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  • Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^–3 kg s^–1 , what is the pressure difference

Q 10.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10-3 kg s-1 , what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

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The volumetric flow rate of glycerine flow would be given by

\\V=\frac{Amount\ of\ glycerine\ flow\ per\ second\ }{Density\ of\ glycerine}\\ V=\frac{M}{\rho }\\ V=\frac{4\times 10^{-3}}{1.3\times 10^{3}}\\ V=3.08\times 10^{-6}ms^{-1}

The viscosity of glycerine is \eta =0.83\ Pa\ s

Assuming Laminar flow for a tube of radius r, length l, having pressure difference P across its ends a fluid with viscosity \eta would flow through it with a volumetric rate of

\\V=\frac{\pi Pr^{4}}{8\eta l}\\ P=\frac{8V\eta l}{\pi r^{4}}\\ P=\frac{8\times 3.08\times 10^{-6}\times 0.83}{\pi \times (10^{-2})^{4}}\\ P=9.75\times 10^{2}\ Pa

Reynolds number is given by

\\R=\frac{4\rho V}{\pi d\eta }\\ R=\frac{2\rho V}{\pi r\eta }\\ R=\frac{2\times 1.3\times 10^{3}\times 3.08\times 10^{-6}}{\pi \times 0.01\times 0.83}\\ R\approx 0.3

Since Reynolds Number is coming out to be 0.3 our Assumption of laminar flow was correct.

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