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
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
@@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
@@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)
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
Same doubt, sir please clear it.
Nice explanation...
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
GATE METALLURGY BY C PATEL Thank you Sir.
great video Sir...:)
Sir am getting h1 = 0.1652m, h2 = 0.25 + h1, accordingly A2 as 1.387 m^2 , kindly verify this approach.
Thank you sir!
I agree with you
Don't you take initial velocity through A1 as 0m/s when it should have some velocity there?
There will be some velocity at A1 and it has been taken care of.
If I have misunderstood then please ask again.
@@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
@@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)
Sir what about the height above the sprue