Given a pipe system shown in the diagram here, we're left with a problem and we've got some missing information. We've got four parts, A runs into B, B runs into C and D. We've got some information listed, the velocity of the flow at I, we've got the velocity of the flow at B. We've given the flow rate at C, we don't have the flow rate at D. We're also given the diameter at B and the diameter at C and D. The flow rate is calculated using a formula which is Q equals A times V, where Q is the flow rate, A is the cross-sectional area of the pipe and V is the velocity of the fluid in the pipe. In any given system with fluid running through it, we consider that the flow rate is the same right throughout the system, that's because you consider water to be non-compressible. It's different for a gas system, but taking this formula Q meaning flow rate equals area times velocity, we can say the flow rate at I, Q, I equals the flow rate at B, which is also equal to the flow rate at the sum of the flow rate at C and the flow rate at D. So we have the formula Q, I equals QB equals QC plus QD, which reads as the flow rate at I, is the same as the flow rate at B, which is the same as the flow rate at the sum of C and D. In this information in the diagram, we are asked to find the diameter at A, which is this dimension across here. We are asked to find the velocity at C, so we can write the L equals for velocity, and we are asked to find the flow at D, so we need to find the flow rate at D. Using our formula, the flow rate at A equals the flow rate at B, we need to find the diameter at A. So let's write that formula over to the side here. The flow rate at A equals the flow rate at B, and we've said that the flow rate equals the area of the pipe times the velocity, so here we can write the flow rate equals the area times velocity at A. So we've got area of at A times the velocity at A equals the area of B times the velocity of B. So remember, we are looking for the diameter at A. We are asking to find the area at A, so we need the substitute again. The area at A is pi R squared because of circular pipe. So it's pi R squared at A times the velocity at A equals the area at B, pi R squared B times the velocity at B. We know pi, which is 3.14, we don't know the radius. We know the velocity at A, that's 2.83, let's give them here, equals pi times the radius at B. We're given the diameter at A times the diameter at B, so the radius must be 2. Radius squared, 2 times 2. Times the velocity at B is 6.8, so we can write the radius at A times the velocity at B times the radius at B. We're given the diameter at B is 4, so the radius must be 2. Radius squared, 2 times 2. Times the velocity at B is 6.37. So now we can calculate these values and transpose to separate R squared to find the value of R. But what I can see we can do is we've got 3.14 on both sides, so we can just cancel those straight out. And then we left with R squared times 2.83 equals 4, 2 times 2, 4 times 6.37. So I'm going to transpose by taking the 2.83 and putting an underneath that equation, multiply becomes divide, so that becomes R squared equals 4 times 6.37 divided by 2.83. And if I calculate that, it becomes 25.48 divided by 2.83, which is equal to 9. So R squared equals 9, then R equals 3. So what's R? R was the radius at A. Now if R is 3, then the diameter at A is equal to 3. And then R equals 6. The next piece of information I need to find is the velocity of the fluid flowing through the pipe at C. I need to find the velocity at C. So the formula I'm going to use is Q equals A V. So I'll just write that here again for this problem. Q equals A V, which is flow rate equals area of times the velocity. Velocity is what we try to find. We have the flow rate. We have the diameter of the pipe, but we don't have the velocity. So let's write this in this form. The flow rate at C equals the area at C times the velocity at C. We know the flow rate is 2, so we can substitute that information. We don't know the velocity, so we're going to leave that at V. And the area is pi R squared. So we're taking area and writing pi R squared instead. So we can rewrite that as 2 equals V times pi, 3.14, times the radius, which is the diameter to 2 divided by 2 squared. 2 divided by 2 squared is 1. 3.14 times 1 is 3.14. So it's true also to write that 2 equals V times 3.14. So we're going to transpose 3.14 to say V equals 2 divided by 3.14 equals 0.637. So the velocity at C is 0.637. So we're going to write this as a velocity at C. So we're going to transpose city of R end. So we are going to find the velocity at the E. And we are going to name this velocity at V and apply it on the duty. When the velocity isormality, first strain it is De initial strain And we are going to extract Qb equals Qc plus Qd, some of the flow rate of Q, and that's Cnd is the flow rate of B. So the information we've got here for B is that the velocity of B multiplied by the area of B is equal to the flow rate of B. So the flow rate equals area times velocity. And that's equal to the flow rate at C plus D. And we know that the velocity at B is 6.37. Multiply by the area at B, which is pi r squared. So I'll just substitute that there. We know the flow rate at C is 2 plus the flow rate at D. So then we can substitute where we've got pi. We can substitute 3.14 times the radius at B, which is 2. Look at that earlier, 2 times 2, r squared pi r squared equals 2 plus the flow rate at D. So then we can transpose to positive here, so it becomes negative on the other side. And we can work out that calculation. The flow rate at D equals 6.37 times 3.14 times 4. We put some brackets around that, transpose the 2. And we find that equals 6. So the flow rate at D equals 6.