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Question:

a molten plastic flows out of a tube 8 cm long at a rate of 13 cm^3/min?

a molten plastic flows out of a tube 8 cm long at a rate of 13 cm^3/min where the pressure differential between the two ends of the tube is 18 cm of mercury. find the viscosity of the plastic. the i.d of the tube is 1.30 mm. the density of mercury is 13.6 g/ cm^3the answer is 0.097 kg/m.s or 97 cp! help!

Answer:

Hello Cedric Mapula, let us use the expression V/t = pi P a^4 / 8 l eta Here V/t = volume flown in one sec = 13 cm^3 / 60 s = 13/60 cm^3 /s pi = 3.14 P = 18*13.6*980 a = 1.3/ 20 cm = 0.065 cm l = 8 cm eta = ? Plug these and find the required eta i.e. coefficient of viscosity of liquid Ans : 0.969 g/cm*s = 0.969*10^-3/10^-2 kg/m*s = 0.0969 kg/m*s

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