If the thermal efficiency of an engine equals 25% and the heating value of the fuel 36,000 kJ/kg, calculate the Braking Standard Fuel Consumption (BSFC) in kgkW-hr.
Solution:
W = 0.25 * 36,000 = 9,000 kJ/Kg fuel
1 kW - hr = 1 kJ /sec * 3,600 Sec = 3,600 kJ
Therefore the BSFC = 0.4 kg/kW-hr
21 - Question 15 - The triple point of the water - Posted by Kambiz Ehsani - 5/5/2018 - Solved
The temperature of the objects on a winter night reaches zero degree C. The moisture in the air is relatively high. Explain which one of the options below is the right process for the moisture on the objects phase change?
1) From vapor to solid
2) From vapor to solid to liquid
3) From vapor to liquid
4) From vapor to liquid to solid
Solution:
Because the temperature of the objects on the surface are below the temperature of the triple point of the water the vapor will become solid directly.
1) From vapor to solid
2) From vapor to solid to liquid
3) From vapor to liquid
4) From vapor to liquid to solid
Solution:
Because the temperature of the objects on the surface are below the temperature of the triple point of the water the vapor will become solid directly.
Photo source :
https://www.quora.com/What-is-the-meaning-of-triple-point-of-water
20 - Question 14 - The Second Law and Carnot Engine - Posted by Kambiz Ehsani - 5/5/2018 - Solved
Is this possible to build a heat engine that works between 800 and 300 degree K and the heat transfer from the hot source Qh = 500 kJ , Ql = 187.5 kJ and produce 200 kJ? Explain your answer.
Solution:
NO beacuse: the First Law is not met here.
Qh - Ql must be the W => this becomes 500 - 187.5 which is not equal the 200 given in the assumptions of the problem and based on the First Law this is not possible because the delta of the Q must be equal to the work created on the shaft and in this example it is not the case.
Solution:
NO beacuse: the First Law is not met here.
Qh - Ql must be the W => this becomes 500 - 187.5 which is not equal the 200 given in the assumptions of the problem and based on the First Law this is not possible because the delta of the Q must be equal to the work created on the shaft and in this example it is not the case.
19 - Question 13 - Thermodynamics - Post by Kambiz Ehsani - 5/5/2018 - Solved
Consider a balloon with the initial internal pressure of P1 when the outside pressure is P0. If air is pumped so that the internal balloon pressure becomes P2, Find the work of system on the surrounding.
Solution:
Assuming the air is the ideal gas and neglect the heat transfer. Therefore the Pv = m RT, since the P is not changing, the work will be applied to increase the volume from V1 to V2 therefore the work will be :
1W2 = P0 (V2 -V1)
Solution:
Assuming the air is the ideal gas and neglect the heat transfer. Therefore the Pv = m RT, since the P is not changing, the work will be applied to increase the volume from V1 to V2 therefore the work will be :
1W2 = P0 (V2 -V1)
18 - What is the Carnot Cycle? Post by Kambiz Ehsani - 5/4/2018 - Solved
Per Wikipedia:"The Carnot cycle is a theoretical thermodynamic cycle proposed by French physicist Sadi Carnot in 1824 and expanded upon by others in the 1830s and 1840s. It provides an upper limit on the efficiency that any classical thermodynamic engine can achieve during the conversion of heat into work, or conversely, the efficiency of a refrigeration system in creating a temperature difference (e.g. refrigeration) by the application of work to the system. It is not an actual thermodynamic cycle but is a theoretical construct."
Why is this cycle so important to design engineers? The Carnot cycle is a theoretical cycle that gives the maximum thermal efficiencies when the cycle works between two thermal sources (high and low temperatures). For example if the Rankine cycle is considered for the argument, where it operates between the temperature of the boiler feed water at 40 and Steam temperature at 400 degree C. What is the Carnot efficiency of such cycle?
Solution:
Efficiency = 1- Tl / Th = 1- (278/478) = 41% and we know that the thermal efficiency of the Rankine Cycle doesn't exceed high 30%. So Rankine Cycle is a high ceiling for the performance or efficiency of a real cycle.
