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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:

1) The three possible assumptions are

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c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

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c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}}  \right )^{K-1}

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

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