Problem #1: Create a problem that highlights the species mass conservation equation. You can only use concepts of mass or mole fractions and the species and total continuity equations to construct your problem. (suggested completion date: Sep 6).
A container is initially 100 m
2
contains 50% CO
2
and 50% O
2
by mass. The density of the room is the same of that of air at ambient conditions. A scrubber removes all the CO
2
from the container and returns pure oxygen. The scrubber can maintain a flow of 2m/s of fluid.
A.) How big must the duct be to process 1kg/s of fluid?
B.) If the scrubber simply processes the fluid into two containers (one for the oxygen, and one for the carbon dioxide) determine: a.
How long it will take to process the container? b.
How much mass of carbon dioxide is in the container? c.
If the container containing oxygen is at the same density and pressure as the original container, what is the volume of the container?
C.) Now the scrubber is attached in a loop to the container. The carbon dioxide is simply removed and pure oxygen is returned into the room. a.
What must be the exit duct size to have a 2m/s return velocity to the room? b.
How long will it take to process the majority of the CO
2
? Assume the scrubber removes 100% of the carbon dioxide.
D.) Repeat C.) for a scrubber with only 75% efficiency.

Problem #2: Pick a closed system that runs at steady state. The system must involve at least two mechanisms of entropy generation. Now do a parametric study in which you change a parameter, vary it over a wide range, and see its effect on total entropy generation rate, and entropy generation due to individual mechanisms . (suggested completion date: Sep 25).
Suppose we have the system below:
𝑄 𝑜𝑢𝑡
T h
𝑊 𝑖𝑛
𝑄 𝑖𝑛
T c
Figure 1. Diagram of system.
The parametric study completed for this system will first vary the heat transfer into the system. Next, the heat transfer will be constant, but the boundary temperatures will be varied.
First, the energy and entropy equations must be solved: 𝑑𝐸 𝑑𝑡
= 0 = −𝑚̇(∆ℎ) + 𝑄̇
𝐸𝑥𝑡
− 𝑊̇
𝐸𝑥𝑡
Since this is a closed steady system, this reduces to:
(1)
(2) 𝑄̇
𝐼𝑛
= 𝑊̇
𝐸𝑥𝑡
− 𝑄̇
𝑂𝑢𝑡
This is useful as the outputted heat is what will be varied for the parametric study.
Similarly, the entropy equation: 𝑑𝑆 𝑑𝑡
= 0 = −𝑚̇(∆ℎ) +
𝑄̇
𝑇
𝐵
+ 𝑆̇
𝐺𝑒𝑛
(3)

This reduces to:
𝑆̇
𝐺𝑒𝑛
=
𝑄̇
𝑂𝑢𝑡
𝑇
𝐵
−
𝑄̇
𝐼𝑛
𝑇
𝐵
Now that we have an equation to calculate the entropy, it is possible to do the parametric study.
(4)
The first part of the parametric study will be to vary the heat out. This will be varied from 150kW to
720kW. The hot side is maintained at a constant 300K; the cold side is also at a constant 273K. The system has a net work input of 110kW.
Using excel to complete the calculations, the graph of the data can be seen below:
5
4,5
4
3,5
3
2,5
2
1,5
1
0,5
0
0 100 200 300 400 500
Heat Rejection (kW)
600 700 800
Figure 2. Entropy generation vs Heat Rejection.
As the chart shows, it is an extremely linear result. This is due the fact that the relationships that are changing are linear in nature. Because of this, it would be of interest to see how the entropy changes with changing boundary temperature.
The system is now modeled as the same above, however the work, heat in, and heat out are all considered constant at 110kW, 150kW, and 50kW, respectively. For this parametric variance, first the hot temperature will be varied from 300 to 700K with a constant cold temperature of 273K. Then the hot temperature will be a constant 300K with the cold temperature varied from 273 to 3K.

When the hot temperature is varied:
0,75
0,7
0,65
0,6
0,55
0,5
0,45
0,4
0,35
0,3
250 350 450 550
Th (K)
650 750 850
Figure 3. Variance in the rejection temperature.
When the hot temperature side is varied, you notice that it is a nonlinear relationship between the hot side temperature and the entropy generation.
When the cold temperature is varied while holding the hot temperature constant:
20
18
16
14
12
10
8
6
4
2
0
0 50 100 150
Tc (K)
200 250 300
Figure 4. Variance in the inlet temperature.
Notice the difference in relationship between the temperature of the cold side and the hot side. The decrease in temperature of the cold side clearly has a huge effect on entropy generation as the temperature is decreased. This is due primarily to the inverse dependence on temperature.

