Two oil tanks are connected by two 9 -m-long pipes, as in...

Two oil tanks are connected by two 9 -m-long pipes, as in Fig. P6.26. Pipe 1 is $5 \mathrm{cm}$ in diameter and is $6 \mathrm{m}$ higher than pipe $2 .$ It is found that the flow rate in pipe 2 is twice as large as the flow in pipe $1 .(a)$ What is the diameter of pipe $2 ?(b)$ Are both pipe flows laminar? (c) What is the flow rate in pipe $2\left(\mathrm{m}^{3} / \mathrm{s}\right) ?$ Neglect minor losses.

Transcript

00:01 So in this question the fluid flowing in both the pipes is s -a -e 30w oil at 20 degree celsius where density row is equal to 891 kg per meter cube and mew is equal to 0 .29 ns per meter square.

00:39 Now we know that flows are laminar and length density and velocity velocity, viscosity is a constant then if flows are laminar and length density viscosity a constant then we have q is directly proportional to d raise to the power four so initially let us assume that the flow is laminar in both the pipes so we can write q2 divided by q1 is equal to two which is given which will be equals to d2 divided by d1 raised to the power 4 the value of d1 is given to be 2 so we can find the value of d2 which will be this is 5 right so 5 multiplied by okay a little confusion this is d q2 divided by q1 ratio is given to be 2 now on simplifying this we get 2 raised to the power 1 divided by 4 multiplied by value of d1 that is 5.

02:30 Solving this we get 5 .95 cm right first solution coming to next now let us apply the energy equation between the surfaces of both the tanks so here p1 will be equals to p2 which will be p atmospheric v1 will be equals to v2 which will be equal to 0 negligible reduction in levels of tanks, delta z will be equals to 22 .15.

03:13 Next, neglecting minor losses, both the pipes are made up of the same material.

03:21 Now, head loss will be same for both the pipes.

03:30 This is equals to 7 meter.

03:33 Right.

03:34 So, we have delta z, which is equals to 128 mu l, q2, divided by pi row g d2 raised to the power four for laminal head loss now here we have the value of delta z7 mu is known which is 0 .29 l is known which is 9 q2 we have to find pi value is known row is known which is 891 g is known which is standard value 9 .81 and d2 is known which is which is 0 .0595.

04:19 Now substituting all these values and solving, let is write it once here, mu is equal to 0 .29, l is equals to 9, row is equals to 891, g is equals to 9 .81, pi is equal to 3 .22 divided by 7, and d2 is equal to 0 .0595.

04:48 Now substituting all these values in solving, also we have the value of z delta z is equal to 7...

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