1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stira [4]
3 years ago
12

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

circular pipe of 5.0 cm diameter.
a. Determine the outlet volumetric flow rate and the mass flow rate of the air.
b. If the air enters the compressor through an inlet at 20.0 oC and 100 kPa and a velocity of 1.0 m/s, determine the volumetric flow rate of the air entering the compressor and the required cross-sectional area of the inlet.
c. Using your equations, plot the inlet volumetric flow rate and the inlet cross-sectional area as the inlet air velocity varies between 0.25 m/s and 10.0 m/s.
Engineering
1 answer:
ioda3 years ago
6 0

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

You might be interested in
Define and discuss the difference between micronutrients and macronutrients. Also, discuss their importance in the body at rest
almond37 [142]

Answer:

Macronutrients are simply nutrients the body needs in a very high amount e.g Carbohydrate.

MicroNutrients are simply nutrients the body needs but in little amount e.g  Minerals.

Explanation:

So for further breakdown:

What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.

Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.

Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins

In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.

In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.

5 0
3 years ago
Depending on the model, all-electric vehicles can go about _______ miles on a single charge.
Lera25 [3.4K]

Answer:

D

Explanation:

Most electric vehicles can go at least 100.....few, if any, can go 400 or more on a single charge

6 0
2 years ago
A patient has swollen feet, ankles, and legs, and has pain in his lower back. Tests show the level of urea, a waste product, in
myrzilka [38]

Answer: B  Excretory

Explanation: Hope this helps :)

3 0
3 years ago
Let S = { p q |p, q are prime numbers greater than 0} and E = {0, −2, 2, −4, 4, −6, 6, · · · } be the set of even integers. . Pr
Finger [1]

Answer:

prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S

Explanation:

Given that S = { p q |p, q are prime numbers greater than 0}

                    E = {0, −2, 2, −4, 4, −6, 6, · · · }

To prove  by constructing a bijection from S to E

detailed  solution attached below

After the bijection :

<em>prove that | S | = | E |</em> :  every element of S there is an Image on E , while not every element on E has an image on S

∴ we can say sets E and S are infinite sets

7 0
2 years ago
Match the given traits to their definitions.
damaskus [11]

Answer:

where the question

Explanation:

8 0
3 years ago
Read 2 more answers
Other questions:
  • List irreversibilities
    11·1 answer
  • How much work is performed if a 400 lb weight is lifted 10 ft ?
    8·1 answer
  • Breaks do not overheat true false ?
    6·1 answer
  • Refrigerant-134a enters an adiabatic compressor at -30oC as a saturated vapor at a rate of 0.45 m3 /min and leaves at 900 kPa an
    13·1 answer
  • I have a question for you guys
    13·2 answers
  • What is meant by the acronym ISO
    15·1 answer
  • What have you learned from the previous lesson? Let's try to check your prior knowledge
    9·1 answer
  • A thin 20-cm*20-cm flat plate is pulled at 1m/s horizontally through a 4-mm thick oil layer sandwiched between two stationary pl
    15·1 answer
  • Identify renewable energy sources you will propose. Explain the key elements to your solution and the basic technical principles
    5·1 answer
  • Why not just put all the set up steps within each step? it is because we want to keep our code __ ? (3 letters)
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!