Liquid Methylamine solution ( ? = 0.897 g/mL) flows into and out of the rectangu

Question

Liquid Methylamine solution ( ? = 0.897 g/mL) flows into and out of the rectangular tank from circular pipes 1, 2, 3 and 4 shown in figure 5 . The base area of the tank is 2 m 2 . Methylamine enters the tank at sections 1 and 2 and leaves the tank at sections 3 and 4. The height of Methylamine in the tank increases at a rate of dh/dt = 3.5 cm/s

Liquid Methylamine solution ( = 0.897 g/mL) flows into and out of the rectangular tank from circular pipes 1, 2, 3 and 4 shown in figure. The base area of the tank is 2 m2. Methylamine enters the tank at sections 1 and 2 and leaves the tank at sections 3 and 4. The height of Methylamine in the tank increases at a rate of dh/dt = 3.5 cm/s. (a) Determine the mass flow rates at sections 1 and 2, m1 and m2 (b) Determine the mass flow rate at section 4, m4. (c) Determine the volumetric flow rate at section 4, Q4 (d) Determine the velocity at section 4, V4. (e) If dh/dt-0, determine the volumetric flow rate at section 4, Qa V,-1.5 m/s Di-40 cm 2m3-63 kg/s Q2-0.01 mls (2 m's D4-25 cm Rectangular Tank. Not to Scale

Explanation / Answer

a) dm1/dt = (D12/4)v1 = 897 * ( * 0.42 / 4) * 1.5 = 169.1 kg/s

dm2/dt = Q2 = 0.01 * 897 = 9.0 kg/s

b) dm1/dt + dm2/dt - dm3/dt - dm4/dt = 2(dh/dt)

=> 169.1 + 9.0 - 67 - dm4/dt = 2 * 0.035 * 897

=> dm4/dt = 48.3 kg/s

c) Q4 = (dm4/dt)/ = 48.3/897 = 0.054 m3/s

d) v4 = Q4/(D42/4) = 0.054 / ( * 0.252 / 4) = 1.1 m/s

e) dm1/dt + dm2/dt - dm3/dt - dm4/dt = 0

=> 169.1 + 9.0 - 67 - dm4/dt = 0

=> dm4/dt = 111.1 kg/s

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