Why is this cycle so important to design engineers? The Carnot cycle is a theoretical cycle that gives the maximum thermal efficiencies when the cycle works between two thermal sources (high and low temperatures). For example if the Rankine cycle is considered for the argument, where it operates between the temperature of the boiler feed water at 40 and Steam temperature at 400 degree C. What is the Carnot efficiency of such cycle?
Solution:
Efficiency = 1- Tl / Th = 1- (278/478) = 41% and we know that the thermal efficiency of the Rankine Cycle doesn't exceed high 30%. So Rankine Cycle is a high ceiling for the performance or efficiency of a real cycle.
17 - Introduction of a book - Shape and Structure, From Engineering to Nature by Adrian Bejan - 5/4/2018
In this book Dr. Bejan propose a topic that has always been and always will be important. The basic idea that the constrained and purposeful optimizations that engineers perform routinely in the design of thermofluid flow systems can help all of us.
Better sense means a simpler, easier - to - understand, more compact, and general summary of explanations of what we see in nature. Such a summary is called a "Principle or Law".
Dr. Bejan shows that Geometric form (Shape and Structure) springs out of the struggle for better global performance subject to global and local constraints.
The thought that the same objective and constraints principle is also the mechanism that constructs geometry in natural flow systems is called "Constructal theory".
Dr. Bejan says in the preface of the book that "There are three aspects of this idea that I pursue in this book:
Better sense means a simpler, easier - to - understand, more compact, and general summary of explanations of what we see in nature. Such a summary is called a "Principle or Law".
Dr. Bejan shows that Geometric form (Shape and Structure) springs out of the struggle for better global performance subject to global and local constraints.
The thought that the same objective and constraints principle is also the mechanism that constructs geometry in natural flow systems is called "Constructal theory".
Dr. Bejan says in the preface of the book that "There are three aspects of this idea that I pursue in this book:
- To start from principle and to arrive through a mental viewing in the powerful position of predicting geometric forms that appear in nature is to practice theory. The time arrow of theory , from principle to nature, runs against the time arrow of empiricism, which begins with nature - the unexplained observation. Empiricism has been the preferred method in the study of naturally organized systems, from river and lung morphology to turbulent eddies and fractal geometry.
- The Second aspect is useful to us as engineers. Engineering is the science of systems and processes with purpose. By identifying the principle that accounts for geometric form in natural flow we improve our pwn vision as designers, as creator. For example, nature impresses us with a multitude of tree-shaped flow: lungs, vascularized tissues, river basins and deltas, lighting. botanical trees, dendritic crystals, nervous systems, street patterns and urban growth, bacterial colonies, transportation, communication and economic networks, etc. Each tree flow connects an infinity of points (volume, or area). This is beautiful example of how, in the end, the theory returns the favor to the field that created it, to engineering .
- The third aspect has to do with the role of engineering in society. once a noble and revered science (think of Leonardo da Vinci, Sadi Carnot etc.), engineering is now taken for granted. Everywhere we look, from university campus politics to the noble prize, engineering rank either low or not at all on the ladder of respect. The scientist of all time wondering about our origin. This is why Dr. Bejan believes that engineers are destined to play a role in the quest for a rational basis - a principle - for the generation of geometric form in nature.
16 - Question 12 by Kambiz Ehsani - Fluid Mechanics - 4/30/2018 - Solved
Explain what happens when we squeeze the gardening hose tip while the water is running? explain what is changing. What happens to the pressure? what happen to velocity and why? What happens to the water flow?
Solution:
Flow rate will be less due to the added friction and restriction. Velocity will be added due to less flow rate. From the Euler equation of p(term) + v(term) + z(term) = cte
We know that z term will be constant therefore when v term increase we would know that delta p will decrease and therefore p of water before exiting the hose will be increased .
Solution:
Flow rate will be less due to the added friction and restriction. Velocity will be added due to less flow rate. From the Euler equation of p(term) + v(term) + z(term) = cte
We know that z term will be constant therefore when v term increase we would know that delta p will decrease and therefore p of water before exiting the hose will be increased .
15 - Question 11 by Kambiz Ehsani - Thermodynamics - Turbines - 4/27/2018 - Solved
Calculate the hp of a steam turbine with the known inlet exhaust and bleed mass flow rates in pounds per hour.
