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un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti

777dan777 [17]

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7 0

2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s

Explanation:

Given:-

- The compressor exit conditions are given as follows:

Pressure ( Pe ) = 825 KPa

Temperature ( Te ) = 150°C

Velocity ( Ve ) = 10 m/s

Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

Pressure = Pi = 100 KPa

Temperature = Ti = 20.0

Velocity = Vi = 1.0 m/s

Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

Qe = Ae*Ve

Where,

Ae: The exit cross sectional area

Ae = π*de^2 / 4

Therefore,

Qe = Ve*π*de^2 / 4

Qe = 10*π*0.05^2 / 4

Qe = 0.01963 m^3 / s

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

Pe / ρe = R*Te

Where,

Te: The absolute temperature at exit

ρe: The density of air at exit

R: the specific gas constant for air = 0.287 KJ /kg.K

ρe = Pe / (R*Te)

ρe = 825 / (0.287*( 273 + 150 ) )

ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

m^ = ρe*Qe

= ( 6.79566 )*( 0.01963 )

= 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

Pi / ρi = R*Ti

Where,

Ti: The absolute temperature at inlet

ρi: The density of air at inlet

R: the specific gas constant for air = 0.287 KJ /kg.K

ρi = Pi / (R*Ti)

ρi = 100 / (0.287*( 273 + 20 ) )

ρi = 1.18918 kg/m^3

Using continuity expression:

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*1

Ai = 0.11217 m^2

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

Qi = Ai*Vi

Where,

Ai: The inlet cross sectional area

Qi = 0.11217*1

Qi = 0.11217 m^3 / s

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*Vi

Ai ( V ) = 0.11217 / Vi .... Eq 1

Inlet flow rate ( Qi ):

Qi = 0.11217 m^3 / s ... constant Eq 2

6 0

3 years ago

help me please help me please help me please help me please help me please help me please help me please help me please help me

kakasveta [241]

Answer:

With what?

Explanation:

Brainliest?

7 0

3 years ago

Read 2 more answers

How does heat conduction differ from convection?

Helga [31]

Explanation:

Conduction:

Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally, phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.

Convection:

In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.

7 0

2 years ago

Read 2 more answers