All categories

3 years ago

What kinds of problems or projects would a mechanical engineer work on?

Engineering

1 answer:

Mademuasel [1]3 years ago

3 0

Answer:

Design power producing machines like electric generators, internal combustion engines, turbines and power using machines like refrigiration or air conditioning. And alot like elevators or such.

Explanation:

If that is what your asking for? Sorry if its not what you wanted. In genral all hard work science and maths

You might be interested in

Proper handling of blueprints includes which of the following

marta [7]

Answer:

folding plans neatly after use

3 0

3 years ago

Define the Problem

gtnhenbr [62]

Answer:

rbrnrifnfnrfbdjrbfbfjrn

4 0

3 years ago

Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat

Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

//Multiply X0 and X1

// Does not handle negative X1!

// Note : This is an in efficient way to multipy!

SUB SP, SP, 16 //make room for X19 on the stack

STUR X19, [SP, 0] //push X19

ADD X19, X1, XZR //set X19 equal to X1

ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

CBZ X19, mult_eol

ADD X9, X9, X0

SUB X19, X19, 1

B mult_loop

mult_eol:

LDUR X19, [SP, 0]

ADD X0, X9, XZR // Move X9 to X0 to return

ADD SP, SP, 16 // reset the stack

BR X30

Explanation:

6 0

4 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

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

Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.

bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

Given the following data:

Emf = 520 Volts

Speed = 660 r.p.m

Number of armature conductors = 144 slots

Number of poles = 4 poles

Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

$E = \frac{\theta ZN}{60}$ × $\frac{P}{A}$

Where:

E is the electromotive force in the DC generator.

Z is the total number of armature conductors.

N is the speed or armature rotation in r.p.m.

P is the number of poles.

A is the number of parallel paths in armature.

Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

$Z = 144$ × $2$ × $3$

Z = 864

Substituting the given parameters into the formula, we have;

$520 = \frac{\theta (864)(660)}{60}$ × $\frac{4}{2}$

$520 = \theta (864)(11)$ × $2$

$520 = 19008 \theta \\\\\Theta = \frac{520}{19008}$

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

5 0

3 years ago