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3 years ago

What is the ILS stand for

2 answers:

7 0

ILS stands for Instrument Landing System.

It is used to help provide lateral and vertical guidance to the pilots when landing an aircraft.

Alika [10]3 years ago

6 0

Answer:

Instrument Landing System

Explanation:

The ILS works by sending radio waves from the runway to the aircraft. Which is then intercepted and is used to guide the aircraft onto the runway.

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Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an

fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago

Calculate the wavelength of the first line in the Lyman series and show this line in the ultraviolet part of the spectrum?

Anit [1.1K]

Answer:

I am not noobqueen of the square footage

Explanation:

I am looking forward to be a hero bocah kok tolol of G if

7 0

3 years ago

Which tool is used to measure movement automotive

Schach [20]

Answer:

Speedometer

It shows our speed, or how fast we're moving.

3 0

3 years ago

Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of

Korolek [52]

Answer:

a) the mass flow rate of the steam is $\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam is $\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle is $A_2$ = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

$P_1 =$ 5 MPa

$T_1$ = 400° C

Velocity V = 80 m/s

Exit:

$P_2 =$ 2 MPa

$T_2$ = 300° C

From the properties of steam tables at $P_1 =$ 5 MPa and $T_1$ = 400° C we obtain the following properties for enthalpy h and the speed v

$h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg$

From the properties of steam tables at $P_2 =$ 2 MPa and $T_1$ = 300° C we obtain the following properties for enthalpy h and the speed v

$h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg$

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

$m_1=m_2=m_3$

Thus

$m_1 =\dfrac{V_1 \times A_1}{v_1}$

$m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}$

$m_1 =\dfrac{0.4 }{0.057838 }$

$\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam.

Using Energy Balance equation:

$\Delta E _{system} = E_{in}-E_{out}$

In a steady flow process;

$\Delta E _{system} = 0$

$E_{in} = E_{out}$

$m(h_1 + \dfrac{V_1^2}{2})$ $= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})$

$- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})$

$- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$V_2^2 = 316631.29 \ m/s$

$V_2 = \sqrt{316631.29 \ m/s$

$\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

$m = \dfrac{V_2A_2}{v_2}$

making $A_2$ the subject of the formula ; we have:

$A_2 = \dfrac{ m \times v_2}{V_2}$

$A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}$

$A_2$ = 0.0015435 m²

6 0

3 years ago