find the dimension of shorter side of duct.

Mumbai University > Mechanical Engineering > Sem 8 > Refrigeration and air conditioning

Marks: 06M

Year: Dec 2016

The given data is

$\frac{b}{a}=7 and \frac{b}{D}=2.6$

For the given condition, the relation between the two geometric sections is given by

D=1.256$[\frac{ab{^3}}{(a+b)}]^{(1/5)}$

substituting the given values

10=$1.265[\frac{(a×7a)^3}{8a}]^{(1/5)}=1.265×a(\frac{343}{8})^{0.2}=2.68a$

∴a=10/2.68=3.73 cm

b=7×3.73=26.12 cm

as $\frac{b}{D}=\frac{26.2}{10}$=2.61 which satisfes the third given condition

The given data is

$\frac{b}{a}=7 and \frac{b}{D}=2.6$

For the given condition, the relation between the two geometric sections is given by

D=1.256$[\frac{ab{^3}}{(a+b)}]^{(1/5)}$

substituting the given values

10=$1.265[\frac{(a×7a)^3}{8a}]^{(1/5)}=1.265×a(\frac{343}{8})^{0.2}=2.68a$

∴a=10/2.68=3.73 cm

b=7×3.73=26.12 cm

as $\frac{b}{D}=\frac{26.2}{10}$=2.61 which satisfes the third given condition

The given data is

$\frac{b}{a}=7 and \frac{b}{D}=2.6$

For the given condition, the relation between the two geometric sections is given by

D=1.256$[\frac{ab{^3}}{(a+b)}]^{(1/5)}$

substituting the given values

10=$1.265[\frac{(a×7a)^3}{8a}]^{(1/5)}=1.265×a(\frac{343}{8})^{0.2}=2.68a$

∴a=10/2.68=3.73 cm

b=7×3.73=26.12 cm

as $\frac{b}{D}=\frac{26.2}{10}$=2.61 which satisfes the third given condition