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speed of water coming out of hose?
Water flows at a rate of 0.25 L/s through a garden hose of inside diameter 2.0 cm. The nozzle of the hose is a circular opening diameter 0.5 cm. What is the speed of the water when it emerges?
0.25 L/s = 0.00025 m^3/s
r_hose = 0.01 m
r_nozzle = 0.0025 m
(p1)(A1)(v1) = (p2)(A2)(v2)
p1 and p2 cancel because density of water is going to be the same.
v2 = (A1)(v1) / (A2)
v2 = (π0.01^2)(0.00025) / (π0.0025^2)
v2 = 0.004 m^3/s ----> converting to L/s gives 4 L/s
Now what do I do to make this into m/s? I suppose I should be dividing it by an area, but I have no clue what it could be.
4 Answers
- Cultural ParadoxLv 61 decade agoFavorite Answer
Think of the water that travels past a point as having the shape of a cylinder...
its volume would be Pi r*2h...
0.25 L = 250 cc...
250 cm*3 = 3.14(1*2)(h)
h = 80 cm
that amount of water passes by each second so the the velocity would be 0.8 m/s...
The water coming out of the nozzle would be...
v2 = (0.01*2)(0.8)/(0.0025*2) ....Pi's cancel
velocity = 12.8 m/s
Hope that helps...
- JacyLv 71 decade ago
flow rate: Q = vA, has units of [m^3/s]
note: 1L = (10^-3) cubic meters = (10^-3) m^3, so that:
0.25 L/s = 0.25(10^-3) m^3/s and this ='s vA
Now using the continuity eqn.: Q = v1A1 = V2A2
and exit velocity: V2 = (A1/A2) V1
= [(0.02/0.005)^2]V1
= (1.6)V1
But Q = V2A2 = 0.25(10^-3) and simply,
V2 = Q/A2 = 0.25(10^-3) / (pi)(0.005/2)^2
V2 = 12.7 m/s
Note: Your mistake was in confusing flow rate Q with
fluid velocity v. They are not the same. Q has units
of cubic meters/sec, while v has units of meters/sec.
- 1 decade ago
the vel is in terms of volume flow rate. so it seems inappropriate to use the formula as .25 l/s represents the vol of water flowing through an area of pipe with dia 2 cm in 1 sec. so in actuality you have been given A1 v1 = .25 l/s. substitue and find the correct ans directly otherwise calc v1 and then find v2.
