Suggested Response - Q3

**A double-pipe heat exchanger consists of a section of a smaller diameter pipe within a larger diameter pipe, creating two flow paths as shown in the figure below. The outer pipe has an internal diameter of 50 mm. The inner pipe has an external and internal diameter of 33 and 25 mm, respectively.

pipe

Water is pumped through the annular flow path and the inner flow path at the same rate of 50 hL/h. Determine the Reynolds number for both flows and state whether the flows are laminar, transitional or turbulent. (Density and dynamic viscosity are 1,000 kg/m3 and 1 cP respectively).

Response:

The geometry of a pipe-within-a-pipe may seem a bit strange, but it can be quite common in so-called double-pipe heat exchangers. It is important to differentiate between the two flow paths (the inner flow and annular path). To determine the Reynolds number, we use the following equation:

R e space equals space fraction numerator rho u d over denominator mu end fraction

As the two flow paths have differing geometries, with the annular flow path being non-circular, we need to use the equivalent diameter de. Let us deal with the two flow paths separately.

The inner flow path

The flow area is simply that of a circle of diameter 25 mm, therefore the flow area is given by:

A space equals space fraction numerator pi d squared over denominator 4 end fraction space equals space fraction numerator pi space cross times space 0.025 squared over denominator 4 end fraction space equals space 0.00049086 space m squared

The flow velocity is given by rearranging the following equation:

Q space equals space A u

u space equals space Q over A space equals space fraction numerator begin display style fraction numerator 50 space cross times space 100 over denominator 1 comma 000 space cross times space 3600 end fraction end style over denominator 0.00049086 end fraction space equals space 2.83 space m divided by s

 

Hence the Re is:

R e space equals space fraction numerator rho u d over denominator mu end fraction space equals space fraction numerator 1000 space cross times space 2.83 space cross times space 0.025 over denominator 1 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space 70750

As this Re is greater than 4,000, we can state that the flow would be turbulent.

Annular flow path

The flow area is the area of the annulus (a ring-shaped object, structure, or region). This can be determined as the area of a 50 mm diameter circle, from which we deduct the area of a circle of diameter 33 mm (the latter is the area of the inner pipe cross-section using its outer diameter):

A space equals space fraction numerator pi d subscript 50 squared over denominator 4 end fraction space minus space fraction numerator pi d subscript 33 squared over denominator 4 end fraction space equals space fraction numerator pi space cross times space 0.05 squared over denominator 4 end fraction space minus space fraction numerator pi space cross times space 0.033 squared over denominator 4 end fraction space equals space 0.0011 space m squared

The flow velocity is given by rearranging the following equation:

Q space equals space A u

u space equals space Q over A space equals space fraction numerator begin display style fraction numerator 50 space cross times space 100 over denominator 1 comma 000 space cross times space 3 comma 600 end fraction end style over denominator 0.0011 end fraction space equals space 1.25 space m divided by s

The diameter that we need to use is the equivalent diameter of the annulus. This is given by:

d subscript e space equals space 4 space fraction numerator F l o w space c r o s s space s e c t i o n a l space a r e a over denominator W e t t e d space P e r i m e t e r end fraction

We have already calculated the cross-sectional flow area at 0.0011 m2. The wetted perimeter would be the length equivalent to the circumference of the inside of the external pipe summed with the circumference of the outside of the internal pipe (you might need to read the previous sentence again! Note there are two separate perimeters which the flow passes between). Consequently, the wetted perimeter is given by:

W e t t e d space P e r i m e t e r space equals space pi d subscript 50 space end subscript plus space pi d subscript 33

W e t t e d space P e r i m e t e r space equals space pi space cross times space 0.05 space plus space pi space cross times space 0.033 space equals space 0.26 space m

Therefore, de is:

d subscript e space equals space 4 space fraction numerator F l o w space c r o s s space s e c t i o n a l space a r e a over denominator W e t t e d space P e r i m e t e r end fraction space equals space 4 space cross times space fraction numerator 0.0011 over denominator 0.26 end fraction space equals space 0.017 space m

And the Re is:

R e space equals space fraction numerator rho u d subscript e over denominator mu end fraction space equals space fraction numerator 1 comma 000 space cross times space 1.25 space cross times space 0.017 over denominator 1 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space 21 comma 250

As this Re is greater than 4,000, we can state that the flow would also be turbulent.

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