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)
Length of the horizontal tube, l=1.5m Radius of the tube, r=1cm=0.01m Diameter of the tube, d=2r=0.02m Glycerine is flowing at a rate of 4×10−3 kg/ s. M=4.0×10−3 kg/s Density of glycerine, ρ=1.3×103kg/m3 Viscosity of glycerine, η=0.83 Pas Volume of glycerine flowing per sec: =4×10−3/(1.3×103) =3.08×10−6m3/s According to Poisevilles formula, we have the relation for the rate of flow: V=πpr4/8ηl where p is the pressure difference between the two ends of the tube p=V8ηl/πr4 =3.08×10−6×8×0.83×1.5/[π×0.014] =9.8×102 Pa Reynolds number is given by the relation: R=4ρv/πdη =4×1.3×103×3.08×10−6/(0.02×0.83) =0.3 Reynolds number is about 0.3. Hence, the flow is laminar.
