4:34 why are you able to use the value for specific heat at constant pressure for nitrogen, the pressure is changing, no?
@gr_engineering5 күн бұрын
That is an excellent question. Because we are applying the first law of thermodynamics over an open system, we deal with specific enthalpy, as opposed to specific internal energy. The cp constant is a function of the specific enthalpy, while the cv constant is a function of the specific internal energy. Therefore, we had to use the cp constant in this case. The names of these variables can be very confusing, but I hope this explanation helps!
@CamdenBowers8 күн бұрын
Thank you so much! This is a great video
@gr_engineering5 күн бұрын
I'm glad I could help!
@champgameplay48319 күн бұрын
When you used cp Delta t to find h1-h2, I the delta shouldn't be T2-T1? therefore +221hp of power?
@gr_engineering5 күн бұрын
Hello, thanks for watching! Please double check your work, if Ẇ=ṁ(h1-h2) then Ẇ=ṁcp(T1-T2). We cannot rearrange the order of the inlet and exit without adding a negative sign. It is not physically possible to achieve +221hp as this is a compressor, which is a power consumption device. The power of a compressor is always denoted as negative, given the sign convention of -Ẇ=power consumption and +Ẇ=power generation. I hope this explanation helps!
@MuhamadHanifHafizhan10 күн бұрын
thank youu, your explanation is really helpful, keep it up!
@gr_engineering5 күн бұрын
I'm glad I could help, I appreciate you watching!
@WalterCheatleGrant10 күн бұрын
You can correct me if I'm wrong, but I believe that the "32.2" in the unit conversion should be "32.0" since the problem takes g=32.0 ft/s^2
@gr_engineering5 күн бұрын
Great question! Technically, 32.2 would be more accurate, but yes, the problem suggested using 32.0. The difference is definitely negigible though, so you shouldn’t have any issues 🙂
@WalterCheatleGrant7 сағат бұрын
@@gr_engineering That's a great point. Thank you for taking the time to answer!
@Premium-y9m13 күн бұрын
is this 2nd law of thermo? Is tthis part of it?
@gr_engineering5 күн бұрын
This is the 1st Law of Thermodynamics.
@therealfelix661216 күн бұрын
I know this is a long time ago but I'm fairly sure your compressor diagram is incorrect? Shouldn't it go from a larger inlet to a smaller outlet?
@gr_engineering5 күн бұрын
You’re correct, good catch! And thank you for watching!
@panzerkampfwagentigerausf.732818 күн бұрын
I think last answer is -313.7kW(sign was wrong I think) Thanks!
@gr_engineering5 күн бұрын
You are correct. I forgot mistakenly had a positive heat transfer, which actually wouldn’t make sense. The correct answer is indeed -313.7 kW, indicating that heat is transferred from the turbine to the surroundings. I apologize for the mistake, and thank you for watching!
@panzerkampfwagentigerausf.73285 күн бұрын
@@gr_engineering My pleasure! Thanks for your reply😄
@panzerkampfwagentigerausf.732818 күн бұрын
I think your last answer was wrong. That is -413.44kW. I guess that you made just little writing mistake(113=>413) because until last calculating equation is same with mine.
@panzerkampfwagentigerausf.732818 күн бұрын
However, Thank you for providing answer. I assured myself my uncertain answer!
@gr_engineering18 күн бұрын
Thank you for watching! Please consider double checking your calculations. It appears that you forgot to divide the kinetic energy by 1000 as shown in 14:22.
@panzerkampfwagentigerausf.732818 күн бұрын
@@gr_engineering Wow. How can you know my mistake exactly? I was so arrogant. So Sorry. Thank you for your kind and brilliant advice. I learned very important one more thing!
@baileymowrer341429 күн бұрын
hell yeah bro
@HemanthKumar-ls9lxАй бұрын
Firstly, thank you for the solution . Really helped me a lot. Correct me if i am wrong. For the caluculation of The mass flow rate at the exit we have to use the steam table again to find the specific volume of vapour since the temeperature at Exit is 320 degrees which is less than that of inlet. So we simply cannot half the Mass flow rate at inlet and write it as the answer for mass flow rate at exit. Thank you . Please let me know if i am wrong. I got mass flow rate at exit as 5.26 kg/s
@gr_engineeringАй бұрын
You’re very welcome! If the properties at the exits were different, you would be correct that ṁ2 does not equal ṁ3. However, because the properties are equivalent at both exits, there is validity in saying that ṁ1=ṁ2+ṁ3 and that ṁ2=ṁ3=(ṁ1)/2. Let me know if this makes sense!