The inverse dependence of the Entropy generation causes spikes in entropy generation as the temperature decreases on either side of the system. This effect is much larger than when the heat traversing the system is changing. Clearly, temperature control is of upmost importance in decreasing entropy generation.
Problem #3: Pick an open steady device(compressor, turbine, nozzle, diffuser, mixing chamber, etc.). Select (or create) a problem similar to the ones posted in TEST>Problems>Chapter 4 on that that runs at steady state. You want to modify and restate that problem in order to study entropy generation (similar to what we did in the class for the compressor and mixing chamber). The problem must involve at least one parametric study. Please post your problem on the blog and wait for an okay from me before you work out the problem and perform the parametric study.
A Brayton Cycle operates at steady state. The cycle has a pressure ratio of 18:1 with the inlet to the compressor being at ambient conditions (100kPa and 300K). Assume that the entire boundary of the device is at 300K. The temperature of the air at the inlet to the turbine is 700C. Determine the following: a.) what is the net-work output of the device. b.) What is the entropy generation for this cycle? c.) What is the thermal efficiency of the cycle? d.) Do a parametric study in which the pressure ratio is varied from 1:1 to 30:1. What is the trend of the entropy generation as the pressure ratio increases? What happens with the net-heat, work, and BWR? Why would a higher pressure ratio be seen as beneficial?
Assume the turbine is fully reversible.
3
4 T
2
1 s
Figure 1. The process lines for a completely reversible Brayton Cycle.
In order to solve this problem, we must first solve for the temperature of each of the four states shown above. The problem provides two:
T
1
= 300K
T
3
= 973K
(1)
(2)

From the Perfect Gas model:
T
2
= T
1
∗ (
P
2 )
P
1 k−1 k
.4
= 300 ∗ (18)
1.4
= 685K = T
2
T
4
= T
3
∗ (
P
4 )
P
3 k−1 k
.4
= 973 ∗ (
1
18
)
1.4
= 426K = T
4 a.) To calculate the net-work out:
(3)
(4)
Ẇ net
= Ẇ turbine
− Ẇ compressor
= ṁ ∗ (∆h turbine
− ∆h compressor
) (5)
Ẇ net
= ṁ ∗ c
P
∗ (∆T turbine
− ∆T compressor
) (6)
Ẇ net
= 5 ∗ 1.005 ∗ (973 − 426 − 300 + 685) = 814.6kW [ kg s
∗ kJ kg∗K
∗ K = kW] (7)
The turbine will have a net output of 814.6kW. b.) To determine the entropy generation, we need to use the entropy rate equation: dS dT
= ṁ ∗ (s in
− s out
) +
Q̇
T
B
+ Ṡ gen
Since the turbine is operating at steady state: dS dT
= 0
Which implies (assuming a constant boundary temperature):
(8)
(9)
Ṡ gen
= ṁ ∗ (s out
− s in
) +
Q̇ out
T
B
−
Q̇ in
T
B
= ṁ ∗ (s out
− s in
) +
Q̇ net
T
B dE dT
= 0 = −ṁ(∆h) + Q̇ net
− W net
(10)
In order to solve this we need to first use the energy rate equation to determine the net heat of the system:
(11)
Q̇ net
= Ẇ net
+ ṁ ∗ c
P
∗ (∆T) = 814.8 + 5 ∗ 1.005 ∗ (426 − 300) = 1448kW (12)
Next is to solve for the ∆𝑠 = 𝑠 𝑜𝑢𝑡
− 𝑠 𝑖𝑛
:
∆s = c p
∗ ln (
T out ) − R ∗ ln (
T in
P out
P in
)
Since the entrance and exit pressures are the same:
(13)
∆s = 1.005 ∗ ln (
426
300
) = .35188
kJ kg
(14)

Solving, numerically:
Ṡ gen
= 5 ∗ .35188 +
1448
300
= 6.586
kJ kgK c.) Determine the thermal efficiency using the following:
(15) 𝜂 =
𝑊𝑎𝑛𝑡
𝑁𝑒𝑒𝑑
=
𝑊̇ 𝑡𝑢𝑟
−𝑊̇ 𝑐𝑜𝑚𝑝
𝑄̇ 𝑖𝑛
= 1 −
𝑇
4
−𝑇
1
𝑇
3
−𝑇
2
= 1 −
426−300
973−685
= 56.3% (16) d.) The parametric study is best done using a program such as excel. The above equations are solved for each pressure ratio from 1:1 to 30:1. The results can be seen following the completion of the problem.
Below are the results of the study plotted:
14
12
10
8
20
18
16
6
4
2
0
0 5 10 15 20
Pressure Ratio
25 30 35
Figure 2. Entropy Generation vs. Pressure Ratio. The Pressure Ratio appears to be having a strong ability to make the entropy generation decrease.
Notice the primary trend of the entropy generation decreases as the pressure ratio increases. This would imply why there is a trend to have a higher pressure ratio in Brayton Cycles.