Solution:
In Turbines with the Bleed the Turbine work is :
Wt = (Hinlet - Hbleed) + (1 - y) Hbleed - (1 - y) Hexhaust
y = bleed ratio = m dot bleed / mdot inlet
Wt = (Hinlet - Hbleed) + (1 - y) (Hbleed - Hexhaust )
Wt will be in Btu/hr to hp multiply by 3.939 E-4 hp / (Btu/hr)
Solution:
In Turbines with the Bleed the Turbine work is :
Wt = (Hinlet - Hbleed) + (1 - y) Hbleed - (1 - y) Hexhaust
y = bleed ratio = m dot bleed / mdot inlet
Wt = (Hinlet - Hbleed) + (1 - y) (Hbleed - Hexhaust )
Wt will be in Btu/hr to hp multiply by 3.939 E-4 hp / (Btu/hr)
14 - Question 10 by Kambiz Ehsani - General physics - 4/27/2018 - Solved
At what degree Rankine the value for the degree C and F will be equal?
Solution:
C = (5/9) * ( F - 32)
F = (9/5) * C + 32
(5/9) * ( T - 32) = (9/5) * T + 32
T = - 40 deg C or F = 420 R
Solution:
C = (5/9) * ( F - 32)
F = (9/5) * C + 32
(5/9) * ( T - 32) = (9/5) * T + 32
T = - 40 deg C or F = 420 R
13 - Question 9 by Kambiz Ehsani - First Lat of the Thermodynamics - Pressure washer problem - 4/27/2018 - Solved
What is the increase of water temperature in the water being pumped by a pressure washer pump at a car wash? The inlet water temperature at the inlet of the pump is 70 degree F. We know that the pressure of the pump is 3000 psig. Please elaborate on your assumptions.
Solution:
In the absence of other information, it is common sense that the pump is adiabatic and isentropic. For isentropic compression and no change in Kinetic and Potential, the steady flow energy equation (SFEE) gives:
Wpump = m * (h2 -h1) = m * v * (v2 - v1)
Where :
v = specific volume of inlet water
v = 0.01672 ft^3 / lbm (Sat. H20 at 1 atm)
and :
p2 = 3000 lbf/ft^2
p1 = 14.7 lbf/inch^2
Then:
from the definition of c_p Delta _T
Delta _T = (9.2 Btu/lbm) / (1 Btu/lbm) = 9.2 Degree
Tout = 70 F + 9.2 F = 79.2 Degree F
Solution:
In the absence of other information, it is common sense that the pump is adiabatic and isentropic. For isentropic compression and no change in Kinetic and Potential, the steady flow energy equation (SFEE) gives:
Wpump = m * (h2 -h1) = m * v * (v2 - v1)
Where :
v = specific volume of inlet water
v = 0.01672 ft^3 / lbm (Sat. H20 at 1 atm)
and :
p2 = 3000 lbf/ft^2
p1 = 14.7 lbf/inch^2
Then:
from the definition of c_p Delta _T
Delta _T = (9.2 Btu/lbm) / (1 Btu/lbm) = 9.2 Degree
Tout = 70 F + 9.2 F = 79.2 Degree F
12- Question 8 by Kambiz Ehsani - Second Law of the Thermodynamics - 4/27/2018 - Solved
The question was adopted from the Van Wylen Textbook 3rd edition:
A cyclic machine is used to transfer heat from a higher to a lower temperature reservoir, as show in the picture below.Determine if this machine with the energy transfer,values as shown , is reversible or irreversible or impossible.
A cyclic machine is used to transfer heat from a higher to a lower temperature reservoir, as show in the picture below.Determine if this machine with the energy transfer,values as shown , is reversible or irreversible or impossible.
Solution;
You have to examine the cycle for the Carnot cycle. For the given TH , TL
Then the Carnot cycle efficiency will be = ( TH - TL ) / TH = 1000 - 400 / 1000 = 0.60
Efficiency of the given system = Wnet / Qh = 0.615 > Carnot Efficiency or 0.6
Impossible
11- Question 7 by Kambiz Ehsani - Machine Design - 4/26/2018 - Solved
Design the K factor of the old revolver so that the initial speed of the bullet before it exit the gun become Vb2.
