The flowmeter shown in the figure measures the flow rate of water in a solar col

Question

The flowmeter shown in the figure measures the flow rate of water in a solar collector system. The flowmeter is inserted in a pipe with inside diameter 1.7 ; at the constriction the diameter is reduced to 0.68. The thin tube contains oil with density 0.88 times that of water.

If the difference in oil levels on the two sides of the tube is 1.7 , what is the volume flow rate? Express your answer to two significant figures and include the appropriate units.

Q=

Explanation / Answer

Venturi is a horizontal pipe narrower at middle (constriction = obstruction), and wider at ends. Let left side, from where water enters, the pressure be (P1), cross-section area (A) and speed (v1). At middle pressure (P2) cross-section area (a) and speed (v2) > equation of continuity for water gives v2 = (A/a) v1 --------- (1) using Bernoulli> P1 + 0.5 rho v1^2 + 0 (horizonal flow) = P2 + 0.5 rho v21^2 + 0 (P1 – P2) = 0.5 rho * v1^2 [(A/a)^2 – 1] >>>>>>>>> using (1) --------------------------------------… since we also introduce a manometer (FILLED WITH OIL, which measures Pressure) > one leg at left (inlet) and other leg at constriction (middle) >> we get pressure difference P1 – P2 = rho (oil) * g * h Where h = difference of oil in 2 legs rho (oil) * g * h = 0.5 rho * v1^2 [(A/a)^2 – 1] v1^2 = 2gh [rho (oil) / rho (water)] / [(A/a)^2 – 1] for volume flow rate, say at middle, V-dot = (a) * (v1) > rho (water) = 1 g/cm^3, rho (oil) = 0.82 g/cm^3, h = 1.4 cm, g = 980 cm/s^2, (A/a) = [pi D^2/4 / pid^2/4] = [D^2/d^2] (A/a)^2 = [D/d]^4 = [1.9/0.5]^4 = 208.51 v1^2 = 2*980*1.4[0.82/1] / [208.51 – 1] v1^2 = 10.84 v1 = 3.293 cm/sec volume flow rate = V-dot = (a)v1 = [pi d^2/4] [3.293] = [3.14*0.5*0.5/4]3.293 V-dot = 0.646 cm^3 /sec

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