0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
0 5 10 15
Pressure Ratio
20 25 30
Figure 3. Efficiency vs. Pressure Ratio. The Pressure Ratio appears to have a drastic effect on the Efficiency.
4000
3500
3000
2500
2000
1500
1000
Net Work
Net Heat
500
0
0 5 10 15
Pressure Ratio
20 25 30
Figure 4. Net-work and net heat vs. Pressure Ratio. Note that the two are converging together. This implies that the engine is getting more efficient as the pressure ratio increases; however the net power output is decreasing.

0,4
0,3
0,2
0,1
0
0
0,9
0,8
0,7
0,6
0,5
5 10 15 20
Pressure Ratio
25 30 35
Figure 5. Back Work Ratio (BWR) vs. Pressure Ratio. The increase of the BWR is to be expected as the pump has to work more to compress the air.
It is easy to see from the graphs above why a higher pressure ratio is desired for the Brayton Cycle. Not only does the entropy generation decrease as the ratio increased, but, necessarily, the efficiency increases. The Net Heat of the system also decreases. Unfortunately, as the pressure ratio increases, so does the BWR, decreasing the outputted work. Because of both the desirable and undesirable consequences of increasing the Pressure Ratio, optimization should be performed to determine the proper pressure ratio to use. This would have the optimum efficiency and power output while minimizing the entropy generation, heat into the system, and BWR.
Problem #4: Pick a hydrocarbon fuel (post your fuel on the blog under Problem 4 topic so that no one else can pick the same fuel). You are to study how the flame temperature changes as this fuel is burned with air and pure oxygen.
You are going to vary the equivalence ratio over a wide range (rich to lean) and see how the flame temperature changes for (a) PG model (two lines on one plot Tf vs phi), (b) IG model, and (c) equilibrium model. This will generate three plots each with two curves (for air vs. pure oxygen as the oxidizer).
Suppose that an equal ratio of C and CH
4
will combust. We wish to see the effect of the equivalence ratio, phi, on the adiabatic flame temperature for combustion with air as well as pure oxygen. For this problem we will plot the adiabatic flame temperature vs phi for (a) PG model, (b) IG model, and (c) the flame temperature as the ratio of fuels changes.

Methane +
Carbon at
298K
Air at 298K
Products
Figure 1. Combustion process being modeled.
First we must determine the chemical balance for theoretical air. This is done below: 𝑎CH
4
+ 𝑏C + 𝑑(O
2
+ 3.76N
2
) → 𝑒CO
2
+ 𝑓H
2
O + 𝑔N
2
(1) 𝑎 = 𝑏 = 0.5
𝑒 = 1 𝑓 = 1 𝑑 = 1.5
(2)
(3)
(4)
(5) 𝑔 = 5.64
(6)
It is necessary to note that for this equation to be usable, it is on a mole basis of one mole of fuel. This is why a and b are equal to 0.5, so that they will combine to make one kmol of fuel.
The above implies that the theoretical reaction is:
0.5CH
4
+ 0.5C + 1.5(O
2
+ 3.76N
2
) → CO
2
+ H
2
O + 5.64N
2
(7)
The distinction has to be made that there is a difference between a phi less than on and a phi greater than one. For a phi less than one, there is excess oxygen produced:
0.5CH
4
+ 0.5C + 𝜆1.5(O
2
+ 3.76N
2
) → CO
2
+ H
2
O + 𝜆5.64N
2
+ 𝑧O
2
(8) where α is the theoretical air percentage. z can be determined from another chemical balance:
2 ∗ 𝜆 ∗ 1.5 = 3 + 2𝑧 (9) 𝑧 = 1.5𝜆 − 1.5
(10)
For a phi less than one we get the following chemical balance:

