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Artist 52 [7]
3 years ago
13

A jet engine for a supersonic transport (SST) propels the airplane at Mach 3 at an altitude of 50,000 ft where the temperature i

s 217°K and the ambient pressure is 0.115 atm. The maximum temperature that the advanced turbine blades can handle is 1900°K. The engine operates at a compressor pressure ratio that allows the engine to achieve maximum work. The temperature at the entrance to the compressor is 595.7K. Assume that γ = 1.4 and Cp = 1004 J/kg-°K throughout the engine. If the engine inlet isentropically slows down the incoming freestream air such that the Mach number of the flow just upstream of the compressor is 0.3,
a) How many stages should the aircraft engine compressor have if the compression pressure ratio of each stage is 1.2? b) Please plot the cycle on a T-s diagram labeling only the temperature at each point? c) What is the thermal efficiency of the engine? d) What is the absolute highest possible thermal efficiency one can expect from a machine operating within the range of temperatures involved in this problem? e) What is the amount of heat per unit mass of working fluid that is needed to be generated by the combustion of fuel in the engine burner?

Engineering
1 answer:
deff fn [24]3 years ago
7 0

Answer:

See attached file

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Hot air at an average temperature of 100 degC flows through a 3 m long tube with an inside diameter of 60 mm. The temperature of
Arte-miy333 [17]

Answer:

\dot Q = 912.017\,W

Explanation:

The heat transfer rate due to convection is calculated by this formula:

\dot Q = h_{air}\cdot A_{tube, in}\cdot (T_{air} - T_{tube})

The output is determined by replacing terms:

\dot Q = (20.1\,\frac{W}{m^{2}\cdot ^{\textdegree}C})\cdot (\pi )\cdot (0.06\,m)\cdot (3\,m)\cdot (100^{\textdegree}C-20^{\textdegree}C)

\dot Q = 912.017\,W

6 0
3 years ago
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The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra
Lina20 [59]

Answer:32.4m/s^2

Explanation:

Given data

radius\left ( r\right )=0.4m

Intial angular velocity\left ( \omega_0\right )=4rad/s

angular acceleration\left ( \alpha\right )=5rad/s^2

angular velocity after 1 sec

\omega=\omega_0+\alpha\times\t

\omega=4+5\left ( 1\right )

\omega=9rad/s

Velocity of point on the outer surface of disc\left ( v\right )=\omega_0\timesr

v=9\times0.4 m/s=3.6m/s

Normal component of acceleration\left ( a_c\right )=\frac{v^2}{r}

a_c=\frac{3.6\times3.6}{0.4}=32.4m/s^2

3 0
4 years ago
What is the aim of reviewing a research paper?
qaws [65]

Answer:

Purpose of review papers

They carefully identify and synthesize relevant literature to evaluate a specific research question, substantive domain, theoretical approach, or methodology and thereby provide readers with a state-of-the-art understanding of the research topic.

3 0
3 years ago
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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
To convert a whole number to a fraction the number can be written over a denominator of
exis [7]
One

When you convert a whole number into a fraction, you put the whole number as the numerator and the denominator as 1.

Such as 12=12/1 or 5=5/1
7 0
3 years ago
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