10.13 Glycerin flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerin 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 glycerin = 1.3 × 10³ kg m^–3 and viscosity of glycerin = 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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Vishal Baghel | Contributor-Level 10

5 months ago

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02m

Glycerine mass flow rate, M = 4 $\times 10^{- 3}$ kg/s

Density of glycerine, $ρ$ = 1.3 $\times 10^{3}$ kg

Viscosity of glycerine, $η$ = 0.83 Pa-s

Now, volume of glycerine flowing per sec V = $\frac{M}{ρ}$ = $\frac{4 \times 10^{- 3}}{1.3 \times 10^{3}}$ $m^{3}$ /s = 3.08 $\times 10^{- 6}$ $m^{3}$ /s

According to Poiseville’s formula, we know the flow rate

V = $\frac{π \times p \times r^{4}}{8 \times η \times l}$ where p is the pressure difference between two ends of the tube

p = $\frac{V \times 8 \times η \times l}{π \times r^{4}}$ = $\frac{3.08 \times 10^{- 6} \times 8 \times 0.83 \times 1.5}{π \times {(0.01)}^{4}}$ = 976.47 Pa

Reynolds’s number is given by the relation

Re = $\frac{4 ρ V}{π d η}$ = $\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{- 6}}{3.1416 \times 0.02 \times 0.83}$ = 0.3

Since the Reynolds&r

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Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02m

Glycerine mass flow rate, M = 4 $\times 10^{- 3}$ kg/s

Density of glycerine, $ρ$ = 1.3 $\times 10^{3}$ kg

Viscosity of glycerine, $η$ = 0.83 Pa-s

Now, volume of glycerine flowing per sec V = $\frac{M}{ρ}$ = $\frac{4 \times 10^{- 3}}{1.3 \times 10^{3}}$ $m^{3}$ /s = 3.08 $\times 10^{- 6}$ $m^{3}$ /s

According to Poiseville’s formula, we know the flow rate

V = $\frac{π \times p \times r^{4}}{8 \times η \times l}$ where p is the pressure difference between two ends of the tube

p = $\frac{V \times 8 \times η \times l}{π \times r^{4}}$ = $\frac{3.08 \times 10^{- 6} \times 8 \times 0.83 \times 1.5}{π \times {(0.01)}^{4}}$ = 976.47 Pa

Reynolds’s number is given by the relation

Re = $\frac{4 ρ V}{π d η}$ = $\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{- 6}}{3.1416 \times 0.02 \times 0.83}$ = 0.3

Since the Reynolds’s number is 0.3, the flow is laminar

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