Example from Lecture 6A stoichiometric mixture of air and gaseous methane at 54oC and 2 bar is buried in a 0.1 m3 rigid vessel. The temperature of the products is measured to be 1,529oC. Given that the internal energy of combustion ∆Uo = - 802,310 kJ/kmol at 25oC, calculate the amount of heat rejected to the environment.

Hence, information given:

To = 25oC = 298 K ∆Uo = -802,310 kJ/kmol

T1 = 54oC = 327 K P1 = 2 bar V = 0.1 m3

T2 = 1,529oC = 1,802 K

Air is provided: Use 79% N2 , 21% O2

Solution

We want to find

Q = Uo + (U0 – U1) + (U2 – U0)

To do this we need the mass m of each component and Cv for each at the average temperature (from tables).

For m we need to find the total number of moles, N:

Recall: n = mol/m3 = P1 / R T1

= 2 bar / [(8.314 J/mol-k) * (327 K)] = 2 x 105 / (8.314 * 327) = 73.57 mol/m3

Hence, N = 0.1 * 73.57 = 7.357 mol

R P

This consists of CH4 (methane), O2, N2 and so we now need to know how much O2 we have (for stoichiometric conditions). The reaction is:

?

For 1 mol of CH4 we have 2 mol of O2 and also 0.79/0.21 * 2 of N2 (because air is provided). So

NCH4 * ( ? ) = ? mol

Hence, NCH4 = 0.699 mol

So, NO2 = 2 NCH4 = 1.398 mol NN2 = 5.529 molNH20 = 1.398 mol NCO2 = 0.699 mol

The heat given off by the reaction at 25oC is

Uo = - 802,310 kJ/kmol * 0.699 mol * 10-3 kmol/mol = - 560.8 kJ

Next we need to determine the internal energy change required to bring the reactants to the “tabulated” reaction temperature and then the products back up to the measured exit temperature. From tables we find

Units: kJ/kmol-K CH4 O2 N2

Reactants Cv [at (25+54)/2] 27.8 21.1 20.8

CO2 H2O N2

Products Cv [at (25+1529)/2] 46.6 33.5 24.6

Hence, for the reactants (R) and products (P):

(U0 – U1) = NCH4 * CvCH4 * (T0 – T1) + N02 * Cv02 * (T0 – T1) + NN2 * CvN2 * (T0 – T1)

= [ NCH4CvCH4 + N02Cv02 + NN2CvN2 ] * (T0 – T1) = [0.699 * 27.8 + 1.398 * 21.1 + 5.529 * 20.8]

* (25 – 54) * 10-3 = - 4.75 kJ

(U2 – U0) = NC02 * CvC02 * (T2 – T0) + NH20 * CvH20 * (T2 – T0) + NN2 * CvN2 * (T2 – T0)

= [ NC02CvC02 + NH20CvH20 + NN2CvN2 ] * (T0 – T0) = [0.699 * 46.6 + 1.398 * 33.5 + 5.529 * 24.6]

* (1529 – 25) * 10-3 = 324.3 kJ

R

P

So the overall balance is

Q = Uo + (U0 – U1) + (U2 – U0)

= -560.8 kJ – 4.75 kJ + 324.3 kJ - 241 kJ

R P

Example from Lecture 7

A combustion process with propane fuel is said to be 98% efficient. Determine the composition of the product gases and exit temperature if the combustor loses 3% of the heat of the combustion to the environment. The following technical data are available:

mf = 0.1 kg/s (at 25°C, 1 atm, gaseous)Excess air: 20%Inlet velocity = 30 m/sOutlet velocity = 300 m/sInlet condition of the reactants 25°C, 1 atmOutlet pressure: 1 atmAssume all products are gaseous

Solution strategy outline:

1. Set up the energy balance (enthalpy) equation2. Assume air is 79% N2 and 21% O2

3. Write out and compute ideal chemical reaction equation4. Modify this for the fact that the process is 98% efficient5. Use ideal reaction value for the enthalpy of formation (call it h’0 for the fuel), from earlier in these notes, and h’’0 for CO (from tables), to find the overall value h0

6. Calculate all the masses (for the reactants and products)7. Check that mass in = mass out ! [should be 1.97 kg/s]8. Calculate H0, Q and K.E. for the energy equation9. Using known inlet conditions T1 = T0 = 25oC use the energy equation to iterate for T2. Since most mass flow is nitrogen use the Cp of nitrogen at 25oC as first guess for T2

10. Use this T2 to calculate Cp at (T2+T0)/2 for all products and calculate the m Cp and keep iterating to a solution

.

