Converging and diverging nozzles

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In summary, the air at this stagnation point has a static pressure of 518.6 K and a temperature of 323.5 K.
  • #1
tsukuba
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Homework Statement


air flows through a device such that the stagnation pressure is 0.6 MPa, the stagnation temperature is 400° C, and the velocy is 570 m/s.
Determine the static pressure and temperature of the air at this state.

Homework Equations


Tt=T + V2 / 2 Cp where Cp is 1.005 Kj/ Kg* ° K

The Attempt at a Solution


what I did was Isolate for T since that is what I am looking for. [/B]

2 Cp x Tt - v2 = T
somewhere in here I know I have to throw in 1000m2/s2/ Kj/kg °k
when I do that I get -323.5 ° K

the answer is 518.6 ° K

Someone please let me know where I have gone wrong. Thank you
 
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  • #2
tsukuba said:

Homework Statement


air flows through a device such that the stagnation pressure is 0.6 MPa, the stagnation temperature is 400° C, and the velocy is 570 m/s.
Determine the static pressure and temperature of the air at this state.

Homework Equations


Tt=T + V2 / 2 Cp where Cp is 1.005 Kj/ Kg* ° K

The Attempt at a Solution


what I did was Isolate for T since that is what I am looking for. [/B]

2 Cp x Tt - v2 = T
somewhere in here I know I have to throw in 1000m2/s2/ Kj/kg °k
when I do that I get -323.5 ° K

the answer is 518.6 ° K

Someone please let me know where I have gone wrong. Thank you

Your algebra is a little dodgy in solving for T.

The correct equation is:
Tt = T + [V2 / (2 Cp)]

which means your equation for T is incorrect. :frown:
 
  • #3
brackets or no brakets, wouldn't T stay the same when isolating it?
 
  • #4
tsukuba said:
brackets or no brakets, wouldn't T stay the same when isolating it?

Tt = T + [V2 / (2 Cp)]

T = Tt - [V2 / (2 Cp)]

But you had

2 Cp x Tt - V2 = T

which is not the same. Apparently, you multiplied only one side of the equation by (2 Cp).
 
  • #5
hmm, not sure how you get that equation but I'm still not getting the right answer. I will have to ask the professor tomorrow
 

1. What is the purpose of a converging nozzle?

A converging nozzle is designed to increase the velocity of a fluid or gas by decreasing its cross-sectional area. This is achieved by gradually narrowing the nozzle from its inlet to its outlet.

2. How does a converging nozzle work?

A converging nozzle works on the principle of conservation of mass and Bernoulli's principle. As the cross-sectional area decreases, the fluid or gas is forced to accelerate, resulting in an increase in velocity. This increase in velocity also leads to a decrease in pressure, which creates a pressure difference that helps to drive the flow through the nozzle.

3. What is the difference between a converging and a diverging nozzle?

A converging nozzle decreases the cross-sectional area from its inlet to its outlet, while a diverging nozzle increases the cross-sectional area. As a result, a converging nozzle accelerates the fluid or gas, while a diverging nozzle decelerates it.

4. What are some practical applications of converging and diverging nozzles?

Converging and diverging nozzles are commonly used in propulsion systems, such as jet engines and rockets, to increase the velocity of exhaust gases. They are also used in various industrial processes, such as spray painting and water jet cutting, to create high-velocity streams of fluids or gases.

5. What factors affect the performance of a converging or diverging nozzle?

The performance of a converging or diverging nozzle is influenced by factors such as the shape and angle of the nozzle, the fluid or gas properties, and the inlet pressure and temperature. Other factors, such as the presence of shock waves and boundary layers, can also affect the performance of the nozzle.

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