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kaheart [24]
3 years ago
11

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

as pressure at the inlet is 5.5 psig, what is the absolute pressure, in psia, of the gas at the exit? Atmospheric pressure is 14.5 lbf/in.2
Engineering
1 answer:
olga55 [171]3 years ago
7 0

Answer:

P_2_{abs}=160\ psia (absolute).

Explanation:

Given that

Pressure ratio r

r=8

r=\dfrac{P_2_{abs}}{P_1_{abs}}

  8=\dfrac{P_2_{abs}}{P_1_{abs}}                                  -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure  + Gauge  pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5  psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

8=\dfrac{P_2_{abs}}{20}

P_2_{abs}=8\times 20\ psia

P_2_{abs}=160\ psia

Therefore the exit gas pressure will be 160 psia (absolute).

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Describe the first case where the power of synthesis was used to solve design problems.
Minchanka [31]

The first case where the power of synthesis was applied was in Chile when they had to put 100 families in houses around 40m².

It should be noted that the power of architecture is the fact that it can synthesize a complex entry to a problem.

Chilean architect Alejandro used this in building more than several houses for the poorest communities in Chile. His rigorous and innovative design approach used a social framework that laid a precedent within the profession.

During this period in Chile, a normal middle-class home was about 80m² but he built his in 40m² and it was a good home that managed the resources that were available.

Learn more about synthesis on:

brainly.com/question/884041

6 0
2 years ago
. Carly's Catering provides meals for parties and special events. In Chapter 2, you wrote an application that prompts the user f
klio [65]

Answer:

Explanation:

public class Event

  public final static int PRICE_PER_GUEST = 35;

  public final static int CUT_OFF = 50;

 

  //Attributes

  private String eventNum;

  private int noOfGuest;

  private int price;

 

  /**

  * param eventNum the eventNum to set

  */

  public void setEventNum(String eventNum)

      this.eventNum = eventNum;

 

 

  /**

  * param noOfGuest the noOfGuest to set

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  public void setNoOfGuest(int noOfGuest)

      this.noOfGuest = noOfGuest;

      this.price = this.noOfGuest * PRICE_PER_GUEST;

 

 

  /**

  * return the eventNum

  */

  public String getEventNum()

      return eventNum;

 

 

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  public int getNoOfGuest()

6 0
3 years ago
An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs
Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h=110\times \frac{5}{18}=30.56 m/s

Where 1 km/h=5/18 m/s

Height , h_2=40 m

g=9.81 m/s^2

Final speed, v=0

Change in kinetic energy,\Delta K.E=\frac{1}{2}m(v^2-u^2)

\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J

\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ

Where 1 KJ=1000 J

Change in potential energy,\Delta P.E=mgh(h_2-h_1)

Initially height, h1=0

Using the formula

\Delta P.E=1100\times 9.81(40-0)

\Delta P.E=431640J

\Delta P.E=431.64KJ

6 0
3 years ago
. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe
Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider  all standard measures of standard single engine propeller plane

as

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient  =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42  lb/hp/hr

=0.42  lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2  -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2  -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

7 0
3 years ago
A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei
miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

4 0
3 years ago
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