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2 years ago

12

Two technicians are discussing cylinder honing technician a says a good cross hatch helps to trap the oil and retain it in the c

ylinder bore where it is needed

1 answer:

aleksley [76]2 years ago

8 0

er:

Explanation:Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A

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A torsion member has an elliptical cross section with major and minor dimensions of 50.0 mm and 30.0 mm, respectively. The yield

Nata [24]

Answer:

What do i have to do

Explanation:

what do i do

4 0

3 years ago

Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using

VashaNatasha [74]

Answer:

For any string, we use $s = xyz$

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

Consider any string a i b j c k in the language. If i = 1 or i > 2, we take $x = \epsilon$ and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.

For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
Finally, for the case i = 0, we take $x = \epsilon$ , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.

8 0

3 years ago

The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( $d^{2}$ - $d_{0} ^{2}$ = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = $d^{2} - d^{2} _{0}$ = kt

d = final grain size

$d_{0}$ = initial grain size

k = constant ( temperature dependent )

t = 0

8 0

3 years ago

Write down one metal or alloy that is best suited for each of the following applications:

777dan777 [17]

Answer:

Explanation:

a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.

b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.

c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.

d. lead can be used as a container for strong acids because of its anti-corrosive properties

e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.

f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.

6 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago