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yan [13]
2 years ago
10

A helicopter is hovering in a steady cross wind at a gross weight of 3,000 lb (1,360.8 kg). This helicopter has 275 hp (205 kW)

delivered to the main rotor shaft. The tail rotor is 2.3 ft (0.701 m) and has an induced power factor of 1.15. The tail rotor is located 15.3 ft (4.66 m) from the main rotor shaft. Determine the crosswind conditions (velocity and direction) in which the tail rotor effectiveness may be reduced or lost. If the center of gravity is assumed to lie on the rotor shaft axis, determine the feasible yawing angular velocity that the pilot can demand that may also result in a loss of tail rotor effectiveness
Engineering
1 answer:
damaskus [11]2 years ago
6 0

Answer: Answer Bellow

Explanation:The helicopter has 275 hp (205 kW) delivered to the main rotor shaft. The tail rotor radius is 2.3 ft (0.701 m) and has an induced power factor of 1.15.

(I would appreciate brainiest)

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Answer:

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Your manager has asked you to research and recommend a writing guide that examiners in your digital forensics company can use fo
vivado [14]

Answer:

Kindly Check the explanation

Explanation:

Report

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Guidelines  

These are the set of rules and written statement which are used to present the important information in a structured and clear format. The main motive behind to define the guidelines is to achieve standardization in the reports. It also improves the readability of the report.  

Guidelines for report  

Abstract:  

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Index:  

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2. The explanation must be justified

3. As possible divide the long paragraph. if possible explain in points.

4. The page must have border on all sides

5. Each pages must have numbering and footer

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The body of the report referred to as a content written inside the report. In order to write an impressive report, it must include all these things:

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8 0
3 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

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3 years ago
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Cloud [144]

Answer:

b

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bcus it is

7 0
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