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Oksana_A [137]
3 years ago
12

1. A hydro facility operates with an elevation difference of 50 m and a flow rate of 500 m3/s. If the rotational speed is 90 RPM

, find the most suitable type of turbine and estimate the power output of the arrangement
Engineering
1 answer:
zloy xaker [14]3 years ago
8 0

Answer:

a) Pelton Turbine

b) P=2.42*10^{5}KW

Explanation:

From the question we are told that:

Height h=50

Flow Rate R= 500 m^3/s

Rotational speed \omega=\90 RPM

Let

Density of water

\rho=1000

Generally the equation for momentum is mathematically given by

P=\rho gRh

P=1000*9.81*500*50

P=2.42*10^{5}KW

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Technician B is correct because the way aluminum collapses can be predicted.

Hardened steel and aluminum are two metals used for different purposes including:

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These two materials have slightly different features in terms of resistance, flexibility, etc.

In the case of hardened steel, this is considered to be malleable but strong. This means it is possible to change its shape under some conditions but it can resist great forces and pressure. Moreover, if the hardening process is carried out properly all the areas should be equally strong.

On the other hand, aluminum is recognized due to its durability and for being lighter than other materials. Despite this, aluminum is more flexible than steel and collapses under weaker forces. This has been widely studied because aluminum collapse shows a predictable pattern.

Based on this, only technician B is correct.

Learn more in: brainly.com/question/24043240

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The water level in a tank z1, is 16 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end
Volgvan

Answer:

40.7 m

Explanation:

Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:

z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g}  +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g}  +z_1\\\\

z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} )  +20\\\\z_2=20.7+20\\\\z_2=40.7\ m

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3 years ago
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