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:

0.0432 m^3/s

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 = [tex]\pi Dt[/tex] = [tex]\pi ( 0.025 ) ( 0.003 )[/tex]  = 0.00023571 m^2

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

A = [tex][\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})[/tex]

   = [tex][ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )][/tex]

   = [ 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

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