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2 years ago

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It is an important part of the differential maintenance which purpose is to make smoother the differential operation by lubricat

ing all components, and to prevent metal to metal contact to reduce wear.
Fueling
Furnishing
Greasing
Lubrication

1 answer:

masha68 [24]2 years ago

4 0

Answer:

lubrication

Explanation:

MARK ME BRAINLEIST

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Two players find themselves in a legal battle over a patent. The patent is worth 20 for each player, so the winner would receive

Alborosie

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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3 years ago

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

3 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a

AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2) + x(2) * h(fg2)

so

exit quality = $\frac{h1 - h(f2)}{h(fg2)}$

exit quality = $\frac{332.17- 9.04}{1382.73)}$

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0

2 years ago

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s 2 , d

sattari [20]

Answer:

a = 1.68m/ $S^{2}$

Explanation:

Please kindly find the attached file for explanations

3 0

3 years ago

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar

Saturation temperature corresponding to saturation pressure =40°C

$h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}$

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

$Total\ heat\ =C_p\Delta T+\Delta h$

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

$h_1=\dfrac{\Delta h}{\Delta T}$

$h_1=\dfrac{2406}{20}$

$h_1=120.3\frac{KJ}{kg-m^2K}$

Now film coefficient after inclusion of sensible heat

$h_2=\dfrac{total\ heat}{\Delta T}$

$h_2=\dfrac{2,544}{20}$

$h_2=127.2\frac{KJ}{kg-m^2K}$

$So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100$

$=\dfrac{127.2-120.3}{120.3}\times 100$

=5.75 %

So Percentage change 5.75 %.

3 0

2 years ago