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Vanyuwa [196]
2 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?
Engineering
1 answer:
Nastasia [14]2 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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2 years ago
A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.
AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass m_{in}+m_{out}=\bigtriangleup m_{system}

so, m_e=m_1-m_2

Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1

#Also, given the properties of water as;

(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg

#We assume constant properties for the steam at average temperatures:h_e=\approx(h_1+h_2)/2

#Replace known values in the equation above;h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg

#Using the mass and energy balance relations;

m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg

#We have Q_i_n+m_eh_e=m_2u_2-m_1u_1: we replace the known values in the equation as;

Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0
3 years ago
How can I draw this image in 2D form
Ket [755]

Answer:

no it is not 2D

Explanation:

it is 3D

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