@SigninPurposesАй бұрын
Hi! would this still be possible to solve if only FA and its angle were the only given?
@gr_engineeringАй бұрын
Hello, at a minimum you would need more information, such as the magnitude and direction of the resultant force or the second force.
@joeyni2861Ай бұрын
Thank you so much for this video. I was struggling so much trying to figure out what angle goes where you’re explanation is so much better than my own professors.
@gr_engineeringАй бұрын
I’m glad I could help. Thanks for watching!
@davidperez730Ай бұрын
A little extra work, using law of cosines. Its easier, to just add -24.808degrees to 180 degress
@gr_engineeringАй бұрын
Are you obtaining the -24.808° using the law of cosines?
@tofftu2 ай бұрын
I think I'll watch your whole channel to understand the subject :)
@tofftu2 ай бұрын
Thank you very much
@gr_engineeringАй бұрын
You are welcome! Thank you for watching!
@tofftuАй бұрын
@@gr_engineering :D and if you could record and solve more problems from the book, I would appreciate even more
@karimkemo57862 ай бұрын
What if fa is unknown and fr is still known and u need fb to be minimum how do u make sure that fb has the minimum value ?
@gr_engineering2 ай бұрын
@@karimkemo5786 Do you have the angles between Fa/FR and Fb/FR?
@karimkemo57862 ай бұрын
Only for Fa which is 30 degree
@gr_engineeringАй бұрын
@@karimkemo5786I apologize for the delayed response. If you still are stuck please submit a request in the link in the description.
@emitre08Ай бұрын
@@karimkemo5786sample problem 2.2 from vector mechanics for engineers?
@gr_engineering2 ай бұрын
CORRECTION: Please note that at 4:22 I wrote down and said to ADD (2)(3)(1.615)cosθ. This expression is SUBTRACTED, NOT ADDED. Therefore, when you plug in (-2)(3)(1.615) into the denominator, you are automatically yielded with the final answer shown here, 38.26°. I apologize for any confusion.
@smartmoneymoves82013 ай бұрын
Been waiting on another video. Nice breakdown!
@gr_engineering3 ай бұрын
Thanks for the kind words!
@karz3213 ай бұрын
Awesome problem solving skills and technique, very thorough.
@zainabreda44353 ай бұрын
What's your reference book
@gr_engineering3 ай бұрын
Check out the description for the referenced textbook.
@zainabreda44353 ай бұрын
Can I know the reference book ?
@gr_engineering2 ай бұрын
Please see the textbook referenced in the description
@LuckyOmogbae4 ай бұрын
So I guess this is why we can't just assume the resultant to be the y or x axis as what most of us do🤦? Your response?
@gr_engineering4 ай бұрын
You have to remember that each force contains 2 important pieces of information: 1. Magnitude 2. Direction Your resultant force has to consider both of these pieces of information for any forces you are trying to add. Let me give you an analogy to make things clearer. If you have a kayak at the origin (the center of the coordinate system) and a you paddle it northeast (such as the direction of force F in this problem) and I paddle it southwest (such as the direction of the 700N force in this problem), the resultant force acting on the kayak is going to take both of our paddling into account. While the resultant force COULD line up with the X or Y axis, it would only occur if we paddled in a certain direction and with a certain amount of strength (magnitude) If you paddled directly west with a paddling force of 100N, and I paddled north with a paddling force of 0N, surely, the resultant force on the kayak will be directly to the west. Now if I paddled north with a paddling force of 100N while you paddled west with 100N, the resultant force on the kayak would be directly between us, in the northwest direction. Finally, if you paddled west with 100N and I paddled north with only 50N, the resultant force on the kayak wouldn’t quite be directly between us anymore in the northwest direction. The resultant force would be skewed more towards you because your magnitude is greater than mine.
@Myllys24 ай бұрын
Thank you
@gr_engineering2 ай бұрын
You're welcome!
