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2 years ago

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Imagine yourself as an Engineer. How would you design your own Wind Turbine to generate electricity (at least 40 watts) at your

home or in some town area. Consider all the elements of the Engineering Design Process.
*Please write detailed answer*

1 answer:

topjm [15]2 years ago

6 0

Answer: You need metal and other stuff

Explanation:

You might be interested in

A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I

Liula [17]

Answer:

fire brick / common brick : 1218 °C
common brick / magnesia : 1019 °C
magnesia / steel : 90.06 °C
heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

R = d/k

so the thermal resistances of the layers of furnace wall are ...

R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

What’s cat plus cat <br><br><br><br><br><br><br><br> ?

iris [78.8K]

Uhmmmmm, a kitten...? Lol

5 0

3 years ago

Read 2 more answers

A timing light checks the ignition timing in relation to the ____ position.

kogti [31]

Answer:

The timing light is connected to the ignition circuit and used to illuminate the timing marks on the engine's crankshaft pulley or flywheel, with the engine running. The apparent position of the marks, frozen by the stroboscopic effect, indicates the current timing of the spark in relation to piston position.

Explanation:

:)

8 0

2 years ago

A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T

solmaris [256]

Answer:

Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m

Explanation:

Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.

We know that heat transfer by convection is given by

$\dot{Q}=hA(T-T_{\infty })$

where,

h = heat transfer coefficient = 10.45 $W/m^{2}K$ (free convection in air)

A = Surface Area of the pipe

Applying the given values in the above formula we get

$\dot{Q}=10.45\times \pi DL\times (130+273-(24+273))\\\\\frac{\dot{Q}}{L}=10.45\times 0.15\times \pi \times (130-24)\\\\\frac{\dot{Q}}{L}=521.99W/m$

5 0

3 years ago