Question-210318

Question Number 210318 by peter frank last updated on 06/Aug/24

Commented by mr W last updated on 06/Aug/24

$${what}\:{should}\:{be}\:{reached}\:{with}\:{the} \\ $$$${single}\:{tube}\:{when}\:{it}\:{replaces}\:{the}\:{three} \\ $$$${ones}?\:{the}\:{same}\:{flow}\:{rate}\:{with}\:{the} \\ $$$${single}\:{tube}\:{as}\:{with}\:{the}\:{three}\:{tubes}? \\ $$

Answered by mr W last updated on 06/Aug/24

Commented by mr W last updated on 06/Aug/24

H=30 cm h_1 =0 h_2 =4 cm h_3 =8 cm L_1 =39 cm

$${H}=\mathrm{30}\:{cm} \\ $$$${h}_{\mathrm{1}} =\mathrm{0} \\ $$$${h}_{\mathrm{2}} =\mathrm{4}\:{cm} \\ $$$${h}_{\mathrm{3}} =\mathrm{8}\:{cm} \\ $$$${L}_{\mathrm{1}} =\mathrm{39}\:{cm} \\ $$

Commented by mr W last updated on 06/Aug/24

Commented by peter frank last updated on 07/Aug/24

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Mr}\:\mathrm{W} \\ $$

Answered by BHOOPENDRA last updated on 06/Aug/24

Let say single tube flow of rate is Q and the length of tube is L′ rest of things you can see from above fig.Mr.W has provided the figure . total flow Q=Q_1 +Q_2 +Q_3 where Q_1 ,Q_2 ,Q_3 flow of tube respectivly. Q=Q_1 +Q_2 +Q_3 ((30)/(L′))=((30)/l_1 )+((26)/l_1 )+((22)/l_1 ) ((30)/(L′))=((78)/(39)) L′=15cm Note−P(pressure)=ρgh Rate of flow Q=((πΔPr^4 )/(8ηl))

$${Let}\:{say}\:{single}\:{tube}\:{flow}\:{of}\:{rate}\:{is}\:{Q}\: \\ $$$${and}\:{the}\:{length}\:{of}\:{tube}\:{is}\:{L}'\: \\ $$$${rest}\:{of}\:{things}\:{you}\:{can}\:{see}\:{from}\:{above}\:{fig}.{Mr}.{W}\: \\ $$$${has}\:{provided}\:{the}\:{figure}\:. \\ $$$${total}\:{flow}\:{Q}={Q}_{\mathrm{1}} +{Q}_{\mathrm{2}} +{Q}_{\mathrm{3}} \\ $$$${where}\:{Q}_{\mathrm{1}} ,{Q}_{\mathrm{2}} ,{Q}_{\mathrm{3}} \:{flow}\:{of}\:{tube}\: \\ $$$${respectivly}. \\ $$$${Q}={Q}_{\mathrm{1}} +{Q}_{\mathrm{2}} +{Q}_{\mathrm{3}} \\ $$$$\frac{\mathrm{30}}{{L}'}=\frac{\mathrm{30}}{{l}_{\mathrm{1}} }+\frac{\mathrm{26}}{{l}_{\mathrm{1}} }+\frac{\mathrm{22}}{{l}_{\mathrm{1}} } \\ $$$$\frac{\mathrm{30}}{{L}'}=\frac{\mathrm{78}}{\mathrm{39}} \\ $$$${L}'=\mathrm{15}{cm} \\ $$$${Note}−{P}\left({pressure}\right)=\rho{gh} \\ $$$${Rate}\:{of}\:{flow}\:{Q}=\frac{\pi\Delta{Pr}^{\mathrm{4}} }{\mathrm{8}\eta{l}} \\ $$

Commented by peter frank last updated on 07/Aug/24