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Pachacha [2.7K]
3 years ago
15

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

ter and is situated in a surrounding ambient temperature of 15°C where the natural convection heat transfer coefficient is 12 W/m2·K. If the efficiency of the electrical heater to transfer heat to the plate is 80%, determine the electric power that the heater needs to keep the surface temperature of the plate at 220°C.
Engineering
1 answer:
denis23 [38]3 years ago
5 0

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature T_\infity = 15 degree celcius

T_S =  220 + 273 = 473 K

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

Q = hA(T_S - T_{\infty}) \eta

= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8

Q = 125.538 W

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Carbon dating for archeological materials is based on the fact that a plant, after its death, stops absorbing radioactive C-14 a
Olin [163]

Answer:

 t = 2212 years

Explanation:

In radioactive decay processes it is described by the equation

         N = N₀ e^{-\lambda t}

to calculate the activity

        T_{1/2} = log 2 /λ

        λ = log 2 / T_{1/2}

     

        λ = log 2 /5715

        λ = 5.267 10⁻⁵

now the amount of carbon 14 is N₀ = 0.1%, the sample contains an amount of N = 0.089%

          N / N₀ = e^{-\lambda t}

          -λ t = ln N / N₀

           t = - 1 /λ  ln N /N₀

           t = 1 / 5.267 10⁻⁵   ln (0.089 / 0.1)

           t = 2,212 10³ years

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8 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is
wariber [46]

Answer:

Explanation:

From the given information:

Strain fracture toughness K_k= 75 MPa\sqrt{m}

Tensile stress \sigma = 361 MPa

Value of Y = 1.03

Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:

a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\  a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\  a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2  \\ \\ \mathbf{ a_c = 12.95 \ mm}

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3 years ago
A precision miling machince wighing 1000lb is supported on a rubber mount. The force deflection relationship of the rubber mount
Alisiya [41]

Answer:

0.2846 in

Explanation:

The static equilibrium position of the rubber mount ( x^*), under the weight of the milling machine,  can be determined from:

1000=3500(x^*)+ 55(x^*)^3\\\\55(x^*)^3+ 3500x^*-1000=0\\\\Solving\ for\ the\ roots\ using \ a \ calculator\ or\ matlab\ gives:\\ \\x^*_1=0.28535\\x^*_2=-0.14267+7.89107i\\x^*_3=-0.14267-7.89107i\\\\We\ are\ using\ the \ real\ root\ which\ is\ x^*_1=0.28535\\

k=\frac{dF}{dx}|_{x^*}\\ \\k=\frac{d}{dx}(3500x+55x^3)|_{x^*}\\\\k=3500+165x^2|_{x^*}\\\\k=3500+165(x^*)^2\\\\k=3500+165(0.28535)^2=3513.435\ lb/in

The static equilibrium position is at:

x=\frac{F}{k}=\frac{1000}{3513.435}  =0.2846\ in

8 0
3 years ago
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