Answer:
The exact time when the sample was taken is = 0.4167337 hr
Explanation:
The diagram of a sketch of the tank is shown on the first uploaded image
Let A denote the first inlet
Let B denote the second inlet
Let C denote the single outflow from the tank
From the question we are given that the diameter of A is = 1 cm = 0.01 m
Area of A is = [tex]\frac{\pi}{4}(0.01)^{2} m^{2}[/tex]
= [tex]7.85 *10^{-5}m^{2}[/tex]
Velocity of liquid through A = 0.2 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = [tex]\frac{Volume of Inlet }{time}[/tex] = [tex]Velocity * Area[/tex] i.e is [tex]m^{2} * \frac{m}{s} = \frac{m^{3}}{s}[/tex]
= [tex]0.2 *7.85*10^{-5} \frac{m^{3}}{s}[/tex]
The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
= [tex]1039.8 * 0.2 * 7.85 *10^{-5} Kg/s[/tex]
= 0.016324 [tex]\frac{Kg}{s}[/tex]
From the question the diameter of B = 2 cm = 0.02 m
Area of B = [tex]\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}[/tex]
Velocity of liquid through B = 0.01 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = [tex]\frac{Volume of Inlet }{time}[/tex] = [tex]Velocity * Area[/tex] i.e is [tex]m^{2} * \frac{m}{s} = \frac{m^{3}}{s}[/tex]
= [tex]3.14*10^{-4} *0.01 \frac{m^{3}}{s}[/tex]
The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
= [tex]1053 * 3.14*10^{-6} \frac{Kg}{s}[/tex]
= 0.00330642 [tex]\frac{Kg}{s}[/tex]
From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s
And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L
So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
= [tex]13\frac{g}{L} * 0.5 \frac{L}{s}[/tex]
= [tex]\frac{6.5}{1000}\frac{Kg}{s}[/tex] Note (1 Kg = 1000 g)
= 0.0065 kg/s
Considering potassium chloride
Let denote the rate at which liquid flows in terms of mass as as [tex]\frac{dm}{dt}[/tex] i.e change in mass with respect to time hence
Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )
(0.016324 + 0.00330642) - 0.0065 = [tex]\frac{dm}{dt}[/tex]
[tex]\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx[/tex]
=> 0.01313122 t = (m - [tex]m_{o}[/tex])
From the question (m - [tex]m_{o}[/tex]) is given as = 19.7 Kg
Hence the time when the sample was taken is given as
0.01313122 t = 19.7 Kg
=> t = 1500.2414 sec
t = .4167337 hours (1 hour = 3600 seconds)
