Mod-24 Lec-24 Fluid Flow Applications Part - III

[@harindamodara1232]

Last Example 0:41, 5:51, 12:12 upto 17:55 a good concept explained, 17:55 Hydraulic Syphon S1-3 > 41:45, 42:38, 43:38 Q1-3 > 44:38, 45:48, 46:38

[@rishabhsahani8173]

amazing lecture Sir. got the concept of cavitation very clearly.

[@ayushdubey840]

@Rishabh Sahani can u please tell me what r u doing right now..? (In perspective of job) means if u have watched this nptel lectures 5 years ago.. Then just wanted to querry.. Above question.. Please guide

[@kumarrohan6186]

if flow takes place from higher energy to lower energy , then why we equating the total energy between two points. Please explain

[@subhambose934]

when a body is dropped from height, the gain in kinetic energy is equal to the loss of potential energy. In general the energy is conserved. the energy is converting from one form to another.

[@ad8128]

Here we are just applying Bernoulli's equation, not bothering about the flow from higher energy to lower energy.

[@VidhuTrivedi12]

high pe to low pe..pe means pressure enegy plus gravitatational energy

[@apoorboguharoy6577]

same doubt.. an ideal fluid should not flow

[@CEteckgd]

Yeh , in case of energy conserved, if flow is taking place then pressure gradient will be negative in flow direction . You can check it on applying energy equation in any inviscid flow. If energy losses due to viscosity ,it is obvious flow take place from higher energy to lower

[@compassft]

In the problem at the start of the lesson, if previously a vortex flow was defined as a flow where Vr=0, how can there be a Vr in the flow inside the cyllinder?

[@DilipSaini888]

Same doubt here as well, while deriving Vt is just function of 'r' and not the 'angle theta' we assumed Vr to be zero. We did it using continuity equation and placed Vr=0 and found Vt=f(r). Here assuming there is radial velocity Vr and again using the same continuity equation and placing 'dVt/dt=0' (since Vt=f(r)) (t represents angle theta) we find Vr=f(r). This truly doesnt make sense. Please Explain

[@vai_-cn9br]

@@DilipSaini888 this is an example of spiral free vortex flow. in that case, tangential and radial velocity components at any point are inversely proportional to the radial co-ordinate of the point.

[@meetpatel-lf8dh]

@@vai_-cn9br thanks 👍🏽

[@VIJAYKUMAR-ic5jb]

Nice concept explain

[@SourajitandAnsu]

Syphon flow

[@hussainbadshah1816]

problem 1 how option a is write pl. guide

[@fabiozangara1532]

Ciao! Fluid flows from B to A because in the energy balance of Bernoulli Equation the (head) energy in point B is greater than total (head) energy, hence a fraction of initial energy B it's utilized to overcome the frictions. For this reason the equation is Energy in A + head loss (energy dissipated to overcome frictions) = Energy in B

[@sahilgajbhiye0656]

ty very much sir

[@ad8128]

Can anyone plz explain solution of last question(no.3) m

[@fabiozangara1532]

Ciao! You have to write the modified Bernoulli Equation between the point in the entry and the point in the summit of siphon. Keep in mind that all data are given in terms of head, so velocity head is defined as (V^2)/(2*g), pressure head as (p/rho) and head loss is simply h, such that all the terms in equation have the unit's measure of a lenght. Hence, computing p on the summit, we obtain p(head) = (10.3 - 0.5 - 3 - 1.5) m = 5.3 m . The pressure is greater than vapor pressure of water, Cavitation is avoided and engineers are happy! Good luck!

[@aiswaryasanthosh7878]

@@fabiozangara1532 At 45.51 , Question 2 how option d also comes?. How can pressure at summit depend on downstream length? please answer

[@fabiozangara1532]

@@aiswaryasanthosh7878 ciao! I followed this course 2 years ago but I try to help you. Based on the explanation of Prof. SOM, if you write the Bernoulli Equation for ideal fluids, between the top corner of the siphon and the tank downstream, setting the level of the lower tank as a zero potential you obtain: p(summit) + rho*g*h + 1/2*rho*V(summit)^2 = p(atm) + 1/2*rho*V(downstram)^2 , where h is the height of the summit with respect to the lower tank. As you can see the pressure at the summit of the siphon (top corner of the tube) depends on the downstream leg h