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JulijaS [17]
3 years ago
13

The minimum safe working distance from exposed electrical conductors

Engineering
1 answer:
SVEN [57.7K]3 years ago
4 0

Answer:

b

Explanation:

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What is the measurment unit of permeability?​
nikitadnepr [17]

Answer:

Henries :)

Explanation:

I looked it up

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3 years ago
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Marat540 [252]

Answer:

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Explanation:

7 0
2 years ago
A tank contains liquid nitrogen at -190℃ is suspended in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3
Svetach [21]

Answer:

C) 0.182 W

Explanation:

The effective conductive section is:

3 * A = 3 * \frac{\pi * d^2}{4} = 3 * \frac{\pi * (0.008 m)^2}{4} = 1.5e-4 m^2

The thermal conductivity of stainless steel is

k = 18 w/(m * K)

Heat conduction in a rod follows this equation:

q = k * A * \frac{(t2 - t1)}{l} = 18 * 1.5e-4 * \frac{(15 - (-190))}{3} = 0.182 W

7 0
3 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
2 years ago
Glyphicons is mainly used for​
-Dominant- [34]

Glyphicons are icon fonts which you can use in your web projects. Glyphicons Halflings are not free and require licensing, however their creator has made them available for Bootstrap projects free of cost.

7 0
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