@ConfusedClipperButterfly-gh4ch4 ай бұрын
where did the 45 degrees come from
@gr_engineering4 ай бұрын
The explanation starts at 1:30. A parallelogram must add up to 360° and we already have two angles that are 135° each so therefore there’s only 90° left in the parallelogram that must be shared between two equal angles, so they are each 45°.
@cleverowlworks5 ай бұрын
Hey there! I'm a 3rd year Aero Eng Major and your solution really helped me out! Your methodology is great and you helped demystify the process(punny?). Thanks a lot!
@gr_engineering5 ай бұрын
Glad I could help :)
@Statics-Hero5 ай бұрын
Goat talk 🗣️
@gr_engineering5 ай бұрын
Glad I could help!
@michaeladams20746 ай бұрын
thank u, only error is multiplying .02051 by 2 and getting .0412, should be .4102
@gr_engineering6 ай бұрын
Hello, thanks for commenting. I still do not see an error in my calculations. .02051 • 2 = .0412. You may have entered into your calculator incorrectly?
@michaeladams20746 ай бұрын
@@gr_engineering my mistake I commented my correction incorrectly, I meant .04102*
@gr_engineering6 ай бұрын
@@michaeladams2074Good catch, you are absolutely correct! Thanks for pointing that out.
@ReddZepppelin6 ай бұрын
When I rearranged my energy balance equation as in 9:57 I flipped the signs as you did and got -Q= m((h2-h1)+(V2-V1/2)) and worked out -Q= 6.5kJ or Q= -6.5kJ (heat loss from sysem), is this correct? Also when I calculated using your values I got 6.5kJ not 6.82kJ
@gr_engineering6 ай бұрын
Hello, thanks for commenting. When you rearranged the first law equation: 0=Q̇-Ẇ+ṁ[(hᵢ-hₑ)+(vᵢ²-vₑ²)/2+g(zᵢ-zₑ)] you kept Q̇ as negative. Therefore your equation should have kept the enthalpy and velocity as inlet minus exit: -Q̇=ṁ[(hᵢ-hₑ)+(vᵢ²-vₑ²)/2] If you divide both sides by (-1) to get rid of the Q̇’s negative sign, which I decided to do in this video, you will have the exit minus the inlet: Q̇=ṁ[(hₑ-hᵢ)+(vₑ²-vᵢ²)/2] I encourage you to work out the numbers again to see why this is. Also, regarding why you calculated 6.5kJ, I reckon you forgot to square the velocities. The number I calculated is valid. Let me know if you have any other questions!
@miguelalvarez52926 ай бұрын
At 3:30 I’m working out the problem but keep getting 0.5853 for the mass flow rate (m°), how did you get 0.3512?
@gr_engineering6 ай бұрын
Hello, thanks for commenting. The reason for that is that I miswrote what T1 is given as. It should be 500K, I wrote 300K, but I punched 500K into my calculator. The 0.3512 kg/s is correct, I just copied the given incorrectly. I apologize for any confusion!
@miguelalvarez52926 ай бұрын
@@gr_engineering Ah yes. Thank you for the clarification !
@miguelalvarez52926 ай бұрын
I’m having trouble grasping the step here at 1:59 If T1 = 40°C and P1 = 3 bar, why is it a SHV? P1 (or 3 bar) is less than 10.164 bar. Wouldn’t it be a compressed liquid? I’ve been seeing a lot of this: “If T is greater than Tsat, it is a SHV”
@miguelalvarez52926 ай бұрын
When looking at the tables, do I refer to the pressure first and then the temperature to see if its a compressed liquid and a SHV?
@gr_engineering6 ай бұрын
In short, if T is greater than Tsat, yes, the state is superheated. However, if P is greater than Psat, the state is compressed liquid. In this case here P<Psat, so the state is superheated. Think about it like this, if the pressure is greater than Psat, then you have a greater quantity of molecules per unit volume (aka more dense), which requires more energy (temperature) to change state (to superheated vapor, in this case). If you have less pressure than Psat, then it is “easier” or requires less energy (again, temperature) to force the working fluid to change state. Let me know if this helps!
@gr_engineering6 ай бұрын
@@miguelalvarez5292 To answer your second question, if you are given the temperature and pressure, it is arbitrary which one you refer to “first”. Phase state is a function of both temperature, and pressure. To identify the state of your fluid when given temperature and pressure: 1. check the corresponding Psat to the temperature you are given OR check the corresponding Tsat to the pressure you are given. 2. @Tgiven: if Psat<Pgiven you have a compressed liquid. If Psat>Pgiven you have a superheated vapor. @Pgiven: If Tsat<Tgiven you have a superheated vapor. If Tsat>Tgiven you have a compressed liquid.
