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Unformatted text preview: 41 Equations
Set 8: 41 The mass flow rate at the inlet of a oneinlet, oneexit control volume varies with time according to: min =100*(1exp(2 [1/hr]*t)) where min is in units of kg/hr and t is in hr. At the exit, the mass flow rate is constant at 100 kg/hr. The initial mass in the control volume is 50 kg. a) Plot the inlet and exit mass flow rates, the instantaneous rate of change of mass and the amount of mass contained in the control volume as functions of time for t ranging from 0 to 3 hr. b) Estimate the time in hrs, at which tank is nearly empty mexit = 100 [kg/h] min = 100 [kg/h] (1  exp (2 [1/hr] t)) mo = 50 [kg] dm/dt = min  mexit $ifnot ParametricTable
4 (1) (2) (3) (4) m = mo +
0 dm/dt dt use EES integration function (5) $integraltable t:.2 m_dot_in, m_dot_exit, m $elseif m = 100 [kg/hr] $endif $TabStops 0.2 3.5 in exp (2 [1/hr] t) 2 [1/hr] analytical integration (6) Solution
dm/dt = 0.03355 [kg/h] t = 4 [hr] m = 0.01613 [kg] mexit = 100 [kg/h] min = 99.97 [kg/h] mo = 50 [kg] Integral Table
Row 1 2 t [hr] 0 0.2 min [kg/h] 0 32.96 mexit [kg/h] 100 100 m [kg] 50 33.52 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 55.06 69.88 79.81 86.47 90.93 93.92 95.92 97.27 98.17 98.77 99.18 99.45 99.63 99.75 99.83 99.89 99.93 99.95 99.97 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 22.47 15.06 10.1 6.767 4.536 3.04 2.038 1.366 0.9152 0.6133 0.4109 0.2752 0.1843 0.1233 0.08244 0.05505 0.03669 0.02438 0.01613 Plot 1 ...
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 Spring '07
 MARTIN
 Derivative, Mass flow rate, Hour

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