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2 years ago

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An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

4/m (i.e., per meter of cablelength). The cable is in an environment having a tem-perature of 30C, and the total coefficient associatedwith convection and radiation between the cable andthe environment is approximately 25 W/m2K.(a) If the cable is bare, what is its surface temperature

1 answer:

KatRina [158]2 years ago

6 0

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + $\frac{q}{h\pi Di}$

Ts = $30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000} }$

$Ts=30+\frac{294}{0.3928} \\$

$Ts =30+748.4\\Ts = 778.4$

The surface temperature is therefore 778.4°C if the cable is bare

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Answer:

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Explanation:

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We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

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h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

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3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

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Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

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$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

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$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

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3 years ago

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Answer: they are connected in series.

Explanation:

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3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

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Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

where D is constant

then

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$x^2_1$ / t₁ = $x^2_2$ / t₂

$x^2_1$ t₂ = t₁ $x^2_2$

t₂ = t₁ $x^2_2$ / $x^2_1$

t₂ = ( $x^2_2$ / $x^2_1$ )t₁

t₂ = $($ $x_2$ / $x_1$ $)^2$ × t₁

so we substitute

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Answer:

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