The known values are as following:
The known values are as following:
- The length of the spring at the normal situation is L1
- The length of the spring at the normal situation is L2
- The Weight of the hammer piece is Mh
- Bullet initial speed is zero or Vb1 = 0
- Vb2 is the velocity of the bullet after the impact is known
- At the time of impact the velocity of the hammer will be Vh1
- Bullet mass is know and that is Mb
Solution: (Parametric)
Start from the conservation of energy for the spring and hammer system:
Potential Energy absorbed in the Spring before pulling the trigger = 0.5 * K* L1^2
Potential Energy absorbed in the Spring after the impact = 0.5 * K * L2^2
Kinetic Energy at the time of the impact = 0.5 * Mh * Vh1^2
Energy Conservation => Total energy before pulling the trigger = Total energy at the time of impact
0.5 * K* L1^2 = 0.5 * K * L2^2 + 0.5 * Mh * Vh1^2 Eq (1)
K = (Mh * Vh1 ^2) / (L2 ^2 - L1^2) Eq (2)
Conservation of the momentum =>
- Vb1 is the velocity of the bullet before the impact was zero
- Vb2 is the velocity of the bullet after the impact is known
- Vh1 is unknown
- Vh2 is assumed to be zero after the impact
Mh * Vh1 = Mb2 * Vb2 Eq (3)
Rearranging :
Vh1 = Mb2 * Vb2 / Mh1 Eq (4)
Substituting Vh1 from Eq (4) in Eq (2)
K = Mh * (Mb2 * Vb2 / Mh) ^ 2 / (L2^2 - L1^2)
10- Question 6 by Kambiz Ehsani - Gravity between two bodies - 4/26/2018 - Solved
Which one of the following statements are true for the a ship passing by another one with little space between them?
A) There is a force pushing them away from each other
B) There is a force pulling them toward eachother
C) Due to the boundary layer the speed of both ships will decrease
D) Due to the boundary layer the pressure on both ships will increase
Explain your answer please.
Solution:
The right answer is "B"
The two mass will be pulled toward each other due to the Newton law of gravity that has inverse relationship with the distance between the two objects. The lesser the distance the more the gravity.
A) There is a force pushing them away from each other
B) There is a force pulling them toward eachother
C) Due to the boundary layer the speed of both ships will decrease
D) Due to the boundary layer the pressure on both ships will increase
Explain your answer please.
Solution:
The right answer is "B"
The two mass will be pulled toward each other due to the Newton law of gravity that has inverse relationship with the distance between the two objects. The lesser the distance the more the gravity.
9 - Question 5 - HP of the escalator - by Kambiz Ehsani - 4/22/2018 - Solved
An escalator must be installed in South Coast Plaza in Orange County California. Design considerations of the escalator are :
1) It moves 30 people in 2 minutes.
2) Each person average weight is 200 pounds
3) Escalator lift is 30 feet
4) Efficiency of the motor or "e" is 0.75
Need to know the horse power of the motor driving the escalator.
Solution:
Delta E = Change in the Potential Energy for all 30 people = 30 * mgh = 30 * 200 * 30 = 180,000 J
Time = t = 2 minutes = 120 seconds
Power (demand) = Pd = Delta E / t = 1,500 lb.ft/sec or 1,500 / 550 = 2.7 hp -demand
Power (motor) = Pm = Pd / e = 2.7 / 0.75 = 3.6 hp
1) It moves 30 people in 2 minutes.
2) Each person average weight is 200 pounds
3) Escalator lift is 30 feet
4) Efficiency of the motor or "e" is 0.75
Need to know the horse power of the motor driving the escalator.
