Question

A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air compressor to the drill has a 40 mm diameter and is smooth. The maximum compressor discharge gauge pressure is 690 kPa. Neglect changes in air density and any effects of hose curvature. Air leaves the compressor at 40° C. What is the longest hose that can be used?

Answer 1

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m  = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

$[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m$

where$h_l = \frac{ flv^2}{2D}$

$h_m minor loss$

density is constant

$v_1 = v_2$

head is same so,$z_1 = z_2$

curvature is constant so$\alpha = constant$

neglecting minor losses

$\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}$

we know$\dot m$ is given as$= \rho VA$

$\rho =\frac{P_1}{RT_1}$

$\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3$

therefore

$v = \frac{\dot m}{\rho A}$

$V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}$

V = 25.90 m/s

$Re = \frac{\rho VD}{\mu}$

for T = 40 Degree, $\mu = 1.91*10^{-5}$

$Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}$

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

$\frac{P_1}{\rho} -\frac{P_2}{\rho} = \frac{ flv^2}{2D}$

$L = \frac{(P_1-P_2) 2D}{\rho f v^2}$

$L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}$

L = 46.35 m