This question is incomplete, the complete question as well as the missing diagram is uploaded below;
Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.
Take [tex]p_o[/tex] = 880 kg/m³ and [tex]p_{glycerol[/tex] = 1260 kg/m³
Answer:
the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
Explanation:
Given that;
Inlet velocity of Glycerin, [tex]V_A[/tex] = 6 m/s
Inlet velocity of oil, [tex]V_B[/tex] = 3 m/s
Density velocity of glycerin, [tex]p_{glycerol[/tex] = 1260 kg/m³
Density velocity of glycerin, Take [tex]p_o[/tex] = 880 kg/m³
Volume of tank V = 4 m
from the diagram;
Diameter of glycerin pipe, [tex]d_A[/tex] = 100 mm = 0.1 m
Diameter of oil pipe, [tex]d_B[/tex] = 80 mm = 0.08 m
Diameter of outlet pipe [tex]d_C[/tex] = 120 mm = 0.12 m
Now, Appling the discharge flow equation;
[tex]Q_A + Q_B = Q_C[/tex]
[tex]A_Av_A + A_Bv_B = A_Cv_C[/tex]
π/4 × ([tex]d_A[/tex])²[tex]v_A[/tex] + π/4 × ([tex]d_B[/tex] )²[tex]v_B[/tex] = π/4 × ([tex]d_C[/tex])²[tex]v_C[/tex]
we substitute
π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²[tex]v_C[/tex]
0.04712 + 0.0150796 = 0.0113097[tex]v_C[/tex]
0.0621996 = 0.0113097[tex]v_C[/tex]
[tex]v_C[/tex] = 0.0621996 / 0.0113097
[tex]v_C[/tex] = 5.5 m/s
Now we apply the mass flow rate condition
[tex]m_A + m_B = m_C[/tex]
[tex]p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C[/tex]
so we substitute
1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5
1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335
59.3712 + 13.27 = 0.06220335p
72.6412 = 0.06220335p
p = 72.6412 / 0.06220335
p = 1167.8 kg/m³
Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