Solution:
Delta E = Change in the Potential Energy for all 30 people = 30 * mgh = 30 * 200 * 30 = 180,000 J
Time = t = 2 minutes = 120 seconds
Power (demand) = Pd = Delta E / t = 1,500 lb.ft/sec or 1,500 / 550 = 2.7 hp -demand
Power (motor) = Pm = Pd / e = 2.7 / 0.75 = 3.6 hp
8- Question 4 - Energy burned in uphill biking - by Kambiz Ehsani - 4/22/2018 - Solved
You are biking uphill with a constant speed of 15 miles per hour. The weight of the bike and you together will be 200 pounds.Assume the slope 5%. How many Kilo Calories will you pedal in 10 minutes? is that the same as Kilo Calories burned by your body? please explain all of your assumptions:
The question was designed by myself as one day I thought, how the apps on my smart phone calculate my Kilo Calories burned :-)
Solution:
Writing the free body diagram of the mass of you and bike as a rigid body on the inclined surface:
f = Friction force = m*g* Sin (Teta)
Power of pedaling = m*g* Sin (Teta) * v = hp = horse power
v = assumed constant velocity
Teta = the angle for the slope
hp = 200 * 5/(10025)^0.5* 15 * 0.44704/0.3048 = 0.4 = 294 watts = 70 cal/sec
Pedaling Power = Pp = 70 cal / sec = in 10 minutes or 600 seconds = 42,000 cal = 42 kcal
This is the pedaling Kilo Calories. The body has an efficiency in converting the food intake energy to exercise energy or useful work. This efficiency is different from person to person. for every 1 Calorie we eat only 0.75 calories turns into useful work. So the efficiency of the body is approximately 25%. So the pedaling calorie and burning calories are not the same. So the burning calories are:
Burned Power = Pb = Pp / 0.75 = 56 Kcal
By the way all the units on the Fitness machines should be expressed as Kcal and it is as we know expressed as Cal.
The question was designed by myself as one day I thought, how the apps on my smart phone calculate my Kilo Calories burned :-)
Solution:
Writing the free body diagram of the mass of you and bike as a rigid body on the inclined surface:
f = Friction force = m*g* Sin (Teta)
Power of pedaling = m*g* Sin (Teta) * v = hp = horse power
v = assumed constant velocity
Teta = the angle for the slope
hp = 200 * 5/(10025)^0.5* 15 * 0.44704/0.3048 = 0.4 = 294 watts = 70 cal/sec
Pedaling Power = Pp = 70 cal / sec = in 10 minutes or 600 seconds = 42,000 cal = 42 kcal
This is the pedaling Kilo Calories. The body has an efficiency in converting the food intake energy to exercise energy or useful work. This efficiency is different from person to person. for every 1 Calorie we eat only 0.75 calories turns into useful work. So the efficiency of the body is approximately 25%. So the pedaling calorie and burning calories are not the same. So the burning calories are:
Burned Power = Pb = Pp / 0.75 = 56 Kcal
By the way all the units on the Fitness machines should be expressed as Kcal and it is as we know expressed as Cal.
UPDATE on 7- How I lost 10 pounds in two weeks? by Kambiz Ehsani - 4/21/2018 update on 5/18/2018
First of all I am 54 years old. I weighed 209 pounds on April 02 this year and I needed to lower my weight. My doctor has been telling me to lower my weight in order for the blood pressure and blood sugar to adjust. I always started the diet hard and gave up soon. I have been on diet since April 02 and I am 199 pounds now. This is 10 pounds. My first short term goal is to reach 195 pounds and then the longer term goal is to get to 190 pounds and ultimate goal is to get 185 pounds. The original BMI was 30. It is 28.55 and my BMI goal is 26.5.
Around late March I was surfing internet when I noticed there is a MD talking about guaranteed diet that people normally lost weight within the first two weeks.
The diet was nothing but the Low Carb Atkins type diet, but he (MD) was selling a powder that goes with the diet. I don't intend to act as his marketing tool and really to this date don't know if the powder has any role for me to handle the Low Carb diet for more than two weeks and be very strict about it.
He called his powder Pre-Biotics. He explains that due to so many things such as Anti-biotics our body loses it's good bacteria in the gut system and that causes some GI issues including cravings. He explained craving comes from the bad bacteria asking for more food (I don't know if this is medically proven).
He called his powder Pre-Biotics. He explains that due to so many things such as Anti-biotics our body loses it's good bacteria in the gut system and that causes some GI issues including cravings. He explained craving comes from the bad bacteria asking for more food (I don't know if this is medically proven).
