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Doss [256]
3 years ago
5

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

12.5 hp for 16 h/day, five days per week. Compute the energy used by the mo- tor for one year. Express the result in ft.lb and W.h
Engineering
1 answer:
ehidna [41]3 years ago
3 0

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

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Comparación de hipotecas Los Chos
aleksklad [387]

Answer:

I don't understand the language French sorry can't answer

3 0
2 years ago
Give the approximate temperature (in K) at which creep deformation becomes an important consideration for each of the following
andrezito [222]

Answer:

691K, 543K, 725K, 1473K, 240K, 373K

Explanation:

Creep deformation of any metal is the transformational tendency of a metal to distort rapidly or slowly when attacked by any form of mechanical stress. The temperature significant for a metal to deform is gotten by the division of the actual temperature of the metal by its melting point. This is termed homologous temperature which is 0.4 or higher. It is calculated by the equation:

0.4Tm

Therefore for the listed metals...

For Nickel, 0.4Tm = 0.4 ×(1455 + 273) = 691 K

For Copper, 0.4Tm = 0.4 ×(1085 + 273) = 543 K

For Iron, 0.4Tm = 0.4 ×(1538 + 273) = 725 K

For Tungsten, 0.4Tm = 0.4 ×(3410 + 273) = 1473 K

For Lead, 0.4Tm = 0.4 × (327 + 273) = 240 K

For Aluminium, 0.4Tm = 0.4 ×(660 + 273) = 373 K

5 0
3 years ago
In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
GrogVix [38]

Answer:

The shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

Explanation:

Given:

Temperature T = 1273.15 K

Initial Pressure P_{1} = 1.8 MPa

Final pressure P_{2} = 0.1 MPa

From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

 W = h_{f} - h_{i}

 W = 2706.54 - 4635

 W = -1928.46 \frac{kJ}{kg}

But here efficiency is 0.56,

So work generated per kg is,

Work = 0.56 \times(- 1928.46)

Work = -1.3 \frac{MJ}{kg}

Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
3 years ago
A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50
anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure  = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r =  120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0
3 years ago
Select three ways an engineer can create a view of a design.
Angelina_Jolie [31]

Sketching projection

  • Sketching will give the brief planning of the engineering work to be done so it's important

Isometric projection

  • Yes doing a work in 3D factor out the mistakes which can be optimised

Third angle projection:-

  • Viewing angles matter a lot
  • So it will help the engineer to make it perfect from all angles
5 0
2 years ago
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