Question

A fresh milk flows in a pipe at a packaging plant with a mass flow rate that would fill 180 1.00-L milk packages per minute. At point 2 in the pipe, the gauge pressure p_2 is 175 kPa and the cross-sectional area A_2 is 10.00 cm^2. At point 1, 1.85 m above point 2, the cross-sectional area A is 3.50 cm^2. The density of milk rho_milk = 1035 kg/m. a) Find volume flow rate V_rate. b) Find the mass flow rate M_rate. c) Find the flow speed v_1 at point 1. d) Find the flow speed v_2 at point 2. e) Find the gauge pressure p_1 at point 1.

Answer 1

a). Volume flow rate will be

   $V'= 180 \times \frac{1.0 L}{1 min}= \frac{180 \times 10^{-3}m^3}{60s}= 0.003 m^3/s$ (ANS)

b). Density of milk given, $\rho_{milk}= 1035 kg/m^3$ (Units of density is misprinted in question)

   Then Mass flow rate will be

   $M'= \rho_{milk }\times V'= 1035 \times 0.003 = 3.105 kg/s$ (ANS)

c). Area at point 1, A₁= 3.50cm²

If "v₁" is the velocity at point 1, then we can write

   $v_1= \frac{V'}{A_1}= \frac{0.003}{3.50 \times 10^{-2 \times 2}}= 8.57 m/s$ (ANS)

d). Similarly Area at point 2, A₂= 10.0cm²

If "v₂" is the velocity at point 2, then we can write

   $v_2= \frac{V'}{A_2}= \frac{0.003}{10.00 \times 10^{-2 \times 2}}= 3 m/s$ (ANS)

e). Pressure at point 2, P₂= 175 kPa

Let Pressure at point 1 be P₁, then following Bernoulli's Theorem,

   $P_1+\frac{1}{2}\rho_{milk}v_1^2+\rho_{milk}gh_1= P_2+\frac{1}{2}\rho_{milk}v_2^2+\rho_{milk}gh_2$

where h₁ and h₂ are the heights of point 1 and 2 above the ground. Its given

  

   $P_1= P_2+\frac{1}{2}\rho_{milk}(v_2^2-v_1^2)+\rho_{milk}g(h_2-h_1)$

then using given values in above,

$P_1= (175 \times 10^3)+\frac{1}{2}(1035)((3)^2-(8.57)^2)+(1035)(9.8)(-1.85)$

$P_1= (175 \times 10^3)+(517.5)(9.00-73.44)-(1.876 \times 10^4)$

$P_1= (17.5 \times 10^4)-(3.334 \times 10^4)-(1.876 \times 10^4)$

$P_1= (17.5-3.334-1.876 )\times 10^4= 122.9 \times 10^3 Pa= 122.9kPa$ (ANS)