02 3.90E-05 -3.10E-08 8.61E-12 C 1 -296.6

CH4 3.43E-02 5.47E-05 3.66E-09 -1.10E-11 C 1 -74.85

Into Stack Heater

n T H_hat n*H_hat

(kmol/hr) ( C) kJ/mol kJ/hr

CH4 7.89E+01 25 -74.85 -5.90E+06

O2 1.74E+02 25 0 0.00E+00

N2 6.53E+02 25 0 0.00E+00

H2O(g) 1.31E+01 25 -241.83 -3.17E+06

CO2 1.01E+04 53 -3.92E+02 -3.98E+09

H2O 1.03E+04 53 -2.41E+02 -2.49E+09N2 5.10E+04 53 8.15E-01 4.16E+07

SO2 1.74E+01 53 -2.95E+02 -5.15E+06

O2 1.77E+03 53 8.27E-01 1.46E+06

Sum= -6.44E+09 kj/hr

Out of Stack Heater

n T H_hat n*H_hat

(kmol/hr) ( C) kJ/mol kJ/hr

CO2 1.023E+04 80 -3.91E+02 -4.00E+09

H2O 1.051E+04 80 -2.40E+02 -2.52E+09

N2 5.166E+04 80 1.60E+00 8.28E+07

SO2 1.742E+01 80 -2.94E+02 -5.13E+06

O2 1.781E+03 80 1.63E+00 2.91E+06

Sum 7.42E+04 Sum= -6.44E+09 kJ/hr

Solver Diff= 0.00E+00

( c) How much more coal?

Assumption: Approximate this by Figuring out how much heat was used to raise T of flue gas,

and then simply divide by the heating value of the coal.

CH4 flow rate= 78.87 kmol/hr

DelH_comb= -890.36 kJ/mol (high heating value)

Heat rate= -7.022E+07 kJ/hr

Change in coal feed rate = 2281.3 kg dry coal/hr

Alternatively, we could do the formal calculation with enthalpy in and out.


9/11

14.15Sum of molar flow rates coming into stack = 7.42E+04 kmol/hr (from 14.14)

Temperature= 80 C 353 K

Pressure= 1 atm

Volumetric flow rate= 2.15E+09 L/hr

597.0 m^3/s

Velocity (m/s) Diameter (m)5 12.33

6 11.26

7 10.42

8 9.75

9 9.19

10 8.72

11 8.31

12 7.96

13 7.65

14 7.37

15 7.12

Stack Diameter vs. Veloci ty

6

7

8

9

10

11

12

13

5 6 7 8 9 10 11 12 13 14 15

Stack Veloci ty (m/s)

StackD

iameter(m)


10/11

14.16 (Cooling water flow rate)

Steam flow rate (from 14.11d)= 1.474E+06 kg steam/hr

Pressure in= 6.55 kPa 0.0655 bar

Inlet Steam properties: (Table B.6)

Sat'd Steam Sat'd Liquid Temperature

Pressure (bar) Enthalpy (kJ/kg Enthalpy (kJ/kg) ( C)

0.06 2567.5 151.5 36.2 2570.305

0.07 2572.6 163.4 39

0.0655 2570.3 158.0 37.5

% Liquid= 27.5%

Net Enthalpy= 1906.9 kJ/kg

Outlet water properties:

Temperature= 38 C

Sat'd P= 0.0662 bar

H (sat'd liq)= 159.1 kJ/kg

Q_steam = -2.577E+09 kJ/hr

For cooling water, assume saturated properties

T_in= 25 C

H_in= 104.8 kJ/kg (Table B.5)

T_out= 28 C

H_out= 117.3 kJ/kg

Mass flow rate of water = 2.061E+08 kg/hr

2.061E+08 L/hr

14.17 (pump requirement)P_in= 0.0662 bar 6.62 kPa

0.00662 MPa

P_out= 24.1 MPa

V_hat= 0.001007 m^3/kg

dens= 993.0 kg/m^3 (Table B.5)

Efficiency= 80%

Flow Rate= 1.474E+06 kg Water/hr (from 14.11d)

Work= 1.242E+07 W

1.242E+04 kW

1.665E+04 hp


11/11

14.18 Efficiency of steam

Delta_H for steam across turbine

H_in= 3313.8 kJ/kg (from 14.7)

H_out= 1906.2 kJ/kg (from 14.16)

Delta_H_steam= -1407.6 kJ/kg

Mdot_steam= 1.474E+06 kg/hr (from 14.11d)

Ws= 2.075E+09 kJ/hr

5.764E+02 MW

Electricity= 500 Mwe

Efficiency= 86.7%

14.19 Why the regulations?

You can't have it on the amount of sulfur emitted because you could have a high percentage of emissions fro

You could have it on the % of sulfur, but this reqards the high sulfur coal. Therefore the mix of amount of sul

energy input makes a lot of sense.

Problem 14.20

Emission limit 520 ng/J

Coal feed rate= 1.60E+05 kg dry coal/hr (from 14.11)

Energy content= 30,780 kJ/kg dry coal

SO2 emission= 2.55E+03 kg SO2/hr

6.13E+04 kg SO2/day

chemical process and principles ch. 14 answer key

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