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Sergio [31]
3 years ago
6

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

he wheel is of length , diameter D, and friction factor f. Minor losses are negligible. Show that the power developed by the turbine is maximum when the velocity head at the nozzle exit is 2H/3. Note: The result of Problem 12.61 may be of use.
Engineering
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

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7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
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Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

V = 28 KN

From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

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