Question

The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three smaller pipes is 2.5 cm, and the pipe wall thickness is 3 mm. The inside diameter of the larger pipe is 8 cm. If the velocity of the fluid in the region between the smaller pipes and larger pipe is 10 m/s, what is the discharge in m^3/s?

Answer 1

0.0432 m^3/s

Step-by-step explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A =
`\pi Dt` =
`\pi ( 0.025 ) ( 0.003 )` = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A =
`[(\pi )/(4) D^(2) _(L) ] - 3(A_(s))`

=
`[ (\pi )/(4) (0.08 )^2 - 3 ( 0.00023571 )]`

= [ 0.00502857 - 0.00070713 ]

= 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

= 0.00432144 * 10

= 0.0432 m^3/s