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4 years ago

Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?

Engineering

1 answer:

Firdavs [7]4 years ago

6 0

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

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A thick steel sheet of area 150 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

Bingel [31]

Answer:

Rate of corrosion = 24.95 mpy

Rate of corrosion = 0.63 mm/yr

Explanation:

given data

steel sheet area = 150 in²

weight loss = 485 g

density of steel = 7.9 g/cm³

time taken = 1 year

to find out

rate of corrosion in (a) mpy and (b) mm/yr

solution

we get here the rate of corrosion that is express as

rate of corrosion = (k × W) ÷ (D × A × T) ..................1

here k is constant and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel

so put her value in equation 1 we get

Rate of corrosion = $\frac{534*485*1000}{7.9*150*24hr*365day/yr}$

Rate of corrosion = 24.95 mpy

and

Rate of corrosion = $\frac{87.6*485*1000}{7.9*150*2.54^2*24hr*365days/yr}$

Rate of corrosion = 0.63 mm/yr

5 0

3 years ago

A thermodynamicist claims to have developed a heat pump with a COP of 1.7 when operating with thermal energy reservoirs at 273 K

Vinil7 [7]

Answer:

cfcghvjvct

Explanation:

3 0

3 years ago

Air enters a constant-area combustion chamber at a pressure of 101 kPa and a temperature of 70°C with a velocity of 130 m/s. By

Norma-Jean [14]

Answer:

451 kj/kg

Explanation:

Velocity = 139m/s

Temperature = 70⁰C

T = 343K

M1 = v/√prt

= 130/√1.4x287x343

= 130/√137817.4

= 130/371.2

= 0.350

T1/To1 = 0.9760

From here we cross multiply and then make To1 the subject of the formula

To1 = T1/0.9760

To1 = 343/0.9760

To1 = 351.43

Then we go to the rayleigh table

At m = 0.35

To1/To* = 0.4389

To* = 351.43/0.4389

= 800k

M2 = 1

Maximum amount of heat

1.005(800-351.43)

= 450.8kj/kg

= 452kj/kg

8 0

3 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

6 0

3 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

$\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}$

Original length $L_(i}$

$\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)$

Where $d_{f}$ is final diameter, $d_{i}$ is original diameter, $L_{f}$ is final length and $L_{i}$ is original length.

$\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}$

$\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}$

Therefore, the original length is 105.70 mm

7 0

4 years ago