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Zolol [24]
2 years ago
15

Periodic lubrication and oil changes according to manufacturer’s recommendations extend the life of your vehicle, and allow you

to avoid costly repairs, and prevent dangerous breakdowns. Generally, owner's manuals suggest that you change your oil and oil filter every ........... miles. Oil change specialists. however, suggest every ...... miles or ........ months. As you know, most of us do a lot of long distance driving during the summer months when an engine is can overheat. So, you should at least check your oil before you head out on any road trip.
1000, 3000, 2
7,500, 3000, 3
2000, 6000, 4
500,1500, 1
Engineering
1 answer:
Vladimir79 [104]2 years ago
6 0

There are different kinds of rules that applies to engine oils.  Generally, owner's manuals suggest that you change your oil and oil filter every <u>7,500 </u>miles. Oil change specialists. however, suggest every <u>3000 </u>miles or <u>3 </u>months.

<h2>What is Periodic lubrication and oil changes </h2>

Periodic lubrication and oil changes in vehicles is usually done on the advice made by manufacturer. This is often done so as;

  • To make the shelf life of vehicles to be longer .
  • To allow you to not spend your money on unnecessary or costly repairs
  • To prevent dangerous breakdowns on the highways.

It is advisable to always check your owner's manual for the type and when to use lubrications and oil changes and make sure that the right oil/fluid is used for all component.

Learn more about lubrication from

  • brainly.com/question/1142755
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Identify factors that can cause a process to become out of control. Give several examples of such factors.
Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

  • Gradual deterioration of lubrication and coolant.
  • change of environmental condition such as temperature, humidity, moisture, etc.
  • Change in the properties of incoming raw material
  • An increase or decrease in the temperature of the heat treating operation
  • Debris interfering with the manufacturing process.
4 0
3 years ago
At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t
nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0
3 years ago
11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t
lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

Substituting values:

Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

5 0
3 years ago
PLEASE HURRY!!!
Naily [24]

Answer:

A

Explanation:

He should get a job in engineering to see what it's like to work in the field.

3 0
2 years ago
Read 2 more answers
Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
SSSSS [86.1K]

Answer:

The solution is attached in the attachment.

3 0
3 years ago
Read 2 more answers
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