0.5CH
4
+ 0.5C + 𝜆1.5(O
2
+ 3.76N
2
) → CO
2
+ H
2
O + 𝜆5.64N
2
+ (1.5𝜆 − 1.5)O
2
(11)
It is worth noting that 𝜆 is the inverse of phi.
For combustion with phi greater than one, we have incomplete combustion. This can be modeled several ways, but for the purpose of this study, it will be modeled as producing carbon monoxide in addition to carbon dioxide. This can be seen below:
0.5CH
4
+ 0.5C + 𝜆1.5(O
2
+ 3.76N
2
) → aCO
2
+ 𝑏H
2
O + 𝜆5.64N
2
+ 𝑐CO (12)
Once again, doing a chemical balance, a, b, and c can be determined: 𝑏 = 1 𝑎 = 3𝜆 − 2 𝑐 = 3 − 3𝜆
(13)
(14)
(15)
Now that we have a case for both a phi less than one, and greater than one, it is possible to determine the adiabatic flame temperature. This is the temperature of the exit when there is no heat transfer to the surroundings. This necessitates that:
∑ 𝛾 𝑝
∗ ℎ 𝑝
= ∑ 𝛾
𝑅
∗ ℎ
𝑅
(16)
For the PG model, this simplifies to:
𝑇 𝑎𝑓
= 𝑇 𝑖
+
−∆ℎ 𝑜 𝑐 𝑐 𝑝
[1+𝐴𝐹]
(17)
This can be said because the c p
value is constant for PG gases. The numerator can be expressed as follows:
∆ℎ 𝑜 𝑐
=
∑ 𝛾 𝑝
̅̅̅̅−∑ 𝛾 𝑟
̅
𝐶+𝐶𝐻4 𝑜
𝑅 (18)
With a phi less than one, this is constant as there is no enthalpy of formation changing in the products or reactants:
∆ℎ 𝑜 𝑐
= −42708.2
kJ kg
With a phi greater than one, this is not the case, and ∆ℎ 𝑜 𝑐
must be calculated at each phi.
(19) a.) Plot the adiabatic flame temperature for the fuel reacting with Air and pure oxygen over a range of phi’s
For air, this can be directly calculated using the above derivation. With oxygen, the only difference is the nitrogen term drops out of the chemical balance. The main effect this has is to drastically increase the

temperature by decreasing the AF ratio. The graph can be seen below with the two temperature ranges super imposed:
1 12000
10000
8000
6000
4000
2000
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
Taf for Air
Taf for Pure O2
Percentage of CO
0
0,4 0,6 0,8 1
Equivalence Ratio
1,2 1,4
0
Figure 1. Temperature vs. Equivalence Ratio for Air b.) Do the same as in a, but us the IG model.
The reason the simplification is so simple for the PG model is that the Cp is assumed to be constant both across temperature ranges and species. For the IG model, this assumption cannot be made. Thus equation (16) must be solved.
Hence, and iterative method muse be employed as the exit temperature is not known, and what we are trying to solve for. Using TEST, we can determine the adiabatic flame temperature. For the IG model, it is plotted below:

1 6000
5000
4000
3000
2000
1000
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
Air Mixture
Pure O2
Percentage of CO
0
0,4 0,6 0,8
Phi
1 1,2 1,4
0
Figure 2. Adiabatic Flame Temperature vs. Equivalence Ratio
We can see that the temperature is less than the PG model, which is to be expected. It also appears to be much less linear than the PG model, also to be expected. Note also, it is much more accurate as the lines are no longer linear, but rather nonlinear in nature. c.) Plot the adiabatic flame temperature as the ratio of methane to carbon changes from zero methane to pure methane. Do this with the PG model with theoretical air.
This problem is a great way to determine the temperature of various mixtures of methane and carbon and what output temperature they could produce. This could be used for a variety of applications. If coupled with a cost of the fuel, it could easily be used to gauge what the optimal ratio of the fuels would be for max temperature output.
Once again, equation (17) is utilized.
However, this time the ∆ℎ 𝑜 𝑐
is not constant. The chemical balance will have to be calculated at each iteration: 𝑎CH
4
+ 𝑏C + 𝑑(O
2
+ 3.76N
2
) → 𝑒CO
2
+ 𝑓H
2
O + 𝑔N
2
(20)

2760
2740
2720
2700
2680
2660
2640
0
Where a varies from 0 to 1, b varies from 1 to 0, and the remaining coefficients can be determined from below: 𝑎 + 𝑏 = 1 = 𝑒
2𝑎 = 𝑓 𝑑 = 1 + 𝑎
(21)
(22) 𝑔 = 𝑑 ∗ 3.76
(23)
(24)
Looping over a and b, we can determine the adiabatic flame temperature. The results are plotted below:
2780
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8
Percentage of Methane in Mixture of Methane and Carbon
0,9 1
Figure 3. The adiabatic flame temperature with various mixture percentages.
From Figure 3 it is clear to see that Methane has a higher adiabatic flame temperature than that of
Carbon. It is interesting to note that the relationship between the two is not linear in nature. The difference in the minimum temperature and max temperature is only 141K, or approximately 4%.