I am Dr. Behrang Asgharian, P.E., a Scholarly Associate Professor at Washington State University. This channel is dedicated to creating high quality videos of solving Thermofluids problems (Thermodynamics, Heat Transfer, Fluid Mechanics, Thermal System Design) for undergraduate students. I also offer online classes in thermofluids (Heat Transfer, Fluid Mechanics, and Thermodynamics). If you like my style of teaching, you can contact me at professorbehrang@gmail.com and we can discuss how I can help you.
Assalamualaikum Sir! Hope you are doing well. Thanks a lot for doing such great work. Sir can you plz guide me what software should I use to make these process diagrams
@@qaz655 You’re welcome! I borrowed these process diagrams from the book. I didn’t particularly use any software but I know there are professional commercial codes that do that.
Good explanation sir. I expect a video from you for same parallel flow with valves of different Cv value installed in 2 pipes. Also valves installed in series too if possible.
hello prof! nice video. can you help me understand again in a case where the 2 pipe in path A and B are made of different material and size will have the same head loss again?
I have a question. We know that one formula for calculating the flowrate is: Q=k√p I want to know that if in this equation, P is static pressure or not? Specially when we use this formla for calculating the flow in a sprinkler system that water flows out of sprinkler with a velocity to the atmosphere pressure. I mean if the p that we use in this equation, is the same static pressure that is used in bernouli's equation and we can find the velocity by that formula(bernoli). So the velocity shall be the same if we calculate it by this equation: V=Q/A. And the area is calculated by the orifice size of the sprinkler. I appreciate if you answer my question. Summery: Q=k√p Bernouli (the energy at point 1 can be assumed the pump pressure plus reservoir and the second part, exit point of sprinkler) V=Q/A Is the we that calculated by the last formula and bernouli eq the same?!
Hi, In that formula that "p" in fact the differential pressure (pressure loss) across a valve or a component and in this case a sprinkler. It would not be the regular static pressure. The "k" value in that equation in American system would not be non dimensionalised. The specific gravity should be incorporated in "k".
bonjour docteur je veux vous poser un questionsur cycle de rankine a soutirage comment dessiner sur {property plot} voila le code : // Utilisation du fluide vapeur d'eau $ifnot diagramWindow fluid$='steam' // Déclaration de la fraction de soutirage fraction_soutirage = 0.20 // 20% de la vapeur est soutirée pour le réchauffeur P_haute = 8000 // Pression de la chaudière en kPa P_basse = 20 // Pression de condensation en kPa $endif // Masse de vapeur soutirée m_soutirage = fraction_soutirage // Masse relative soutirée pour le réchauffeur // Masse de vapeur continuant vers la turbine basse pression m_continue = 1 - m_soutirage // Masse relative continuant après la soutirage // Constantes du cycle P_soutirage = 1000 // Pression de soutirage en kPa // État 1: Entrée de la pompe à liquide saturé x[1]=0 h[1] = enthalpy(fluid$, x=x[1], P=P_basse) // Enthalpie du liquide saturé à P_basse s[1] = entropy(fluid$, x=x[1], P=P_basse) // Entropie du liquide saturé à P_basse T[1] = temperature(fluid$, x=x[1], P=P_basse) // Température du liquide saturé à P_basse // État 2: Sortie de la pompe P[2] = P_haute s[2] = s[1] // L'entropie reste constante à travers la pompe (compression isentropique) h[2] = enthalpy(fluid$, P=P[2], s=s[1]) // Enthalpie après la compression T[2] = temperature(fluid$, P=P[2], s=s[1]) // Température après la compression // État 3: Sortie de la chaudière (entrée de la turbine haute pression) x[3] = 1 P[3] = P_haute h[3] = enthalpy(fluid$, x=x[3], P=P[3]) // Enthalpie de la vapeur saturée à P_haute s[3] = entropy(fluid$, x=x[3], P=P[3]) // Entropie de la vapeur saturée à P_haute T[3] = temperature(fluid$, x=x[3], P=P[3]) // Température de la vapeur saturée à P_haute // État 4: Sortie de la turbine haute pression (avant soutirage) P[4] = P_soutirage s[4] = s[3] // Détente isentropique jusqu'à la pression de soutirage h[4] = enthalpy(fluid$, P=P[4], s=s[3]) // Enthalpie à P_soutirage T[4] = temperature(fluid$, P=P[4], s=s[3]) // Température à P_soutirage x[4] = quality(fluid$, P=P[4], s=s[3]) // Qualité de la vapeur à P_soutirage // État 5: Sortie du réchauffeur (entrée de la turbine basse pression) x[5] = 1 P[5] = P_soutirage h_5_prime = enthalpy(fluid$, x=x[5], P=P[5]) // Enthalpie après le réchauffement s_5_prime = entropy(fluid$, x=x[5], P=P[5]) // Entropie après le réchauffement T[5] = temperature(fluid$, x=x[5], P=P[5]) // Température après le réchauffement // Application du bilan de masse pour le mélange à l'état 5 h[5] = (m_soutirage * h[4] + m_continue * h_5_prime) / (m_soutirage + m_continue) s[5] = (m_soutirage * s[4] + m_continue * s_5_prime) / (m_soutirage + m_continue) // État 6: Sortie de la turbine basse pression P[6] = P_basse s[6] = s[5] // Détente isentropique jusqu'à la pression de condensation h[6] = enthalpy(fluid$, P=P[6], s=s[5]) // Enthalpie à P_basse T[6] = temperature(fluid$, P=P[6], s=s[5]) // Température à P_basse x[6] = quality(fluid$, P=P[6], s=s[5]) // Qualité de la vapeur à P_basse // État 7: Entrée dans le condenseur x[7] = 0 P[7] = P_basse h[7] = enthalpy(fluid$, x=x[7], P=P[7]) // Enthalpie du liquide saturé à P_basse s[7] = entropy(fluid$, x=x[7], P=P[7]) // Entropie du liquide saturé à P_basse T[7] = temperature(fluid$, x=x[7], P=P[7]) // Température du liquide saturé à P_basse // Calculs de travail et de chaleur W_pompe = h[2] - h[1] W_turbine_haute = h[3] - h[4] W_turbine_basse = h[5] - h[6] Q_chaudiere = h[3] - h[2] // Travail net produit par l'unité de masse circulant dans la chaudière W_net = W_turbine_haute + (m_continue * W_turbine_basse) - W_pompe // Efficacité thermodynamique ajustée pour le soutirage eta = W_net / Q_chaudiere
If the pump is isentropic, s_2=s_1 and also you know P_2. Knowing two independent properties s_2 and P_2 will give you T_2 if you need it using EES. The thing is if you are not using EES and you are using tables, since this process (1->2) happens in the compressed liquid area, T_2 will be very close to T_1. Maybe just 1 degrees Celsius of difference. So it won't be a big deal for problems like this.
