4 63 the 05 m wide lid on the barbecue grill is being closed...

4-63. The 0.5-m-wide lid on the barbecue grill is being closed at a constant angular velocity of $\omega$ = 0.2 rad/s, starting at $\theta$ = 90°. In the process, the air between A and B will be pushed out in the radial direction since the sides of the grill are covered. Determine the average velocity of the air that emerges from the front of the grill at the instant $\theta$ = 45°. Assume that the air is incompressible. Problem 4-63

Transcript

00:01 So in this question, what we have is that we have air, right, and says that this air assumed incompressible, incompressible, right, and inviscered, right? then it says that this air, it flows into an outdoor cooking grill through nine holes, right? through nine holes right and it says that the diameter is for these holes right so that diameter is equal to 0 .40 inch right as shown in the figure now i'd say that the flow rate the flow rate it's equal to 40 inch cube per second right and we are required to determine the pressure that is whether the grill near the holes right so let's start with the solution of this question now again this question is very simple right now um what we need to do over here the first thing right that is to obtain the magnitude of the velocity of air right with which it is um with with which it flows in the outdoor quick and grill right so the velocity of air is going to be calculated from using the equation of continuity right so let's start now um the velocity it's equal sorry the volume flow rate it's equal to nine because there are nine holes right that is why we have written nine over here it's multiplied with the product of the area of the hole times the velocity right of the hole so you're going to plug in an expression for area right so we have nine into pi by four the square sub hole right and we subhole right now we are going to plug in the value that we have the given values and we are going to find out the velocity for the hole right so just plug in the values that we have right and we get that the velocity of this whole it's equal to 0 .0 2 3 2 feet divided by now this vector that i'm going to these factors that i just plugged in right they are the calculated factors right so you can just calculate them by yourself by just plug in the values in this equation right so when you calculate it you get that the velocity it's equal to 2 .95 feet per second right now so since we have the velocity we just need to plug it in the bernoulli's equation right that is between the outside between the outside of grill and the and any point right any point on any of the whole right so there's equation right so that would be p sub 1 plus as we know right the density of air times the velocity plus pressure energy times z sub 1 then again pressure plus density times velocity times the pressure energy times z sub 2 right now over here this pressure at this point 1 it can be considered to be equal to zero in terms of the gorge pressure right in terms of gorge pressure and the charge pressure and the velocity we sub 1 right um it can also be approximately it can also be considered to be approximately equal to zero right and so is the height term right so why is that happening because we are going to consider this both z sub one and z sub two to be equal to zero is that because they lie on the same plane on the same horizontal plane right so for that um it's safe to say that all of of this left hand side right it's equal to zero right and we are only left with these two values right so we have zero is equal to p sub two plus half of the product of density of air times the velocity of the whole square right so we are required to find out the pressure right so it means that we need to so rearrange this equation and plug in the value.

05:39 So we get that the pressure at point 2, right? so it's equal to negative 0 .014 pounds per feed square into feet square divided by 144 n square...

Question

Please give Ace some feedback