Parts a and b please (from expert TA) (4%) Problem 17: A nozzle with a radius of

Question

Parts a and b please (from expert TA)

(4%) Problem 17: A nozzle with a radius of 0.25 cm is attached to a garden hose with a radius of 0.88 cm. The flow rate through hose and nozzle is 0.35 L/s Randomized Variables n0.25 cm rh0.88 cm Q=0.35 L/s 50% Part (a) Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m. Grade Summary Deductions Potential 0% 100% Submissions cosO cotan) asin) sin Attempts remaining: 7 acos) sinhO cotanhO % per attempt) detailed view cos 0 END Degrees -Radians Submit Hint I give up! Hints: 0% deduction per hint. Hi ints remaining: 4 Feedback: 0%-deduction per feedback. maximum height ul uirted wit Wi zzle remov

Explanation / Answer

Given

radius of nozzle is r1 = 0.25 cm = 0.0025 m,

radius of hose is r2 = 0.88 cm = 0.0088 m,

rate of flow is Q = 0.35 L/s

Q = 0.35 L/s * m^3/1000s = 0.35*10^-3 m^3/s

we know that the rate of flow is Q = A*v

given Q = A1*v1 = A2*v2  

where A1 = pi*r1^2 , A2 = pi*r2^2

from Bernouli's equation

P1+0.5*rho*v1^2+ rho*g*h1 = P2+0.5*rho*v2^2+ rho*g*h2

here when the water reaches the maximum height the velocity become zero that is v2 = 0 m/s , and h1=0 m , and the pressures P1= P2 ==> difference is P2-P1 = dP = 0 Pa

h1 = v1^2/2g

to calculate v

from Q = A*v ==> v1 = Q/A = 0.35*10^-3 / (pi*0.0025^2) m/s = 17.83 m/s

for hose

Q = A2*v ==> v1 = Q/A2 = (0.35*10^-3 / (pi*0.0088^2) m/s = 1.438 m/s

a)

now h2 = v1^2/2g = 17.83^2/(2*9.8) m = 16.22 m

the maximum height to which water could squirted with the hose if it emerges from the nozzle is 17.83 m/s

b)

and for the hose if the nozzle is removed then  

from

h2 = v1^2/2g = 1.438^2/(2*9.8) m = 0.10550 m = 10.55 cm

the maximum height to which water could squirted with the hose if it emerges from the nozzle removed is 10.55 cm

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