Fitting Pressure Loss – Worked Example

Let us look at a worked example where we calculate a fitting pressure loss.

Example:

A fitting with a loss coefficient of 0.9 is present within a 125 mm internal diameter wort main. If the wort flow rate is 1,000 hL/hr, what will be the pressure loss across the fitting? Wort density is 1,050 kg/m³.

Stating the fitting pressure loss expression:

F i t t i n g space p r e s s u r e space l o s s space left parenthesis N divided by m to the power of ² right parenthesis equals left parenthesis k u to the power of ² over 2 right parenthesis rho

Hence, we need to determine the average flow velocity u from continuity considerations, as we have been given all other values to solve this equation.

Q equals A u

So, rearranging for u:

u equals Q over A equals fraction numerator begin display style fraction numerator begin display style fraction numerator 1 comma 000 cross times 100 over denominator 3 comma 600 cross times 1 comma 000 end fraction end style over denominator pi cross times 0.125 to the power of ² end fraction end style over denominator 4 end fraction equals fraction numerator 0.028 over denominator 0.012 end fraction equals 2.26 m divided by s

And substituting into the fitting pressure loss expression:

F i t t i n g space p r e s s u r e space l o s s equals left parenthesis k u to the power of ² over 2 right parenthesis rho equals left parenthesis 0.9 cross times fraction numerator 2.26 to the power of ² over denominator 2 end fraction right parenthesis cross times 1 comma 050

F i t t i n g space p r e s s u r e space l o s s equals 2.41 space k P a

Pipeline systems with multiple fittings

Practical pipeline systems are generally not restricted to a single fitting in the line, but rather multiple fittings. So, what do we do to determine the pressure loss through a number of fittings which are in a series within a pipeline? Serial fittings in a pipeline are probably the most common arrangement, for example, of moving fluid from tank A, through a pipeline to tank B. Well, we simply determine the pressure drop through each fitting and then sum them up.

Pipelines, with parallel arms and fittings in each arm, are a bit more complicated and we will not consider these further. If you are presented with these in practice, it is time to call a friendly engineer!

In a pipeline with a number of fittings in a series, we simply determine the pressure drop through each fitting and then sum the pressure drops up.

Let us look at a worked example to clarify this.

Example:

A bulk caustic tank is connected to a supply main where the mean flow velocity is constant throughout the main at 1.9 m/s. The main consists of an exit from the bulk tank: a half open gate valve and six threaded 90° elbows.

What would be the pressure drop along the length of the main due to fitting losses only? The caustic density is 1,200 kg/m³.

The loss coefficients are as follows: k=0.5, exit from the tank; k=0.9, 90° elbow; and k=5.6, 50% open gate valve.

The first thing to do is sketch the tank/main set-up:

caustic tank

The total fitting pressure loss will be given by:

increment P subscript բ equals increment P subscript e x i t end subscript plus 6 cross times increment P subscript e l b o w end subscript plus increment P subscript v a l v e end subscript

Where each term is given by:

increment P subscript բ equals k subscript e x i t end subscript left parenthesis fraction numerator u to the power of ² rho over denominator 2 end fraction right parenthesis plus 6 cross times k subscript e l b o w end subscript left parenthesis fraction numerator u to the power of ² rho over denominator 2 end fraction right parenthesis plus k subscript v a l v e end subscript left parenthesis fraction numerator u to the power of ² rho over denominator 2 end fraction right parenthesis

Note that in this case, the average flow velocity is the same in all cases (we were told this in the question), so we can group all the u² ρ/2 terms thus:

increment P subscript բ equals left parenthesis fraction numerator u to the power of ² rho over denominator 2 end fraction right parenthesis left square bracket k subscript e x i t end subscript plus 6 cross times k subscript e l b o w end subscript plus k subscript v a l v e end subscript right square bracket

increment P subscript բ equals left parenthesis fraction numerator 1.9 to the power of ² cross times 1 comma 200 over denominator 2 end fraction right parenthesis space left square bracket 0.5 plus 6 cross times 0.9 plus 5.6 right square bracket

increment P subscript e x i t end subscript equals 24.9 k P a

Now that we have covered how to determine the pressure loss due to fittings, and previously covered pressure loss due to friction, we can put both elements together to examine pressure loss along a pipeline due to the losses incurred from fittings and friction. Let us look at that next.

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