He blames some material called Lectins that can be found in the tomato skin and beans etc to be the cause for the GI issue such as leaky gut and due to the leaky gut he was blaming that the cause for the arthritis pains and skin issues etc. He says stay away from the Lectins get the Low Carb Atkins type diet and take the powder once a day (dissolve in 8 oz of water). However the powder hardly dissolves and you feel you are drinking the dissolved soil.Taste weird -:(
What I eat on a daily basis is:
8 am a couple of eggs (sunny side up)
10 am a protein bar or shake (Kind brand full of nuts and covered with the Dark chocolate)
12 noon a Green chicken salad with the Balsamic Vinaigrette dressing from Habit
2 pm a little bag of raw nuts (no cashew) that I bring from home
6 pm a warm grilled chicken or small steak (MD suggested that your protein portion shouldn't be larger than your palm size) plus a large portion of salad, I love the asparagus with my dinner too.
On top of that I walk between 5 to 10 thousands steps daily and once a week I play friendly raquet ball with my friend at gym. I will report to you guys in a week or so. I hope I can keep going and reach my goal (185 pounds)!
If you are interested to see the details of the MD recommendation and the name of his book, please contact me via my email: kbig104.3@gmail.com
Update on 5//8/2018
Started April 4/2/2018 (a month and 6 days ago) or 5 weeks ago I was 209 pound. Today I was 196.3 pounds. The last two days I mixed the powder from the doctor with the protein powder in the morning. So it is not that bad anymore!
Update on 5/18/2018
Today is almost a month and half since I started. I weighed 195.8 pounds today lost already 13.1 pounds). I was happy today since this is the first time in months that I see 195 pounds on my scale. :-) Another update is that my waste size came down from 36 to 34 few days ago. This is after more than a decade!!
8 am a couple of eggs (sunny side up)
10 am a protein bar or shake (Kind brand full of nuts and covered with the Dark chocolate)
12 noon a Green chicken salad with the Balsamic Vinaigrette dressing from Habit
2 pm a little bag of raw nuts (no cashew) that I bring from home
6 pm a warm grilled chicken or small steak (MD suggested that your protein portion shouldn't be larger than your palm size) plus a large portion of salad, I love the asparagus with my dinner too.
On top of that I walk between 5 to 10 thousands steps daily and once a week I play friendly raquet ball with my friend at gym. I will report to you guys in a week or so. I hope I can keep going and reach my goal (185 pounds)!
If you are interested to see the details of the MD recommendation and the name of his book, please contact me via my email: kbig104.3@gmail.com
Update on 5//8/2018
Started April 4/2/2018 (a month and 6 days ago) or 5 weeks ago I was 209 pound. Today I was 196.3 pounds. The last two days I mixed the powder from the doctor with the protein powder in the morning. So it is not that bad anymore!
Update on 5/18/2018
Today is almost a month and half since I started. I weighed 195.8 pounds today lost already 13.1 pounds). I was happy today since this is the first time in months that I see 195 pounds on my scale. :-) Another update is that my waste size came down from 36 to 34 few days ago. This is after more than a decade!!
6- Question 3 - Second Law - by Kambiz Ehsani - 4/20/2018 - Solved
We want to build a 500 MW steam power plant. The condensers will be cooled with river water. The maximum steam temperature is 540 degree C. The pressure of the Condenser will be 11 kPa. You are asked to estimate the temperature rise of the river.
