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3 years ago

5

An industrial system is rated at 600 v the system includes a partially exposed terminal block thats mounted on a bulkhead a new

electrical apprentice who isnt considered to be qualified according to OSHA's standards approches the energized block the closest the apprentice should come to the block is?

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

10 ft or 3.048 m (for unqualified person)

1.09 ft or 0.33 m (for qualified person)

Explanation:

OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.

For a unqualified person:

The minimum approach distance is 10 ft or 3.048 m

For a qualified person:

For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m

Since 600V lies in the above range hence it is applicable on 600V.

OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)

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Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,

Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take $p_o$ = 880 kg/m³ and $p_{glycerol$ = 1260 kg/m³

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

Explanation:

Given that;

Inlet velocity of Glycerin, $V_A$ = 6 m/s

Inlet velocity of oil, $V_B$ = 3 m/s

Density velocity of glycerin, $p_{glycerol$ = 1260 kg/m³

Density velocity of glycerin, Take $p_o$ = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, $d_A$ = 100 mm = 0.1 m

Diameter of oil pipe, $d_B$ = 80 mm = 0.08 m

Diameter of outlet pipe $d_C$ = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

$Q_A + Q_B = Q_C$

$A_Av_A + A_Bv_B = A_Cv_C$

π/4 × ( $d_A$ )² $v_A$ + π/4 × ( $d_B$ )² $v_B$ = π/4 × ( $d_C$ )² $v_C$

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)² $v_C$

0.04712 + 0.0150796 = 0.0113097 $v_C$

0.0621996 = 0.0113097 $v_C$

$v_C$ = 0.0621996 / 0.0113097

$v_C$ = 5.5 m/s

Now we apply the mass flow rate condition

$m_A + m_B = m_C$

$p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C$

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p

72.6412 = 0.06220335p

p = 72.6412 / 0.06220335

p = 1167.8 kg/m³

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

4 0

3 years ago

A metal alloy has been tested in a tensile test with the following results for the flow curve

MA_775_DIABLO [31]

Answer:

dhjdjs

Explanation:

ehejsikajaudiieisisjsjsjsjsjjsjsjsjsjs

4 0

3 years ago

Viteza unui mobil care se deplasează cu accelerație constantă crește de la 3,2 m/s la 5,2 m/s în timp de 8 s. Accelerația mobilu

Verizon [17]

Answer:

$a=0.25\ m/s^2$

Explanation:

Initial speed of the mobile = 3.2 m/s

Final speed of the mobile = 5.2 m/s

Time, t = 8 s

We need to find the acceleration of the mobile. It can be given by the change in velocity divided by time. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(5.2-3.2)\ m/s}{8\ s}\\\\=0.25\ m/s^2$

So, the acceleration of the mobile is $0.25\ m/s^2$ .

3 0

3 years ago

A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p

Cerrena [4.2K]

Answer:

4.26

Explanation:

The wavelength λ is given by:

$\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\$

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

$Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j$

$\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\ \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26$

6 0

3 years ago

Determine ten different beam loading values that will be used in lab to end load a cantilever beam using weights. Load values sh

nasty-shy [4]

Answer:

1st value = 1.828 * 10 ^9 gm/m^2 ------- 10th value = 7.312 * 10^9 gm/m^2

Explanation:

initial load ( Wp) = 200 g

W1 ( value by which load values increase ) = 100 g

Ten different beam loading values :

Wp + w1 = 300g ----- p1

Wp + 2W1 = 400g ---- p2

Wp + 3W1 = 500g ----- p3 ----------------- Wp + 10W1 = 1200g ---- p10

x = 10.25" = 0.26 m

b = 1.0" = 0.0254 m

t = 0.125" = 3.175 * 10^-3 m

using the following value to determine the load values at different beam loading values

attached below is the remaining part fo the solution

5 0

3 years ago