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Gelneren [198K]
3 years ago
14

A single phase molor is located

Engineering
1 answer:
belka [17]3 years ago
5 0

Answer:

hyguhttbjjgccd h jb

Explanation:

hcufugibibuguguv

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One of the disadvantages of the test of the hypothesis is that the final decision cannot be said to be completely correct eviden
vfiekz [6]

Answer:

A. True

Explanation:

4 0
3 years ago
In a short essay, discuss the question, "How are you an innovator?"
iragen [17]

Answer:

Being innovative means doing things differently or doing things that have never been done before. An innovator is someone who has embraced this idea and creates environments in which employees are given the tools and resources to challenge the status quo, push boundaries and achieve growth.

Explanation:

Hope it helps..

But it's a little bit long..

Correct me if I'm wrong..

7 0
2 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
3 years ago
Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0
3 years ago
Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon
balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

torque =\frac{power}{2\pi N}

power=2\pi N\times T

P=2\times \pi \times2500 \times 4.5

P=70,685W

P=70.685 KW

power=\frac{work done}{time}

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ

7 0
3 years ago
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