@@professorbehrang Thanks a lot. I am watching recent videos also, I want to understand gas cycles and using EES. i am taking a course on Exergy and my lack of a good background in Thermodynamics is making it so difficult despite reading and listening in class, I am glad the videos can help.
Not at all ME student, I want to go to school for CE. Yet here I am, with having watched two videos on Rankine, following along and writing down each problem. Thank you for being a good teacher and continuing to do this for us
The principle is no matter how many pipes you have in parallel after the whole reaches becomes steady state, the head loss in every pipe becomes the same.
That's true but after the process becomes steady state . Adding another branch will change the total steady state solution if the upstream pumping power stays the same. A new branch after becomes steady state will change the total head loss. In other words, without an added branch you have one flowrate in the pipe. Once you add the branch you won't have the same flowrate in the same pipe. You might think it won't matter because you are looking for the total flowrate. But if the pumping power upstream stays the exact same, adding another branch will also change the total steady state flow rate of the system.
Professor Behrang, I have a 24" line that I need to measure flow. The pipe is liquid full, the pressure varies but is measured and always known, temperature is pretty constant but is also measured and known. The current method being used to measure the flow is an ultrasonic meter which is not working properly due to the fluid causing internal scaling. The scaling has a negative effect on the meter. Taking the line out of service to install an inline meter such as a mag meter or an orifice plate is not an option. I am contemplating hot tapping the pipe so I can install a small 2" lined pipe with a Coriolis meter. My question is can I simply tap and install a 2” line in parallel to the 24” line, and measure the flow in the 2” and be able to determine the total flow in the 24” line?DeDear Professor Behrang, I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it. Can I measure the flow in the 2" line and determine the total flow in the 24" line? Thank you.ar Professor Behrang, I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it, then measure the flow in the 2" line and determine the total flow in the 24" line? Thank you.
If I measure the flow in the smaller line (Qa) then I should be able to determine the total flow by using the following equation Qa + Qa/J. WOuld this be the correct understanding?
Hi Mr. Professor! Thank you for your useful videos. I don't know how to write exp(....) and ln(...) on EES. Could you please help me? From where can i learn writing these
Sorry Sir! I thought ln(...) may be written like "naturallogaritma" and exp(...) may be written like "e^". But, after asking you this question, i realized that how to write functions is shown in "function info" part of the ees. So now, there is no problem. Thank you for your help.
I have the same question as one of the other commenters. To calculate J, you need to know both FA and FB. You glossed over determining this in the video. Unless I am mistaken determining these values require you to know the velocity in each branch. But we don't know the velocity in each branch because we don't yet know the flow in each branch. Seems like a chicken and egg problem. Can you explain what I am missing?
No, You are not missing anything. We need to know friction factors to be able to solve this problem. However, there are 2 ways to solve the problem if you don't know the friction factors. Case 1: sometimes you can guess the values of friction factors correctly. For example, if the pipes are really rough and the flow is fully turbulent f values on the moody chart become independent of Reynolds number and can be picked from the Moody chart. Case 2: You will need to add 2 equations for friction factor (Colebrook equation) to your set of equations in a code or something and have it iterate until it converges.
Hi Mr. Professor Behrang Your videos are very helpful, thank you very much. I need to ask a question. In some articles, writers say "the value of k (thermal conductivity is taken from EES library). Can EES specify the value of k itself according to case. Up to what should we choose the value of k
Why I'am asking this question is the temperature on one side of the metal material is known, but on another side, the temperature is not known. To calculate this unknown temperature, we should know k value.
Thanks for the nice explanation! One thing I do not understand is why the pressures (P2) in the two branches are equal when you write the bernoulli equation for the two branches?
Thank you Prof. Thats really a nice video. Just curious to know the gadgets you have used for the video. They are really awesome. If its sharable, what kind of hardware and software you are using for this?
After watching your video I am totally able to drive the equation for total drag force. Professor, would you please clarify near the wall(surface) what would be the behaviour of the local and integral drag friction, both increasing or one increasing the other is decreasing?
The local drag always decreases until the flow pattern goes from laminar to turbulent. There is an increase in local friction coefficient during the transition. Then it will decrease again. The total drag force always increases because it’s the integral of all local forces. The bigger the surface, the bigger the drag force.
