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IRISSAK [1]
3 years ago
7

Air at 7 deg Celcius enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to engine). Air is hea

ted in the combustion chamber and it leaves the engine at 427 deg Celcius. Determine the thrust produced by this turbojet engine.
Engineering
1 answer:
Savatey [412]3 years ago
4 0

To solve this problem we will start using the given temperature values and then transform them to the Kelvin scale. From there through the properties of the tables, described for the Air, we will find the entropy. With these three data we can perform energy balance and find the speed of the fluid at the exit, which will finally help us find the total force:

V_i = 300m/s

T_1 = 7\°C = 280K

T_2 = 427\°C = 700K

\dot{m} 16kg/s

\dot{Q}_{in} = 15000kJ/s

Using the tables for gas properties of air we can find the enthalpy in this two states, then

T_1 = 280K \rightarrow h_1 = 280.13kJ/Kg\cdot K

T_2 = 700K \rightarrow h_2 = 713.27kJ/Kg\cdot K

Applying energy equation to the entire engine we have that

\frac{\dot{Q}_{in}}{\dot{m}}+h_1 +\frac{1}{2} V^2_{in} = h_2 + \frac{1}{2} V^2_{out}

\frac{15000}{16}*10^3+280.13*10^3+\frac{1}{2} 300^2 = 713.27*10^3+\frac{1}{2} V^2_{out}

V_{out} = 1048.198m/s

Finally the force in terms of mass flow and velocity is

F = \dot{m} (V_{out}-V_{in})

F = 16(1048.198-300)

F= 11971.168N

Therefore the thrust produced by this turbojet engine is  11971.168N

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3 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

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Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

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here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

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and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

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Vs= 4/(1/195+1/190+1/185+1/180)

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2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

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Answer:

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Explanation:

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x2 = mean(grade);

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x3=std(grade);

fprintf('The standard deviation is %5.2f \n',x3);

x4 = median(grade);

fprintf('The median grade is %5.2f \n',x4);

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3 years ago
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