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3 years ago

Why do engineers (and others) use the design process?

Engineering

1 answer:

Naily [24]3 years ago

4 0

Explanation:

Engineering design is an iterative process used to identify problems and develop and improve solutions. The engineering design process can be extremely useful to any individual trying to solve a problem.

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Technician A says that the most commonly used combustion chamber types include hemispherical, and wedge. Technician B says that

Inessa05 [86]

Answer:

Technician A and Technician B both are correct.

Explanation:

Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.

Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.

6 0

3 years ago

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

aksik [14]

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!

4 0

3 years ago

21. How long can food that requires time-temperature control be left in the danger zone?

dimulka [17.4K]

Answer: A maximum of 1 hour

Explanation:

Read your lesson buddy!!

8 0

3 years ago

Read 2 more answers

. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C

MrRissso [65]

Answer:

The inside temperature, $T_{in}$ is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×( $T_s$ - $T_{\infty$ ) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

$P = \dfrac{K \times A \times \Delta T}{L}$

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

$2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}$

$T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C$

The inside temperature, $T_{in}$ = 247.99 °C ≈ 248 °C.

3 0

3 years ago