@miguelalvarez52926 ай бұрын
@@gr_engineering Your first comment answered everything perfectly for me. I aspire to be a mechanical engineer grad and as knowledgeable as you. Thanks for the help!
@gr_engineering6 ай бұрын
@@miguelalvarez5292 You’re too kind :) Thank you for watching, you got this!
@ar-rayyanronie5876 ай бұрын
where did you get the 497
@ar-rayyanronie5876 ай бұрын
ahh okok i get it never mind
@GooseHonkHonkHonk6 ай бұрын
It looks like your first ideal gas equation - p(V-dot)=mRT should be p(V-dot)=(m-dot)RT. Thank you for these videos! These are a lifeline for me when I get stuck.
@gr_engineering6 ай бұрын
Thank you for the catch, yes you are correct, V̇ must be accompanied by ṁ. I’m glad you find these videos helpful!
@cole_solomon7 ай бұрын
This video helped a lot. It's still the same exact material in 2024.
@boomermydog41987 ай бұрын
thermodynamics is hard. For that, I sincerely appreciate you. Bless your soul.
@gr_engineering7 ай бұрын
I’m glad I could help, thanks for the kind words!
@matthewpeters65767 ай бұрын
The area for a circle is not (pi)d^2 but rather (pi)r^2
@gr_engineering7 ай бұрын
The area for a circle is πr^2 or (πd^2)/4, as shown at 1:21.
@TylerCheng-o4i7 ай бұрын
hi,teacher. If we know the weight of piston, when we evaluate the piston work, W=(Fg-Fatm-mg)d, am i right?
@gr_engineering7 ай бұрын
Thanks for asking such a great question. Had the mass of the piston been told, we would have included it in our calculation for work. However, because the mass was not given, it is implying that we neglect the weight of the piston.
@Kohlmam8 ай бұрын
awesome video
@gr_engineering8 ай бұрын
Thanks!
@pawan1209898 ай бұрын
Great
@gr_engineering8 ай бұрын
Glad you liked it!
@TheRealGr33nGuy8 ай бұрын
How do you find the molar mass when calculating the gas constant?
@gr_engineering8 ай бұрын
Check out Table A-1 in your property table.
@aaronschueler31018 ай бұрын
Bro is single-handedly carrying my thermo career
@gr_engineering8 ай бұрын
I’m glad I could help, definitely give yourself some credit :)
@WattBuzz-ce4be8 ай бұрын
You are a lifesaver!!!!
@gr_engineering8 ай бұрын
I'm glad I can help!
@boomermydog41988 ай бұрын
i love you
@gr_engineering8 ай бұрын
Thanks for the kind words!
@kaitlynhawkins82938 ай бұрын
Why is the force from Patm not considered a force doing work? Why consider gravity to be the only force doing work?
@gr_engineering8 ай бұрын
The force from atmospheric pressure, Patm was in fact used in this calculation. Patm was subtracted from Pgas and then multiplied by the difference in volume to find the work. Gravity was not considered in this calculation (mass was not given, assumed to be negligible). Let me know if this is still unclear to you.
@vivelz9 ай бұрын
yessir thank you for the help ❤️
@gr_engineering9 ай бұрын
Happy to help!
@dihanmahmood1329 ай бұрын
why is it inlet minus outlet and not outlet minus inlet since all formulas are outlet minus inlet.
@gr_engineering9 ай бұрын
The energy balance equation over an open system at steady state is always: 0=Q̇-Ẇ+ṁ[(hᵢ-hₑ)+(vᵢ²-vₑ²)/2+g(zᵢ-zₑ)] I am guessing you are used to seeing Q̇= hₑ-hᵢ or -Ẇ= hₑ-hᵢ If that is the case, notice it is just rearrangement Q̇ or Ẇ to the other side of the equation.
@dihanmahmood1329 ай бұрын
understood, thank you so much sir!@@gr_engineering
@DHRUVKUNVARANI10 ай бұрын
Why did we use ideal gas model for first part and not Pv^n = constant tho its adiabatic process?