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1 year ago

Technician A says that in a recirculating-ball steering box when the sector gear and nut teeth

Engineering

2 answers:

anyanavicka [17]1 year ago

5 0

Answer:

It's Technician B. A is not a good grade,

Explanation:

NISA [10]1 year ago

3 0

The second one I believe

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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

2 years ago

Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please

kompoz [17]

Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

Input

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

Processing

Assign variable to the input number (x)

Calculate the square (x = 5 * 5)

Display the result (25) ----> This will be passed to the output module

Output

Display 25

5 0

2 years ago

A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP

Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

a) Determine the pressure at inlet to expansion process

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp) $\frac{r-1}{r}$ ]

0.569 = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569 = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

= 5.2 kPa

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

= 0.493 = 49.3%

4 0

2 years ago

ILL GIVE YOU BRAINLIEST PLEASE HELP PLEASE

Lesechka [4]

Answer:

try to pop it back in good luck im scared for you

4 0

2 years ago

A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com

Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For vapor-compression refrigeration cycle

P1 = P4 ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

From pressure table of R 134a refrigerant

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

3 0

2 years ago