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@m.ehsaan. 3 күн бұрын
Can I do it without the cosine rule?
@youssefmostafa2063 5 күн бұрын
thanks man! you are the savior.
@gr_engineering 5 күн бұрын
You’re welcome, happy to help!
@anawaz189 6 күн бұрын
4:34 why are you able to use the value for specific heat at constant pressure for nitrogen, the pressure is changing, no?
@gr_engineering 5 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!
@CamdenBowers 8 күн бұрын
Thank you so much! This is a great video
@gr_engineering 5 күн бұрын
I'm glad I could help!
@champgameplay4831 9 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_engineering 5 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!
@MuhamadHanifHafizhan 10 күн бұрын
thank youu, your explanation is really helpful, keep it up!
@gr_engineering 5 күн бұрын
I'm glad I could help, I appreciate you watching!
@WalterCheatleGrant 10 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_engineering 5 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 🙂
@WalterCheatleGrant 7 сағат бұрын
@@gr_engineering That's a great point. Thank you for taking the time to answer!
@Premium-y9m 13 күн бұрын
is this 2nd law of thermo? Is tthis part of it?
@gr_engineering 5 күн бұрын
This is the 1st Law of Thermodynamics.
@therealfelix6612 16 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_engineering 5 күн бұрын
You’re correct, good catch! And thank you for watching!
@panzerkampfwagentigerausf.7328 18 күн бұрын
I think last answer is -313.7kW(sign was wrong I think) Thanks!
@gr_engineering 5 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.7328 5 күн бұрын
@@gr_engineering My pleasure! Thanks for your reply😄
@panzerkampfwagentigerausf.7328 18 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.7328 18 күн бұрын
However, Thank you for providing answer. I assured myself my uncertain answer!
@gr_engineering 18 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.7328 18 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!
@baileymowrer3414 29 күн бұрын
hell yeah bro
@HemanthKumar-ls9lx Ай 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_engineering Ай 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!
@SigninPurposes Ай бұрын
Hi! would this still be possible to solve if only FA and its angle were the only given?
@gr_engineering Ай бұрын
Hello, at a minimum you would need more information, such as the magnitude and direction of the resultant force or the second force.
@joeyni2861 Ай 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_engineering Ай бұрын
I’m glad I could help. Thanks for watching!
@davidperez730 Ай бұрын
A little extra work, using law of cosines. Its easier, to just add -24.808degrees to 180 degress
@gr_engineering Ай бұрын
Are you obtaining the -24.808° using the law of cosines?
@tofftu 2 ай бұрын
I think I'll watch your whole channel to understand the subject :)
@tofftu 2 ай бұрын
Thank you very much
@gr_engineering Ай бұрын
You are welcome! Thank you for watching!
@tofftu Ай бұрын
@@gr_engineering :D and if you could record and solve more problems from the book, I would appreciate even more
@karimkemo5786 2 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_engineering 2 ай бұрын
@@karimkemo5786 Do you have the angles between Fa/FR and Fb/FR?
@karimkemo5786 2 ай бұрын
Only for Fa which is 30 degree
@gr_engineering Ай бұрын
@@karimkemo5786I apologize for the delayed response. If you still are stuck please submit a request in the link in the description.
@emitre08 Ай бұрын
@@karimkemo5786sample problem 2.2 from vector mechanics for engineers?
@gr_engineering 2 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.
@smartmoneymoves8201 3 ай бұрын
Been waiting on another video. Nice breakdown!
@gr_engineering 3 ай бұрын
Thanks for the kind words!
@karz321 3 ай бұрын
Awesome problem solving skills and technique, very thorough.
@zainabreda4435 3 ай бұрын
What's your reference book
@gr_engineering 3 ай бұрын
Check out the description for the referenced textbook.
@zainabreda4435 3 ай бұрын
Can I know the reference book ?
@gr_engineering 2 ай бұрын
Please see the textbook referenced in the description
@LuckyOmogbae 4 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_engineering 4 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.
@Myllys2 4 ай бұрын
Thank you
@gr_engineering 2 ай бұрын
You're welcome!
@ConfusedClipperButterfly-gh4ch 4 ай бұрын
where did the 45 degrees come from
@gr_engineering 4 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°.
@cleverowlworks 5 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_engineering 5 ай бұрын
Glad I could help :)
@Statics-Hero 5 ай бұрын
Goat talk 🗣️
@gr_engineering 5 ай бұрын
Glad I could help!
@michaeladams2074 6 ай бұрын
thank u, only error is multiplying .02051 by 2 and getting .0412, should be .4102
@gr_engineering 6 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?
@michaeladams2074 6 ай бұрын
@@gr_engineering my mistake I commented my correction incorrectly, I meant .04102*
@gr_engineering 6 ай бұрын
@@michaeladams2074Good catch, you are absolutely correct! Thanks for pointing that out.
@ReddZepppelin 6 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_engineering 6 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!
@miguelalvarez5292 6 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_engineering 6 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!
@miguelalvarez5292 6 ай бұрын
@@gr_engineering Ah yes. Thank you for the clarification !
@miguelalvarez5292 6 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”
@miguelalvarez5292 6 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_engineering 6 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_engineering 6 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.
@miguelalvarez5292 6 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_engineering 6 ай бұрын
@@miguelalvarez5292 You’re too kind :) Thank you for watching, you got this!
@ar-rayyanronie587 6 ай бұрын
where did you get the 497
@ar-rayyanronie587 6 ай бұрын
ahh okok i get it never mind
@GooseHonkHonkHonk 6 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_engineering 6 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_solomon 7 ай бұрын
This video helped a lot. It's still the same exact material in 2024.
@boomermydog4198 7 ай бұрын
thermodynamics is hard. For that, I sincerely appreciate you. Bless your soul.
@gr_engineering 7 ай бұрын
I’m glad I could help, thanks for the kind words!
@matthewpeters6576 7 ай бұрын
The area for a circle is not (pi)d^2 but rather (pi)r^2
@gr_engineering 7 ай бұрын
The area for a circle is πr^2 or (πd^2)/4, as shown at 1:21.
@TylerCheng-o4i 7 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_engineering 7 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.
@Kohlmam 8 ай бұрын
awesome video
@gr_engineering 8 ай бұрын
Thanks!
@pawan120989 8 ай бұрын
Great
@gr_engineering 8 ай бұрын
Glad you liked it!
@TheRealGr33nGuy 8 ай бұрын
How do you find the molar mass when calculating the gas constant?
@gr_engineering 8 ай бұрын
Check out Table A-1 in your property table.
@aaronschueler3101 8 ай бұрын
Bro is single-handedly carrying my thermo career
@gr_engineering 8 ай бұрын
I’m glad I could help, definitely give yourself some credit :)
@WattBuzz-ce4be 8 ай бұрын
You are a lifesaver!!!!
@gr_engineering 8 ай бұрын
I'm glad I can help!
@boomermydog4198 8 ай бұрын
i love you
@gr_engineering 8 ай бұрын
Thanks for the kind words!
@kaitlynhawkins8293 8 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_engineering 8 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.
@vivelz 9 ай бұрын
yessir thank you for the help ❤️
@gr_engineering 9 ай бұрын
Happy to help!
@dihanmahmood132 9 ай бұрын
why is it inlet minus outlet and not outlet minus inlet since all formulas are outlet minus inlet.
@gr_engineering 9 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.
@dihanmahmood132 9 ай бұрын
understood, thank you so much sir!@@gr_engineering
@DHRUVKUNVARANI 10 ай бұрын
Why did we use ideal gas model for first part and not Pv^n = constant tho its adiabatic process?