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2 years ago

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Engineering

2 answers:

inn [45]2 years ago

8 0

Thank youuuu!!!!!!!!!!!!

kobusy [5.1K]2 years ago

6 0

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

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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

gayaneshka [121]

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

7 0

3 years ago

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,

Scorpion4ik [409]

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

4 0

3 years ago

The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric mo

ludmilkaskok [199]

Answer:

Explanation:

This is a two sample t-test and not a matched pair t-test

null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors

alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.

So, option D is rejected

The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.

Option A and C are also rejected

5 0

3 years ago

The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th

Alja [10]

Answer:

Explanation:

$y'''+y=0---(i)$

General solution

$y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)$

$y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1$