Asked by kelsi
You need to siphon water from a clogged sink. The sink has an area of 0.36 and is filled to a height of 4.0 . Your siphon tube rises 45 above the bottom of the sink and then descends 85 to a pail as shown in the figure. The siphon tube has a diameter of 2.4
A)Assuming that the water level in the sink has almost zero velocity, estimate the water velocity when it enters the pail.
B)Estimate how long it will take to empty the sink
A)Assuming that the water level in the sink has almost zero velocity, estimate the water velocity when it enters the pail.
B)Estimate how long it will take to empty the sink
Answers
Answered by
kelsi
sorry, the units didn't show up. it's .36 m^2, 4.0cm, 45cm, 85cm, 2.4cm
Answered by
Damon
water .04 m deep.
volume = .04*.36 = .0144 m^3
bottom of sink to pail height difference is .85-.45 = .4 m
We will have to assume no friction in the pipe or entrance or exit losses unless you know those . Thefore use bernoulli
p + rho g * change in h + (1/2) rho v ^2 = constant
p is air pressure, v at start = 0 so
rho (9.81) (.4) = .5 rho v^2
v^2 = 7.848
v = 2.8 m/s
then
radius = .024/2 = .012 m
pi r^2 = .000452 m^2
so
volume flow rate = 2.8*.000452 = .001267 m^3/s
volume of water = .0144 so
time = .0144/.001267 = 11.36 seconds
volume = .04*.36 = .0144 m^3
bottom of sink to pail height difference is .85-.45 = .4 m
We will have to assume no friction in the pipe or entrance or exit losses unless you know those . Thefore use bernoulli
p + rho g * change in h + (1/2) rho v ^2 = constant
p is air pressure, v at start = 0 so
rho (9.81) (.4) = .5 rho v^2
v^2 = 7.848
v = 2.8 m/s
then
radius = .024/2 = .012 m
pi r^2 = .000452 m^2
so
volume flow rate = 2.8*.000452 = .001267 m^3/s
volume of water = .0144 so
time = .0144/.001267 = 11.36 seconds
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