Start with the energy balance equation:

Q = H0 + Cp (T2 – T0) - Cp (T1 – T0) + K.E.

where Q = 0.03 H0 (3% heat loss) H0 = ?h’0 = -2044 MJ/kmolf (enth. of comb of fuel)

Finding H0 (assuming 79% N2 and 21% O2):

We have excess air of 20%. Ideal reaction requires 5O2 and so 20% excess air means and extra 1O2

Ideal reaction is:

C3H8 + 5O2 3CO2 + 4H20

+ excess air of 1 O2 + 6 * (0.79/0.21) N2 = O2 + 22.57 N2

m mP R

Actual reaction gives 98% CO2 yield so: 3 * 0.98 = 2.94 CO2

with the rest given off as CO. Hence, actual reaction is

?

Computing overall enthalpy of combustion:

For C3H8 + 5O2 3CO2 + 4H20 h’0 = -2044 MJ/kmolf

For CO2 CO + ½ O2 h’’0 = 283 MJ/kmolCO (from tables)

So h0 = h’0 – 0.02 * ( 3 h’’0 )

Since 1 - CO = 0.02 and there are 3 CO2 / molf

h0 = -2044 – 0.06 * 283 = -2027 MJ/kmolf

We could also write this as:

h0 = 0.98 h’0 + 0.02 ( h’0 + 3 * h’’0 )

(again because process is 98% efficient and there are3 CO2 3 CO + 3/2 O2)

h0 = 0.98 * -2044 + 0.02 * ( -2044 + 3 * 283 )= -2027 MJ / kmolf

Calculate all the masses and the heat rejection given that the reaction equation is:

?

Since and Mf = 44 kg/kmol (3 MC + 8 MH)s/kg1.0mf

Check:

H0 = h0 = 2.27x10-3 kmol/s * -2027 MJ/kmolf = - 4.60 MW

So heat loss to surroundings is Q = 0.03 H0 = - 0.138 MW

Computing:

where subscripts i = in and o = out, noting that the fuel (0.1 kg/s) is injected into the combustion process so that the mass flow rate of air is 1.87 kg/s.

K.E. = 0.089 MW – 0.00084 MW = 0.088 MW

We know the inlet conditions are T1 = 25oC and T0 = 25oC

and so Cp (T1 – T0) = 0

s/kg97.1ms/kg97.1m outin

fn

2V

m2

Vm.E.K

2i

air

2o

P

So, we need to solve

or, rearranging in terms of T2

We don’t know the outlet temperature T2 and so we need an iteration algorithm to find it:

(1) Since most of the mass flow is nitrogen we can use the value of Cp of nitrogen (at T = 25oC which changes little with T) to get a first guess for T2.

MW374.4.E.KHQTTCpm 002P

CpmMW374.4

25Cpm

E.KHQTT 0

02

(2) Calculate T2 from 1st Law relationship(3) Use this T2 to calculate Cp at (T2+T0)/2 for all products. Call this T2, (T2)old.

(4) Calculate Cp (5) Find new T2 = (T2)new

(6) If (T2)new - (T2)old < tolerance, the iteration is complete. Otherwise use (T2)new as the new guess and start again from step 2. Tolerance depends on the application but 50oC is a reasonable rule-of-thumb. So first iteration (Tave = 25oC) is:

m

Species Mass flow (kg/s) Cp (kJ/kgK)

N2 1.44 1.042

O2 0.075 0.922

CO2 0.29 0.842

CO 0.0038 1.041

H2O 0.164 1.872

Cp = 2.125 kJ/kgK

Giving T2 = 25 + (4.374 MW / 2.125 kJ/kgK) = 2084oC

Now we need to compute Cp values at (T2+T0)/2 = 1054oC

Cp = 2.520 kJ/kgK

Giving T2 = 25 + (4.374 MW / 2.520 kJ/kgK) = 1757oC etc!

m

Species Mass flow (kg/s) Cp (kJ/kgK)

N2 1.44 1.167

O2 0.075 1.090

CO2 0.29 1.234

CO 0.0038 1.185

H2O 0.164 2.440m

Example from Lecture 7

Butane (C4H10) is burned with dry “theoretical” air (21%O2, 79%N2) at an air-fuel ratio of 20. Calculate the percentage of excess air and the volume percentage of CO2 in the products. Take the molecular weight of air as 29 kg/kmol.

What is the reaction equation for theoretical air ?

?

What is the air-fuel ratio AFth for theoretical air ?

AFth = mair = ? = 15.47 kg air mfuel kg fuel

This represents 100% theoretical air. The actual air-fuel ratio is 20 and so what is the percentage of excess air ?

% excess air = ( AFact – AFth ) x 100% AFth

= (20 – 15.47) x 100 = 29.28% 15.47

So, actual reaction with 129.28% theoretical air is:

?

Hence, volume percentage of CO2 using total moles in products of combustion is:%CO2 = ( 4 / 42.5 ) * 100 = 9.41%