A fish tank has dimensions 46 cm wide by 1.0 m long by 0.70 m high. If the filter should process all the water in the tank once every 3.0 h, what should the flow speed be in the 3.0-cm-diameter input tube for the filter?
I tried using Q=v/t and Q=vA...
V=0.46m*1m*0.7m=0.322m^3
t=3hours=10800 seconds
A=3.14*0.03^2=.0028
Q=v/t=0.322/10800=2.96*10^-5/0.0028
Q=vA -> v=Q/A=2.96*10^-5/0.0028= 0.010479
so, v=~0.0105 m^3/s right? but it keeps coming up as wrong...what am i doing wrong??
A=3.14*0.03^2=.0028
!!!!!!!!!!!!!!!!!!!
.03 is DIAMETER
NOT Radius
you want A = pi D^2/4
otherwise you have the right idea
try 4 times that :)
3.3
I need based on necessary steps
You have correctly calculated the volume of the fish tank, the area of the input tube, and the time in seconds. However, there seems to be an error in your calculation for the flow speed (v).
To find the flow speed, you should divide the volume (Q) by the cross-sectional area (A) of the input tube, not the other way around. So the correct calculation should be:
v = Q / A = (0.322 m^3) / (0.0028 m^2) = 115 m^3/s
Therefore, the flow speed should be approximately 115 m^3/s.
2.96*10^-5 /(.0028/4)
= .0423 m^3/s