A cylinder of 0.040-ft radius rotates concentrically inside a fixed cylinder of 0.42-ft radius. Both cylinders are 1.00 ft long. Determine the viscosity of the liquid that fills the space between the cylinders if a torque of 0.650 lb. ft is required to maintain an angular velocity of 60 rpm.
Text Solution
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Sol. The torque is transmitted through the field layers to the outer cylinder. Since the gap between the cylinders is small, the calculations may be made without integration. The tangential velocity v, of the inner cylinder = r ω , where r = 0.40 ft and ω = 2 π rad/s. Hence, v t = (0.40) (2 π ) = 2.51 ft/s. For the small space between cylinders, the velocity gradient may be assumed to be a straight line and the mean radius can be used. Then,
dv/dy = (2.51 – 0)/ (0.42 – 0.40) = 125.5 s –1 . Since applied torque equals resisting torque, applied torque = ( τ ) (area)(arm), 0.650 = τ [(1.00) (2 π ) (0.40 + 0.42)/2] [(0.40 + 0.42)/2], τ = 0.615 lb/ft 2 = μ (dv/dy), 0.615 = ( μ ) (125.5), μ = 0.00490 lb.s/ft 2 .
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