Question

Water flows through a Pipe A with diameter 50mm, which is in series with pipe Bwith 100mm diameter, in which the mean flow is 1.0m/s. Pipe B branches into pipe Cwith diameter dc and mean flow is 1.5m/s, and D with a diameter of 75mm. Flow rateQc in pipe C is twice the flow rate Qd in pipe D. Determine the following:

a) Qa
b) Qb
c) Va
d) Qc
e) Qd
f) Dc​

Answer 1

The flow rate in pipe A and pipe B is 0.00196m³/s. The flow rate in pipe C (Qc) is twice the flow rate in pipe D (Qd), and can be calculated as Qc = π(dc/2)² * 1.5m/s. The velocity in pipe A (Va) is 1.0m/s. The diameter of pipe C (Dc) is equal to dc.

Step-by-step explanation:

The correct answer is option c) Vad) Qce) Qd.

To solve this problem, we can use the principle of continuity, which states that the flow rate of an incompressible fluid remains constant throughout a pipe or series of pipes, as long as there are no leaks or changes in the pipe diameter. In this case, the flow rate in pipe B (Qb) is equal to the flow rate in pipe A (Qa), which can be calculated using the formula Q = Av, where Q is the flow rate, A is the cross-sectional area of the pipe, and v is the mean flow velocity.

Given that the diameter of pipe A is 50mm and the flow velocity is 1.0m/s, we can calculate the cross-sectional area using the formula A = πr², where r is the radius of the pipe. Therefore, the radius of pipe A is 25mm or 0.025m, and the cross-sectional area is 0.00196m². Substituting these values into the equation Q = Av, we get Qa = 0.00196m² * 1.0m/s = 0.00196m³/s.

Since pipe B is in series with pipe A, the flow rate in pipe B is also 0.00196m³/s. Pipe B branches into pipe C, where the flow rate (Qc) is twice the flow rate in pipe D (Qd). Therefore, Qc = 2Qd. To find the flow rate in pipe D, we can substitute the value of Qc into the equation Qc = Avc and solve for Qd. Given that the diameter of pipe C is dc, the radius of pipe C is dc/2, and the cross-sectional area of pipe C is π(dc/2)². Substituting the values into the equation, we get 2Qd = π(dc/2)² * 1.5m/s. Simplifying the equation, we find Qd = (π(dc/2)² * 1.5m/s)/2.

Now we can find the values of Qa, Qb, Qc, and Qd as follows:

Qa = 0.00196m³/s

Qb = 0.00196m³/s

Qc = 2Qd = 2 * (π(dc/2)² * 1.5m/s)/2 = π(dc/2)² * 1.5m/s

Qd = (π(dc/2)² * 1.5m/s)/2

Therefore, the answers to the given questions are:

a) Qa = Qb = 0.00196m³/s

b) Qc = 2Qd = π(dc/2)² * 1.5m/s

c) Va = 1.0m/s

d) Qc = π(dc/2)² * 1.5m/s

e) Qd = (π(dc/2)² * 1.5m/s)/2

f) Dc = dc