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BUTTERFLY VALVE LOSSES STEPHEN “DREW” FORD P6.160 13 FEB 2007.

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Presentation on theme: "BUTTERFLY VALVE LOSSES STEPHEN “DREW” FORD P6.160 13 FEB 2007."— Presentation transcript:

1 BUTTERFLY VALVE LOSSES STEPHEN “DREW” FORD P6.160 13 FEB 2007

2 P6.160 GIVEN: The butterfly valve losses in Fig. 6.19b may be viewed as a Bernoulli obstruction device, as in Fig. 6.39. FIND: First fit the K mean versus the opening angle in Fig6.19b to an exponential curve. Then use your curve fit to compute the “discharge coefficient” of a butterfly valve as a function of the opening angle. Plot the results and compare them to those for a typical flowmeter.

3 View of the butterfly valve from downstream Using the  as the degree of the opening for the variable in out equation.

4 Fig 6.19(b) Using these values as the original K’s

5 ASSUMPTIONS The velocity through the valve is different than the velocity in the entire pipe due to the small silver opening in the valve. Steady Incompressible Frictionless Continuous

6 The problem states using Eq. 104 may be helpful. Where A t is the area of the silver in the valve. This area is Found to be (1-cos(  )) The continuity equation gives us: Q = A p V p = A t V t

7 Manipulation of the continuity equation in terms of the Velocity in the Valve.

8 The minor losses in valves can be measured by finding “the ratio of the head-loss through the device to the velocity head of the associated piping system.”, K is the dimensionless loss coefficient

9 The problem states that using the Velocity through the valve will give a better K value. So multiplying the original K by a value of (V pipe /V valve )2 which is a mathematical “ 1 ” gives a new equation for K. Which yields the new equation for K

10 The calculated values of the K of three different manufactures from the original K values form Fig. 6.19b. Finding the curve fit lines of the original K values and plug in it in to the K optimal equation. These curve fit equations are: K 1 = 1101.9e -0.0978  (1-cos(  )) 2 K 2 = 1658.3e -0.1047  (1-cos(  )) 2 K 3 = 1273.3e -0.1919  (1-cos(  )) 2

11 The graph of all 3 manufacture’s valves with the new K values.

12 Biomedical Application Patients who suffer from valvular diseases in the heart may require replacing the non-functioning valve with an artificial heart valve. These valves keep the large one-dimensional flow going. Fluid mechanics plays a large role in designing these valve replacements, which require minimal pressure drops, minimize turbulence, reduce stresses, and not create flow separations in the vicinity of the valve.


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