· Specify the input and output. For mass balance, list the names of the components and their flow rates; for energy balance, list the name, flow rate, specific enthalpy and enthalpy of each component. In the cases when you need to use results (e.g. the flow rates) calculated in a previous problem, you can just say “The flow rates are calculated in Problem XXX” and then copy them over.

· Show the results for each term in the overall mass/energy balance equation.

· Show the result for the unknown term in the overall mass/energy balance equation.

· Show the method by which you calculate the wanted flow rate, temperature or whatever that is asked by the question.

· Show the final answers.

An example is given below:

PROBLEM 14.6

NOTE: FEED TO EACH SCRUBBER IS 1/2 OF TOTAL GAS FLOW RATE

Part (a)

Do mass balance on the scrubber

For each component: Input + Produced in Reaction = Output

The only constituents of the feed gas whose flow rate changes as they flow through the scrubber are:

• H2O

• SO2 : 90% of the inlet SO2 is absorbed and reacted with CaCO3

• CO2 (because of the formation of CO2 upon reaction of SO2 with CaCO3)

• fly ash: all of the entering fly ash is removed from the gas stream

Since the gas stream leaving the absorber is saturated with water,

pH2O in outlet stream = vapor pressure of water at 53 C = mole fraction water ´ Total Pressure

p* H2O (@53 C) =

0.14305

bar

(from Table B.5)

n_H2O/n_total =

p*H2O/P =

0.14305/1.013=

0.1412

GAS ENTERING SCRUBBER

(Flow rates are calculated in Problem 14.5)

Component

Flow rate

CO2

3.130

kmol/min

137.77

kg/min

H2O

1.689

30.43

15.983

447.74

SO2

0.055

3.50

0.553

17.70

Fly ash

0.0029

For SO2:

Input =

0.055

kmol/min

Produced in Reaction=

-0.055*0.9 =

-0.0495

Output=

0.0055

For CO2:

Input =

3.130

kmol/min

Produced in Reaction=

0.055*0.9 =

0.0495

Output=

3.18

GAS LEAVING SCRUBBER

Component

Flow rate

CO2

3.180

kmol/min

139.93

kg/min

H2O

15.983

447.74

SO2

0.0055

0.35

0.553

17.70

Fly ash

0.00

For H2O:

X/(3.180+X+15.983+0.0055+0.553) = p*H2O/P = 0.1412

Solve the equation:

3.243

kmol/min

58.37

kg/min

About the flow chart in 14.1, how much information about the equipment’s inlet and outlet should I put in there?

You need to show in the flow chart what is fed into and what is released from the equipment, but you don’t have to specify its detailed compositions. For example, you can draw the furnace like this:

Also, you may want to use different colors and different line thickness to highlight the main equipment and the main flow stream.

How do I do 14.2 without knowing how much coal is fed into the furnace?

Right above 14.1, there is a line that says, “Problems 14.2 through 14.10 should be solved using a basis of 100 kg dry coal/min fed to the furnace.” So once again, read the process description and the problems very carefully.

When talking about the combustion of the coal in 14.3, are we expected to factor in the Oxygen within the coal to the combustion equation?

Yes, you should.

What is “the degree of superheat of the wet air” in 14.4 (b)?

The degree of superheat = The Wet Air’s Temperature – The Wet Air’s dew point = T- T_dew

For 14.4 (c), what state of the air should we use to calculate its feed rate?

Use the state before the air enters the heater exchanger, i.e. its temperature is 25 ^oC and pressure is 1 atm.

For 14.6 (a), don’t forget that each scrubber processes only half of the flue gas from the furnace.

Do we need to consider the amount of CaCO₃ and CaSO₃ dissolved in the water when doing the mass balance for Problem 14.6(C)? If yes, what is the solubility for each of them?

Yes, you should consider the dissolved CaCO₃ and CaSO₃for the mass balance. For the solubility, you can use

CaCO₃: 0.002 kg /100 kg water

CaSO₃: 0.003 kg /100 kg water

Some students found this problem particularly confusing and hard to solve. Here is a little help (Of course, you don’t have to follow the steps below. It will be great if you can find other ways to solve this problem.)

· Establish the following system (the figure) to do mass balance on.

· Do mass balance on H₂O and find out the flow rate of water leaving the scrubber.

· Do mass balance on CaCO₃ and CaSO₃, respectively.

· Find a way to calculate the combined mass flow rate of CaCO₃ and CaSO₃×1/2H₂O in the slurry leaving the scrubber. Note that you don’t have to know the flow rate for each of them. Instead, you only need to know their combined flow rate.

· Calculate the solid-liquid ratio in the slurry leaving the scrubber.

For 14.6(d), before you embark on a mass balance on the blending tank for CaCO₃, go back to the process description and look for this sentence: “The fresh ground limestone is fed to the blending tank at a rate that is 5.2% in excess of that required to react with the SO₂from the flue gas.”

For 14.7, the basic energy balance should be (Enthalpy in) – (Enthalpy out) + (Heat generation) – (Heat to steam) = 0.

· Enthalpy in and Enthalpy out need to be calculated for CO₂, H₂O in the air, H₂O in the coal, N₂, O₂, SO₂ and ash (Note that both the states and the flow rates of these components have changed after the combustion.). You need to define a reference state in order to calculate enthalpy. To make things easier, define 25C, 1atm and liquid as the reference state for water and 25C, 1atm for everything else. Then, use Table B.2 to calculate Cp for each component and with that, calculate the enthalpy.

