for the expression in the box, how can mass flowrate equals the leaking rate? I
ID: 2844147 • Letter: F
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for the expression in the box, how can mass flowrate equals the leaking rate? I thought it would be M* = -.6 sqrt(.098h) x 10^3 as in Mass = density x volume or the change in mass = density x change in volume?
I think i overlooked something in the problem?
Illustration 2.2-4 Another Problem Using the Rate-of-Change Form of the Mass Balance An open cylindrical tank with a base area of 1 m2 and a height of 10 m contains 5 m3 of water. As a result of corrosion, the tank develops a leak at its bottom. The rate at which water leaves the tank through the leak is Leak rate (m3/s)= 0 5 where Delta P is the pressure difference in bar between the fluid at the base of the tank and the atmosphere. (You will learn about the origin of this equation in a course dealing with fluid flow.) Determine the amount of water in the tank at any time. Solution Note that the pressure at the bottom of the tank is equal to the atmosphere pressure plus the hydrostatic pressure due to the water above the leak; that is, P = 1.013 bar + rhogh, where rho is the density of water and h is the height of water above the leak. Therefore, Delta P = (1.013 + rhogh) - 1.013 = rhogh and Delta P = 103 kg/m3 times h m times 9.807 m/s2 times 1 Pams/kg times 10 -5 bar/Pa = 0.09807h bar Since the height of fluid in the tank is changing with time, the flow rate of the leak will change with time. Therefore, to solve the problem, we must use the rate-of-change form of the mass balance. The mass of water in the tank at any time is M(t) = rhoAh(t) = 10 3 kg/m3 middot 1 m2 middot h(t) m = 103h(t) kg The mass balance on the contents of the lank at any lime is dM(t)/dt = 103 dh(t)/dt = M = - 0.5 = - 0.1566 where the negative sign arises because the flow is out of the tank. Integrating this equation between t = 0 and any later time t yields 2 - 2 = 2 - 2 = - 0.1566 times 10-3 t or = - (0.1566/2) times 10-3 t which can be rearranged toExplanation / Answer
take the volume of the tank in the cylinder as the system
since in the tank (no mass as per our system) so change is mass is due to so mass coming - mass leaving (but in our case it's only leaking so, mass is reducing
hence mass flowrate equals the leaking rate
in the expression right above the box you have (change of mass in the bucket at any time)
d(M(t)/dt=rho*d(Ah(t))/dt=rho*A*d(h(t))/dt;
mass flow rate= rho*leaking rate;
thus makes it clear leaking rate is not equal to mass flow rate the book has solved it wrongly
(as can also be seen from the units of the leaking rate its m^3/s but mass flow rate has unit kgm^3/s I guess you understand the difference)
if you still bewildered leave a comment on my answer
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