Conservation of Mass - Clarkson University

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π πA ( ) 21= D1 = × 2.00 ft = 3.14 ft4 42 2A π π( ) 22= D2 = × 2.50 ft = 4.91 ft4 42 2Let us find the mass flow rate.⎛ lb ⎞ ⎛ ft ⎞2 lbm2 = m = m1= ρ1V 1A1 = 2.31⎜ 33.7 3.143 ⎟× ⎜ ⎟× ( ft ) = 244⎝ ft ⎠ ⎝ s ⎠sNow, we can calculate the velocity V2fromV( lb s)m244 /= = =22ρ2A21.30 /34.912( lb ft ) × ( ft )38.2ftsExample 2 – Unsteady Mass BalanceDia D=2 m1V1 = 1.5 m/sD = cm15Free surfacehV2 = 1.0 m/sD = cm282Is the liquid level in the above tank rising or falling? How fast?To solve this problem, we must use the unsteady version of the conservation of mass equationapplied to a control volume that is shown in the sketch using a dashed line. The control volumeoccupies the entire interior of the tank, including the inlet and exit pipes. There is an entrance tothe control volume at location 1 and an exit at location 2. Liquid flows into the tank at location 1and flows out at location 2, as shown in the sketch.8
We begin withdMdtCS( )= − ∫ ρ V • nwhere M ( t ) is the mass of the liquid in the control volume, which depends on time t . Thedensity of the liquid ρ can be assumed to be constant.We can write the mass content in the control volume as the sum of the mass of the liquid in the2tank, given by ( π /4) ρ Dht ( ) , and the constant mass in the inlet and outlet pipe sections thatare contained the control volume, termed C . The integral on the right side of the unsteady massconservation equation works out to ρ ( Q1 − Q2). Using this information, the unsteady massconservation equation becomesdAd ⎛π⎞ π dh⎜ ρ ⎟ ρ ρdt ⎝ 4 ⎠ 4 dt( )2 2Dh+ C = D = Q1 −Q2ordhdt=( − )4 Q Q1 22π DTherefore, we must calculate the numerical value of the right side in the above equation. Theareas of the inlet and outlet can be calculated as follows.π π( ) 2−A1 = D1 = × 0.05 m = 1.96 × 104 4mπ π( ) 2−A2 = D2 = × 0.08 m = 5.03 × 104 4m2 3 22 3 2Therefore, the volumetric flow rates are⎛m⎞−Q1 = VA1 1= 1.5 ⎜ ⎟× 1.96 × 10 ( m ) = 2.94 × 10⎝ s ⎠3 2 −3⎛m⎞Q2 = VA2 2= 1.0 ⎜ ⎟× 5.03× 10 ( m)= 5.03×10⎝ s ⎠−3 2 −3Substitute the above information in the differential equation for the height of the liquid in thetank.−−( Q − Q ) ( × − × )3 3 3dh 44 2.94 10 5.03 10 m / s1 2 −4m= = = − 6.66 × 1022dt π D πs( 2.0 m)ms3ms39
Page 2: estimate the rate of change of the
Page 5 and 6: The mass flow rate m = ρ Q so that
Page 10: Because the rate of change of the h

Conservation of Mass - Clarkson University

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