great video Sir...:)

Nice explanation...

Sir in my point of view ACCORDING TO CONTINUITY OF MASS LAW Q=dA1V1=dA2V2 1st step: Q=A1V1 Q=A1Squre root of 2gH1 (4/2.2)^2=2x9.8xH1 H1=0.168 m 2nd step Ht=H1+H2 Ht=0.168+0.25 Ht=0.418 m 3rd step Q=A2V2 Q=A2V2 that means A2xSqure root of 2gHt 4/Squre root of 2x9.8x041=A2 A2=1.39 m^2 A1=AREA AT TOP OF SPRUE, H1=HEIGHT FROM POURING BASIN TO TOP OF SPRUE A2=AREA AT BOTTOM OF SPRUE H2= HEIGHT FROM TOP OF SPRUE TO BOTTOM OF SPRUE Ht= HEIGHT FROM TOP OF POURING BASIN TO BOTTOM OF SPRUE, d=DENSITY

@sangitasonwane

6 жыл бұрын

Same doubt, sir please clear it.

Thank you sir for clearing the concept of air aspiration. I am having doubt about sprue height. Is sprue height the total height from pouring basin that is h2 in your case? Also during solution of problem we didn't require area of top of sprue so is it given for confusion. Kindly help me in clearing my concept. Thank you

@abhishektomar8814

6 жыл бұрын

GATE METALLURGY BY C PATEL Thank you Sir.

Sir what about the height above the sprue

Sir am getting h1 = 0.1652m, h2 = 0.25 + h1, accordingly A2 as 1.387 m^2 , kindly verify this approach. Thank you sir!

@rahulmaji8768

3 жыл бұрын

I agree with you

Don't you take initial velocity through A1 as 0m/s when it should have some velocity there?

@cpatelacademy

4 жыл бұрын

There will be some velocity at A1 and it has been taken care of. If I have misunderstood then please ask again.

@tonda9325

4 жыл бұрын

@@cpatelacademy alright, so then I would assume the V2 is going to be V1+sqrt(2gh). Isn't that the case? I'm probably wrong tbf but it seems clear to me there should be this. (In the calculation, you also didn't use the A1 at all, right?) Cheers

@BrownMInc

2 жыл бұрын

@@tonda9325 I know you're probably way past this but the idea Im getting is that it depends on where you reference. In total V2 = V1 + sqrt(2gh2-h1) but since V2 is the final speed we'd be looking for, we aren't concerned at all with V1, we use the total height, h2, thus being V2 = sqrt(2gh2)