A tube of radius 5 cm is connected to tube of radius 1 cm as shown above. Water

Question

A tube of radius 5 cm is connected to tube of radius 1 cm as shown above. Water is forced through the tube at a rate of 10 liters/min. The pressure in the 5 cm tube is 1105 Pa. The density of water is 1000 kg/m3. Assume that the water is nonviscous and uncompressible.
a) What is the velocity of the water in the 5 cm radius tube?
b)What is the velocity of the water in the 1 cm radius tube?
c) What is the water pressure in the 1 cm radius tube?

Explanation / Answer

sol: suppose that, so we get You have 10 lit/min = (10*10^-3)/60 m³/s = 1.666*10^-4 m³/s (1 litre = 1 dm³ = 10^-3 m³) (1 minute is 60 seconds). In this problems you must use the Bernoulli expression: P/? + V²/2 + g*z = constant. The z position is constant, the same for all pipes so: P/? + V²/2 = constant. V is the velocity in m/s, P is the pressure in Pa (Pascal) and ? is the density 1000 kg/m³. The rate is Q = V*A with A = area. PIPE 5 cm. The area A1 = p*D1²/4 (D1 = diameter in meters) A1 = p*(0.05)²/4 = 0.0019635 m² Q = 1.666*10^-4 m³/s The velocity V1 = Q/A1 = 0.0848 m/s. The energy is (P1 = 10^5 Pa): P1/? + V1²/2 PIPE 1 cm. The area A2 = p*D2²/4 A2 = p*(0.01)²/4 = 7.85398*10^-5 m² Q = 1.666*10^-4 m³/s The velocity V2 = Q/A2 = 2.122 m/s. The energy is: P2/? + V2²/2 the is equal to the pipe of 5 cm. => P2/? + V2²/2 = P1/? + V1²/2 P1 - P2 = ?*(V2²/2 - V1²/2 ) P1 - P2 = 1000*[(2.122)²/2 - (0.0848)²/2] P1 - P2 = 2247.84 Pa P2 = 10^5 - 2247.84 = 97752.16 Pa. Warning! The area is not A1 = p*D1² = 78.54 cm² but A1 = p*R1² and the radius R1 = D1/2 => R1² = D1² /4 A1 = p*R1² = p*D1² /4 = 19.635 cm² = 0.0019635 m². answer

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