$T h e f l o w i n a c h a n n e l i s l a m i n a r a n d R e y n o l d^{'} s n u m b e r i s g i v e n b y 1200 . W h a t i s t h e f r i c t i o n f a c t o r f o r t h e c h a n n e l ?$

0.001

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0.008

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0.023

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0.053

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Solution

The correct option is D $0.053$
$D a t a g i v e n :$

$F l o w i n t h e c h a n n e l i s l a m i n a r .$
$R e y n o l d^{'} s n u m b e r, R_{e} = 1200$
$F o r l a m i n a r f l o w f r i c t i o n f a c t o r i s g i v e n b y f = \frac{64}{R_{e}} = \frac{64}{1200} = 0.053$

$N o t e$ :

$F l o w t h r o u g h p i p e :$

$F o r l a m i n a r p i p e f l o w, R_{e} < 2000$

$F o r t u r b u l e n t p i p e f l o w, R_{e} > 4000$

$F o r t r a n s i t i o n p i p e f l o w, 2000 < R_{e} < 4000$

$F r i c t i o n f a c t o r (f) = 4 \times F r i c t i o n o f c o e f f i c i e n t (f^{'})$

$F r i c t i o n f a c t o r a n d R e y n o l d^{'} s n u m b e r r e l a t i o n s h i p f o r d i f f e r e n t c o n d i t i o n s :$

$1 . f = \frac{64}{R_{e}} : F o r l a m i n a r f l o w$

$2. \frac{0.316}{(R_{e})^{\frac{1}{4}}} : F o r t u r b u l e n t f l o w i n s m o o t h p i p e (4000 < R_{e} < 10^{5})$

$3. f = 0.0032 + \frac{0.221}{(R_{e})^{0.237}} : F o r t u r b u l e n t f l o w i n s m o o t h p i p e (10^{5} < R_{e} < 4 \times 10^{7})$

$4. \frac{1}{\sqrt{f}} = 2 l o g_{10} (\frac{R}{K}) + 1.74 : F o r t u r b u l e n t f l o w i n r o u g h p i p e$

where,

$R = R a d i u s o f P i p e$
$K = A v e r a g e s i z e o f r o u g h n e s = R o u g h n e s s h e i g h t$

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