Problem 2: A nozzle with a radius of 0.35 cm is attached to a garden hose with a

Question

Problem 2: A nozzle with a radius of 0.35 cm is attached to a garden hose with a radius of 0.85 cm The flow rate through hose and nozzle is 25 L/s. Randomized Variables r" = 0.35 cm h 0.85 cm Q = 0.25 L/s Part (a) Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m. Numeric A numeric value is expected and not an expression. Part (b) Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the smumeric value is expecied and not Numerie :A numeric value is expected and not an expression Problem 3: Suppose the horizontal velocity of the wind against a sail is 3. m/s paralel to its front surface and 3,2 m's along its back surface

Explanation / Answer

First, we solve Q = Av for v1 and note that the cross-sectional area is A = r2, yielding

                   v1 = Q/A1 = Q/r12

Substituting known values and making appropriate unit conversions yields

                      v1 = 0.25*10^-3/(*8.5*8.5*10^-6) = 1.1 m/s (velocity of the water in the hose)

We could repeat this calculation to nd the speed in the nozzle v2, but we will use the equation of continuity to give a somewhat dierent insight. Using the equation which states

                         A1v1 = A2v2,

solving for v2 and substituting r2 for the cross-sectional area yields

     v2 = A1*v1/A2 = r12*v1/r22 = (0.85/0.35)^2*1.1 = 6.5 m/s (velocity of the water in the nozzel)

Now a) h1 = v2²/(2g) = 6.5²/(2*9.8) = 2.16 m

b) h2 = v1²/(2g) = 0.062 m

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