Soru
A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm , and it reduces to 0.8 cm at the nozzle exit (Fig. 5-1 O).If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose , and (b) the average velocity of water at the nozzle exit.
Çözüm
4.5
(88 Oylar)
Gürsel
Usta · 5 yıl öğretmeni
Uzman doğrulaması
Cevap
(a) To determine the volume flow rate of water through the hose, we need to calculate the cross-sectional area of the hose and the nozzle exit, and then multiply it by the velocity of water.The cross-sectional area of the hose can be calculated using the formula for the area of a circle: A = πr^2, where r is the radius of the hose. The radius of the hose is half of the diameter, so r = 2 cm / 2 = 1 cm = 0.01 m. Plugging this value into the formula, we get:A_hose = π(0.01 m)^2 = 3.14 x 10^-4 m^2Similarly, the cross-sectional area of the nozzle exit can be calculated as:A_nozzle = π(0.8 cm / 2)^2 = 5.03 x 10^-4 m^2The velocity of water can be calculated using the formula: v = Q / A, where Q is the volume flow rate and A is the cross-sectional area. Since the hose is attached to a nozzle, we can assume that the volume flow rate is constant throughout the hose. Therefore, we can calculate the velocity of water at the hose inlet as:v_hose = Q / A_hoseAnd the velocity of water at the nozzle exit as:v_nozzle = Q / A_nozzle(b) The average velocity of water at the nozzle exit can be calculated using the formula: v_avg = Q / A_nozzle. Since we already know the value of Q from part (a), we can plug it into the formula to get:v_avg = Q / A_nozzle = (3.14 x 10^-4 m^3/s) / (5.03 x 10^-4 m^2) = 6.27 m/sTherefore, the average velocity of water at the nozzle exit is 6.27 m/s.