Question #142985
Two plates are placed at a distance of 0.15 mm apart. The lower plate is fixed while the upper plate having surface area 1.0 m^2 is pulled at 0.3 m/s. Find the force and power required to maintain this speed, if the fluid separating them is having viscosity 1.5 poise.
1
Expert's answer
2020-11-09T20:28:22-0500

Let force required be FF


F=ν(area)dvdzF=\nu\cdot(area)\cdot\dfrac{dv}{dz}

Putting all the values in SI unit


F=1.5×(0.1Pas)(1.0m2)0.3×103m/s0.15×103m=F=1.5\times(0.1Pa\cdot s)\cdot(1.0m^2)\cdot\dfrac{0.3\times10^{-3}m/s}{0.15\times10^{-3}m}=

=0.3N=0.3N

The force is 0.3N.0.3N.


Let power required to maintain this speed be PP


P=FvP=F\cdot v

P=0.3N0.3×103m/s=9×105W=90μWP=0.3N\cdot0.3\times10^{-3}m/s=9\times10^{-5}W=90\mu W

Power required to maintain this speed is 9×105W=90μW.9\times10^{-5}W=90\mu W.



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