Question

A 4 standard type L copper tube conveys propyl alcohol at 151/s. The pipe is laid out horizontally. It is 60 îm long. Calculáte the pressure drop

Answer 1

Answer:

 P₁ - P₂ = 17.18 MPa

Explanation:

Pressure drop along the horizontal pipe = P₁ - P₂

                                                               = $\dfrac{128\mu L Q}{\piD^4}$

length of the pipe  = 60 m

flow rate of propyl alcohol = 15 l/s

                                            = 0.015 m³/s   (∵ 1 l/s  = 10⁻³ m³/s)

for 4 standard pipe,

diameter of the pipe = 8 mm = 0.008 m

viscosity of propyl alcohol  = 0.00192 N-s/m²

$P_1-P_2 =\dfrac{128\mu L Q}{\piD^4}$

                     = $\dfrac{128\times 0.00192 \times 60 \times 0.015}{\pi\times 0.008^4}$

           P₁ - P₂ = 17.18 MPa

hence, pressure drop is equal to 17.18 MPa