· The heat generation can be calculated using the HHV value of dry coal.

· To calculate the rate of steam generation, you need to find out the specific enthalpy of the saturated condensate entering the furnace and the super-heated steam coming out of the furnace. You can find these parameters in the steam table in the book (use linear interpolation if the exact temperature and pressure can not be found) or use the code given by Dr. Davis to calculate them.

How exactly should I do the linear interpolation to find the enthalpy of the 24.1Mpa, 540C steam for 14.7? Also, where can I find the enthalpy of the 38C condensate?

In Table B.7 (pp. 651), you can find the following specific enthalpies (kJ/kg) of the superheated steam: (x1, x2 and x are unknown enthalpies.)

------------------------------------

500C 540C 550

221.2bar | 3210 x1 3370

241 bar | x

250 bar | 3166 x2 3337

------------------------------------

The linear interpolation:

(3370-x1)/(x1-3210)=(550-540)/(540-500) => x1=3338 kJ/kg

(3337-x2)/(x2-3166)=(550-540)/(540-500) => x2=3302.8 kJ/kg

(x1-x)/(x-x2)=(221.2-241)/(241-250) => x=3313.8 kJ/kg

For 38C liquid water, its enthalpy (relative to 0C liquid water) can be found in Table B.5 (pp. 642). It is 159.1 kJ/kg.

Also, it can be calculated using the specific enthalpy of liquid water in Table B.2 (pp. 637): 75.4/1000*(38-0)=2.8652 kJ/mol = 159.18 kJ/kg

So the enthalpy change from the condensate to the superheated steam = 3313.8-159.18=3154.62 kJ/kg

How do I solve 14.9? The unknown temperature of flue gas as it leaves the heat exchanger is needed for the energy balance!

The only way to do this is the trial and error method. You assume a temperature for the flue gas and see if the enthalpy in and enthalpy out are balanced. If not, you'll have to change the temperature and try again. The process continues until you find the right answer. I would suggest you use Excel or even write a small Matlab program to do this. The same method should be used to solve 14.14 (b). Of course, in that case the unknown is the rate at which methane is burned.

For 14.11, do we have to go through all the calculations for the new feeding rate of coal when the efficiency of the power plant is 39%?

I think the best way to do this problem is to first find out the factor by which the feeding rate of coal needs to be increased and then to multiply all the results you’ve got for 100 kg dry coal/min feeding rate by that factor.

What’s the feeding rate of dry coal for 14.14? I can’t find it in the problem description.

For 14.14, you should use the required feeding rate calculated in 14.11. The 100kg dry coal/min is just for 14.2 through 14.10.

How do I solve 14.14(b)? It seems to be a pretty complicated system …

Well, as always, first draw a figure (like the one below) for the system you are studying. Then do the mass and energy balances.

For 14.14(b), where can I find the heat of combustion for methane?

In Table B.1 of the textbook. Note that what you can find there is combustion heat with H₂O(liquid) as a product while what you need is combustion heat with H₂O(gas) as a product. You should find out how to convert one to the other.

When solving 14.14(c), you need to assume that all the heat produced by burning the extra amount of coal is transported to the scrubbed flue gas by the steam. So the energy balance here should be Enthalpy in in flue gas + Enthalpy transported by steam = Enthalpy out in flue gas. Note that when more coal is burned in the furnace, the amount of flue gas will be increased.

In 14.17, how can I calculate the size (kW and horsepower) of the pump without knowing the height by which the condensate needs to be raised?

That’s one of the simplifying assumptions you need to make when using Bernoulli equation (below) to solve this problem.

DU + gDz + D(u²)/2 + D(pv) = Q_e + W_e

Find out what are the physical meanings of these characters and how to simplify this equation by making reasonable assumptions. Note that the pressure and specific volume of the condensate and the superheated steam can be found in the Appendix of the book. For the specific volume of the steam, interpolation method similar to the one in Q&A #10 should be used. Also note that the flow rate of the condensate/steam should be the scaled-up number from 14.11, not from 14.7.

In 14.18, what’s difference between the 39% efficiency and the efficiency asked by this problem?

The 39% efficiency is an overall efficiency of the power plant. It means (in 14.11) that “for each unit of heat released with the combustion of coal, 0.39 unit is converted to electrical energy”. In the following figure, it can be calculated as 500MWe/(Combustion Heat). The efficiency 14.18 asks for can be expressed as for each unit of heat transferred to the power generating system by the steam, what percentage of it is converted to electrical energy. In the following figure, it can be calculated as 500MWe/(Heat taken out by steam).

Can you give us some hint in answering 14.19?

Actually 14.19 itself gives some information for its answer. It asks you to “include in your discussion on the impact of the regulation on the use of clean coal”. Obviously, the power plants can’t use very dirty coals because of the 520 ng SO₂/J HHV criterion. Dirty coal has lower HHV and higher S percentage. As to the 90% SO₂ removal criterion, I think the government just wants the power plants to remove most of the SO2 in their stack gas even though they are using clean coals and are producing very small amount of SO₂. In a word, these two criteria are there to prevent power plants from releasing too much SO₂ into the atmosphere.