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Fluid mechanics problem?
A partly full water tank admits water at 20°C and 85 N/s
weight flow while ejecting water on the other side at 5500
cm3/s. The air pocket in the tank has a vent at the top and
is at 20°C and 1 atm. If the fluids are approximately incompressible,
how much air in N/h is passing through the
vent? In which direction?
1 Answer
- tiggerLv 71 decade agoFavorite Answer
I think this question is simply about relating the volume of inflow to the volume of outflow. Working in SI units, the rate of volume of inflow is -
Vin = m*rho where m is the mass rate, and rho is the density of water (1000kg/m^3)
Vin = 85/(rho*g) = 8.66*10^-3 m^3/s
Outflow = Vout = 5.5*10^-3 m^3/sec
Inflow exceeds outflow by 3.16m^3/s This is the rate at which air leaves the tank. To find the mass flow rate of air you must multiply this volume flow rate by the density of air. (Google 'density of air' to find this). Note that to get an accurate result you must find a value for the density for air of 100% relative humidity - not dry air)
Finally, to find the weight flow rate you multiply mass flow rate by g (9.81m/s^2)
I am tempted to comment that N/sec is a bizarre way of specifying a flow rate. Force per second seems very odd in this context !!
Rate of change of force with time is a quantity that has some meaning, but it does not seem to me to be the best way to characterise a rate of low of a quantity of material with respect to time; mass or volumetric flow rate would be better. Using N/s in this way suggests a re-emergence of that old problem of the confusion of force with mass, and the tendency to treat them as equivalent quantities, for which of course there is no justification at all !!