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3 years ago

6

The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T i

s that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False

1 answer:

Butoxors [25]3 years ago

5 0

The answer is: a. True

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A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

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Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

=65.44 rad/sec

$F_T = mw^2 e$

$F_T = mew^2$

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F_T =428.36 N

Transmission ratio $=\frac{300}{428.36} = 0.7$

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transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

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3 years ago

Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1

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Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2 and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

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y-x = -2

The table will be :

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1 -1

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Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

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4 years ago

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8. When supplying heated air for a building, one often chooses to mix in some fresh outside air with air that has been heated fr

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Answer:

Check the explanation

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Kindly check the attached images below to see the step by step explanation to the question above.

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3 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

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final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

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so gauge pressure is 133 kPa

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3 years ago