Answer:
a) Volume rate of flow = 0.00056 m³/s
b) Shearing stress acting at the bottom = 80 N/m²
c) Velocity along the center line of the channel = 0.211 m/s
Explanation:
Pressure drop per unit length, [tex]\frac{dP}{dl} = 40 kPa/m[/tex]
dp = 40 kPa
The distance between the plates, l = 4 mm = 0.004 m
Distance between the horizontal plate and the center, h = 0.004/2 = 0.002 m
[tex]\mu = 0.38 Ns/m^{2}[/tex]
a) The volume rate of flow, [tex]v = \frac{2h^{3} }{3 \mu}* \frac{\delta p}{\delta l}[/tex]
[tex]v = \frac{2* 0.002^{3} * 40 * 10^{3} }{3*0.38}[/tex]
[tex]v = 0.00056 m^{3} /s[/tex]
b) The magnitude of the shearing stress acting at the bottom
Shear stress, [tex]\tau = h dP[/tex]
[tex]\tau = 0.002 * 40000\\\tau = 80 N/m^{2}[/tex]
c) The velocity along the center line of the channel
velocity, v = [tex]\frac{h^{2} }{2 \mu} * \frac{\delta p}{\delta l}[/tex]
v = [tex]\frac{0.002^{2} }{2 * 0.38} * 40000[/tex]
v = 0.211 m/s
