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ANEK [815]
3 years ago
13

Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

ción de temperatura en función de "x" el cual es la distancia desde la cara izquierda de una pared, está dado por la siguiente función: T(x) = 150x2 -30x, donde x está en metros. Calcúlese el flujo de calor por unidad de área cuando x=0 y x=30, para cada caso menciones si se está enfriando o calentando el sólido.
Engineering
1 answer:
aksik [14]3 years ago
4 0

Answer:

Para x=0:

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-2.4 W/m^{2}  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

\phi=-k\frac{dT}{dx}(1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

\frac{dT(x)}{dx}=300x-30

Remplezando esta derivada en (1):

\phi=-0.04(300x-30)

Para x=0:

\phi=0.04(30)

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-0.04(300*0.3-30)

\phi=-2.4 W/m^{2}    

Espero que te haya ayudado!

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Answer:

You have to know the basic mechanics to help you correctly use the motorcycle, and you also have to have the right mindset - knowing you can ride it, and having determination. All of these factors are the primary mental skills you need for riding a motorcycle.

5 0
2 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
What are the two main what are the two main concerns in the research of fluid power efficiency?
Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on

1) Advances in fluid power

2) Making use of the advantages

3) Making use of the other externally available technological advantages

4) Giving allowance for disadvantages

Areas of interest in advances in fluid power are;

a. Computer optimized flow

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3 0
3 years ago
Read 2 more answers
A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30
hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

V_z=V_z_o+I_zr_z

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

V_z_o=V_z-I_zr_z

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V

The value of I_L is given as 5 mA

Now the expression of current is as

I=I_z+I_L\\I=10mA+5mA\\I=15 mA

Now the resistance is calculated as

R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA

The load voltage is 7.485 V

8 0
3 years ago
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