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3 years ago

What factors determine tire inflation pressure?

Engineering

1 answer:

Virty [35]3 years ago

6 0

Answer:

Tire inflation can be caused by temperature and speed :)

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Energy is generated uniformly in a 6 cm thick wall. The steady-state temperature distribution

mezya [45]

Answer:

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

$T(z)=145+3000z-1500z^2$ -----1

K= 15 W/m.k

As we know that at steady state condition

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$ -----2

Where q is the heat generation per unit volume.

So from equation 1

$\dfrac{dT}{dz}=3000-3000z$

$\dfrac{d^2T}{dz^2}=-3000$

Now from equation 2

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$

$-3000+\dfrac{q}{15}=0$

So q= 45 KW

So heat generation per unit volume is 45 KW.

6 0

3 years ago

Why does my man bun not have its own erodynamics

Aloiza [94]

Answer:

umm okay for starters I have no clue lol.

7 0

3 years ago

Read 2 more answers

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

$h_{in} = 3353.1\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mixture)

$h_{out} = 890.1\,\frac{kJ}{kg}$

The power produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})$

$\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )$

$\dot W_{out} = 3863.98\,kW$

8 0

3 years ago

Only answer this if your name is riley

Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

3 0

3 years ago

Read 2 more answers

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

So the shear angle is 18.03°.

4 0

4 years ago