Question

win to the Problem 2: Selected: 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 p2 is 175 kPa and the cross- sectional area A2 is 10.00 cmr. At point 1, 1.85 m above point 2, the cross-sectional area Al is 3.50 cm The density of milk pr 1035 kg/m. milk a) Find volume flow rate Vr b) the flow rate M, c) Find the flow speed vu at point 1. d) Find the flow speed vn at point 2. e) Find the gauge pressure ph at point 1.


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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'= A_1v_1

  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'= A_2v_2

  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

  h_1-h_2= 1.85m

  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)

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