$A v i s c o u s l i q u i d ρ = 1.2 \times 10^{3} k g / m^{3}, μ = 0.0048 P a . s f l o w s a t a r a t e o f 10 m l / s t h r o u g h a h o r i z o n t a l, 4 m m d i a m e t e r t u b e . T h e p r e s s u r e d r o p a l o n g a 1 m l e n g t h o f t h e t u b e w h i c h i s f a r f r o m t h e t u b e e n t r a n c e s o t h a t t h e f l o w i s f u l l y d e v e l o p e d i s$ _______ $k P a$

7.64

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Solution

The correct option is A 7.64
$R e = \frac{ρ V D}{μ} = ρ (\frac{4 Q}{π}) \times \frac{D}{μ} = \frac{4 ρ Q}{π μ D}$

$= \frac{4 \times 1.2 \times 10^{3} \times 10 \times 10^{- 3} \times 10^{- 3}}{π \times 0.0048 \times 4 \times 10^{- 3}} \approx 796$

$T h u s, f l o w i s l a m i n a r .$

$Δ p = \frac{8 μ L Q}{π R^{4}} = \frac{8 \times 0.0048 \times 1 \times 10 \times 10^{- 3} \times 10^{- 3}}{π \times 2^{4} \times 10^{- 12}} = 7.64 k P a$

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A viscous liquid ρ=1.2×103kg/m3,μ=0.0048Pa.s flows at a rate of 10ml/s through a horizontal, 4mm diameter tube. The pressure drop along a 1 m length of the tube which is far from the tube entrance so that the flow is fully developed is_______kPa 7.64

The correct option is A 7.64Re=ρVDμ=ρ(4Qπ)×Dμ=4ρQπμD =4×1.2×103×10×10−3×10−3π×0.0048×4×10−3≈796 Thus, flow is laminar. Δp=8μLQπR4=8×0.0048×1×10×10−3×10−3π×24×10−12=7.64kPa