At 10:15 minutes for Motor power requirement I have multiplied Efficiency into density*Q*g*H by mistake. Efficiency should be in division (There will be 3 efficiency Pump efficiency, Mechanical efficiency, and motor efficiency. All 3 should be in division to get Motor power requirement.)
hello sir, I watched your pump series, and it helped me to clear campus placement interview. thank you so much for the unparalleled industrial Knowledge and experience.
11:56 Mistake in Affinity laws. Q is proportional to D^3, H is proportional to D^2, P is proportional to D^5. I checked 2 textbooks both mentioned this.
Sir, i watched your video more time because it's knowledgeable so thanks providing such type of knowledge. Sir please tell me how draw curve line where it's come.
In the cavitation and NPSH part 3 . It was mentioned that NPSH req will reduce with increase in suction line which was correct. However in this video with increase in impeller diameter How NPSH req reduces I didn't understand. Could you please provide an explanation for that as well
It is because, when impeller diameter is small, the gap between impeller diameter and pump casing is high, which result into loss of centrifugal force around impeller and hence minor quantity of fluid is circulating around pump casing and impeller. That's why if impeller is large , The gap between impeller and casing is low, lower circulation of flow around impeller, ultimately reducing NPSH r
Brother....@ 9:10 mins in this vedio..the point when your efficiency starts to go down again after reaching the max is the point when ur multiplication factor of Q*H will start to go down...right?? i mean to say...even though ur flow will be rising continously, ur H will go down...and cumulatively ur decrease in head factor will overcome the increase in flow factor(though flow will rise, it will be very less when compared with reduction in H) so ur overall Q*H will start to go down... am i right?? then....it will move towards minimun... thanks
Yes at the left section, as Q increases, the losses dont have much effect on H, so with increase in Q, the decrease in H is less in left. But at right side, with increase in Q, the H loss becomes high because of the dominating friction losses at higher velocities. Thus the decrease in H is more than increase in Q, thus the multiplication result falls and we get the optimum/peak value of efficiency curve.
sir plz explain how at 10.01 min of this video you are saying power required by motor is equal to hydraulic power multiply efficiency.i think its wrong.it should be like efficiency X shaft power= QxdensityxHxg and from this equation we can deduce that efficiency is directly proportional to Q.how come motor requirement power and the formula also seems wrong.Plz correct me if i am wrong
Sir, i watched your video more time because it's knowledgeable so thanks providing such type of knowledge. Sir please tell me how draw curve line where it's come.
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At 10:15 minutes for Motor power requirement I have multiplied Efficiency into density*Q*g*H by mistake. Efficiency should be in division (There will be 3 efficiency Pump efficiency, Mechanical efficiency, and motor efficiency. All 3 should be in division to get Motor power requirement.)
Sir, please make video on G value.
hello sir, I watched your pump series, and it helped me to clear campus placement interview. thank you so much for the unparalleled industrial Knowledge and experience.
Startup procedure hot pump and cold pump per videus banaye sir ji
Sir i have never seen such explanation,
The way you explain is very good ❤❤❤❤❤
11:56 Mistake in Affinity laws. Q is proportional to D^3, H is proportional to D^2, P is proportional to D^5. I checked 2 textbooks both mentioned this.
You are right but for all practical purposes in industry....Q is proportional to D....check Wikipedia
Bahut badia ... 👍
Jab ap kis topic physical chemistry se related krte hai uha pr Bhut maza ata hai
Sir, i watched your video more time because it's knowledgeable so thanks providing such type of knowledge. Sir please tell me how draw curve line where it's come.
Great explanation
Thank you guru ji
Sir plz explain why we use Q on x-axis & H on y-axis?
sir can you tell what will be the characteristics curve of a system looks like having a centrifugal and reciprocating pumps in series
Does Brake Horse formula is same for compressor.
Very important lecture
Proper explained to understand concept
Sir please make video on G value of the bearing
Sir.. pumps ya compressor me shaft ko rotate karane ke liye kha per. motor aur turbine use kerte hain.,?
Please make one videos on Pump mechanical Data sheet
Thank you sir,,
Great video sir 👍😇
Sir ! Some centrifugal pump rotates in a anticlock wise manner why? And what are the comes out of this fashion?
Sir hydrogen generation me high temperature shift converter and low temperature shift converter smjh Dena with related physical chemistry se
How to drawn system curve by equation
Thankyou sir 👍
In the cavitation and NPSH part 3 . It was mentioned that NPSH req will reduce with increase in suction line which was correct. However in this video with increase in impeller diameter How NPSH req reduces I didn't understand. Could you please provide an explanation for that as well
With increase in diameter flow recirculation loss will decrease
It is because, when impeller diameter is small, the gap between impeller diameter and pump casing is high, which result into loss of centrifugal force around impeller and hence minor quantity of fluid is circulating around pump casing and impeller. That's why if impeller is large , The gap between impeller and casing is low, lower circulation of flow around impeller, ultimately reducing NPSH r
Brother....@ 9:10 mins in this vedio..the point when your efficiency starts to go down again after reaching the max is the point when ur multiplication factor of Q*H will start to go down...right??
i mean to say...even though ur flow will be rising continously, ur H will go down...and cumulatively ur decrease in head factor will overcome the increase in flow factor(though flow will rise, it will be very less when compared with reduction in H) so ur overall Q*H will start to go down...
am i right??
then....it will move towards minimun...
thanks
Yes you are right. Generally we operate in the left of best efficiency point.
Yes at the left section, as Q increases, the losses dont have much effect on H, so with increase in Q, the decrease in H is less in left. But at right side, with increase in Q, the H loss becomes high because of the dominating friction losses at higher velocities. Thus the decrease in H is more than increase in Q, thus the multiplication result falls and we get the optimum/peak value of efficiency curve.
@@abhrajitsaha7 Thanks
Horizontal split case pump
Sir I think you are wrong when you were writing affinity law for diameter can you confirm please
sir plz explain how at 10.01 min of this video you are saying power required by motor is equal to hydraulic power multiply efficiency.i think its wrong.it should be like efficiency X shaft power= QxdensityxHxg and from this equation we can deduce that efficiency is directly proportional to Q.how come motor requirement power and the formula also seems wrong.Plz correct me if i am wrong
Efficiency is not proportional to Q, rather Efficiency first increase and then decrease with increasing Q
You are right I have multiplied by efficiency by mistake. Sorry and thanks for correction.
Why H available curve is non linear
Didnt understand why we draw H vs Q
7777777
Sir, i watched your video more time because it's knowledgeable so thanks providing such type of knowledge. Sir please tell me how draw curve line where it's come.