Author will mail a prize to the first correct answer received before Saturday 4/28/18. The correct answer must be emailed to my email kbig104.3@gmail.com
Similar question was found in The Thermodynamics Textbook by Van Wylen and Sontag
Solution:
Wnet = 500 MW * 1,000 kW / MW * 3,412 Btu /kW hr = 1.7 * 10 ^ 9 Btu / kWhr
Th = 1,004 F = 1,464 R
Tl = 101 F = 561 R
Eff th max = Wnet / Qh = (Th - Tl ) / Th = 903 / 1460 = 0.62
Therefore :
Ql = Wnet * (1-Eff th)/Eff th = 1.7 * 10^9 (0.38/0.62) = 1.04*10^9 Btu/hr
m dot H2O = 60 m * 8 m * 10 m/min * 1000 kg/ m^3 * 60 min / hr * 2.2 lb / kg=
m dot H2O = 5.8 * 10^10 lb/hr
Delta T H2O = Ql / [(m dot H2O) * Cp H20) = 1.04*10^9 / 5.8 * 10^10 = 0.017 F
Author will mail a prize to the first correct answer received before Saturday 4/28/18. The correct answer must be emailed to my email kbig104.3@gmail.com
Similar question was found in The Thermodynamics Textbook by Van Wylen and Sontag
Solution:
Wnet = 500 MW * 1,000 kW / MW * 3,412 Btu /kW hr = 1.7 * 10 ^ 9 Btu / kWhr
Th = 1,004 F = 1,464 R
Tl = 101 F = 561 R
Eff th max = Wnet / Qh = (Th - Tl ) / Th = 903 / 1460 = 0.62
Therefore :
Ql = Wnet * (1-Eff th)/Eff th = 1.7 * 10^9 (0.38/0.62) = 1.04*10^9 Btu/hr
m dot H2O = 60 m * 8 m * 10 m/min * 1000 kg/ m^3 * 60 min / hr * 2.2 lb / kg=
m dot H2O = 5.8 * 10^10 lb/hr
Delta T H2O = Ql / [(m dot H2O) * Cp H20) = 1.04*10^9 / 5.8 * 10^10 = 0.017 F
5- Kambiz personal interests and hobbies
•Koi fish
•Eating authentic food
•Traveling
•Watching European Soccer
•Playing Chess
•Photography
•Reading human biology
•Reading human biology
4- About author (Kambiz Ehsani MSc.M.E. PE.)
Life Milestones
·
Born in 1963
·
Finished Mechanical Engineering Degree (ME.) in
1990
·
Started first engineering job in 1990
·
Married in 1994
·
Became father in 1999 and 2002
·
Became Professional Engineer (P.E.) in CA in 2003
·
Finished MSc. in ME. in 2009
3- Exergy by Kambiz Ehsani - 4/21/2018
What is Exergy?
The Exergy method of analysis has been developed and used in the former Soviet Union and Europe, Primarily Germany and Poland. The earliest use of the Word Exergy has been attributed to Rant in 1956 and by Bosnakovic 1960 et al.
The Exergy method of analysis is a particular approach of the second law of thermodynamics of engineering systems. Evolution of the development relating to the second law is mostly is contained in the thermodynamics textbooks that have been written since Carnot, Clausius and Kelvin. Later on, US. scientists like Kennan, Van Wylen and Sonntag included this method and concept in their textbooks.*
In a nutshell, what is exergy for us as field engineers? Exergy for a process from point 1 to 2 is:
Exergy = Available Work = (H2 - H1) - T0 * (S2 - S1)
H = total energy of the matter at any point
T0 = Reference temperature
T0 * (S2 - S1) = Destruction of available work = Lost Work
S = Entropy of matter at any point
In a perfect world with all the processes ideal and reversible (No Entropy), 1 unit of enthalpy is capable to produce shaft work and it becomes Available work!!!
Conserving Exergy should be every design engineers job. How can we conserve Exergy?
Conserving Exergy will be by retrofitting existing energy system in a fashion that in the new system entropy generation is minimized.
*John E. Ahern - The Exergy Method of Energy Systems Analysis
The Exergy method of analysis has been developed and used in the former Soviet Union and Europe, Primarily Germany and Poland. The earliest use of the Word Exergy has been attributed to Rant in 1956 and by Bosnakovic 1960 et al.
The Exergy method of analysis is a particular approach of the second law of thermodynamics of engineering systems. Evolution of the development relating to the second law is mostly is contained in the thermodynamics textbooks that have been written since Carnot, Clausius and Kelvin. Later on, US. scientists like Kennan, Van Wylen and Sonntag included this method and concept in their textbooks.*
In a nutshell, what is exergy for us as field engineers? Exergy for a process from point 1 to 2 is:
Exergy = Available Work = (H2 - H1) - T0 * (S2 - S1)
H = total energy of the matter at any point
T0 = Reference temperature
T0 * (S2 - S1) = Destruction of available work = Lost Work
S = Entropy of matter at any point
In a perfect world with all the processes ideal and reversible (No Entropy), 1 unit of enthalpy is capable to produce shaft work and it becomes Available work!!!
Conserving Exergy should be every design engineers job. How can we conserve Exergy?
Conserving Exergy will be by retrofitting existing energy system in a fashion that in the new system entropy generation is minimized.
*John E. Ahern - The Exergy Method of Energy Systems Analysis
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