Question

A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 8.0 m/s.
A:What is the speed in the second segment? ______m/s
B:What is the speed in the third segment? _______m/s
C:What is the volume flow rate through the pipe? _______ L/s

Answer 1

Answer

given,

diameter of segment 1, d₁ = 1 cm

diameter of segment 2, d₂ = 2 cm

diameter of segment 3 , d₃ = 0.5 cm

speed of liquid in first segment, v₁ = 8 m/s

a) using continuity equation

      A₁ v₁ = A₂ v₂

 $\dfrac{\pi}{4}d_1^2 \times v_1 = \dfrac{\pi}{4}d_2^2 \times v_2$

 $d_1^2 \times v_1 = d_2^2 \times v_2$

 $1^2 \times 8= 2^2 \times v_2$

      v₂ = 2 m/s

speed in the second segment = 2 m/s

b) again, using continuity equation

      A₂ v₂ = A₃ v₃

 $\dfrac{\pi}{4}d_2^2 \times v_2 = \dfrac{\pi}{4}d_3^2 \times v_3$

 $d_2^2 \times v_2 = d_3^2 \times v_3$

 $2^2 \times 2= 0.5^2 \times v_2$

      v₂ = 32 m/s

speed in the second segment = 32 m/s

c) Flow rate in the pipe

    Q = A₁ V₁

   $Q = \dfrac{\pi}{4}d_1^2 \times v_1$

   $Q = \dfrac{\pi}{4}\times 0.01^2 \times 8$

      Q = 6.28 x 10⁻⁴ m³/s

       Q = 62.8 L/s

discharge in the pipe is equal to 62.8 L/s

Answer 2

(a) The speed of the liquid in the second segment is 2 m/s.

(b) The speed of the liquid in the third segment is 32 m/s.

(c) The volume flow rate of the liquid through the pipe 0.6284 L/s.

The given parameters;

• diameter of the first segment, d₁ = 1 cm = 0.01 m
• diameter of the second segment, d₂ = 2 cm = 0.02 m
• diameter of the third segment, d₃ = 0.5 cm = 0.005 m
• speed of the liquid in the first segment, v₁ = 8 m/s

Apply continuity equation to determine the speed of the liquid at each segment.

$A_1v_1 = A_2 v_2 = A_3v_3\\\\\pi \frac{d_1^2 }{4} v_1 = \pi \frac{d_2^2 }{4} v_2 = \pi \frac{d_3^2 }{4} v_3\\\\d_1^2 v_1 = d_2 ^2 v_2 = d_3^2 v_3\\\\v_2 = \frac{d_1^2 v_1}{d_2^2} \\\\v_2 = \frac{(0.01)^2 \times (8)}{(0.02)^2} \\\\v_2 = 2 \ m/s\\\\ d_2 ^2 v_2 = d_3^2 v_3\\\\v_3 = \frac{d_2^2 v_2}{d_3^2}\\\\v_ 3 = \frac{(0.02)^2 \times (2)}{(0.005)^2}\\\\v_ 3= 32 \ m/s$

The volume flow rate through the pipe is calculated as follows;

$Q = Av\\\\Q = \frac{\pi d_1^2}{4} v_1\\\\Q = \frac{\pi \times (0.01)^2}{4} \times 8\\\\Q = 0.0006284 \ m^3/s\\\\Q = 0.0006284 \times 1000 = 0.6284 \ L/s$

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