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1 year ago

Not all projects that engineers work on will have human factors involved.

Engineering

1 answer:

Sliva [168]1 year ago

6 0

Projects that don't have human factors involved are:

The Use of Artificial Intelligence to Power Road Construction Robots;
End to End robotic Automobile production line.

Other projects that require human factors are;

Traditional road construction projects;
Traditional building construction projects.

<h3>What are the human factors for Road Construction?</h3>

The role of human factors includes but is not limited to:

Operation of Heavy-Duty Vehicles
Human Resources (taking care of people)
Finance operations (Finance and Accounting) etc.

At construction sites, excavation is one of the roles of humans. Another instance is injuries, compensation, etc is the work of the Human Resources Officer.

Human factors include environmental, organizational, and job elements, as well as human and individual qualities that influence workplace behavior and can have an impact on health and safety.

Human-factors engineering has been applied to simple equipment such as highway signs, telephone sets, hand tools, and stoves, as well as a wide range of contemporary, sophisticated complexes such as data systems, automated warehouses, and factories. robots, and space vehicles.

Learn more about Road Construction:
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3 years ago

Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg

Dennis_Churaev [7]

Answer:

$h = 1429.74\,\frac{kJ}{kg}$

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The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

$\nu_{f} = 0.001080\,\frac{m^{3}}{kg}$

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Since $\nu_{f} < \nu < \nu_{g}$ , it is a liquid-vapor mixture. The quality of the mixture is:

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$x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }$

$x = 0.392$

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$h_{f} = 589.16\,\frac{kJ}{kg}$

$h_{g} = 2733.5\,\frac{kJ}{kg}$

The specific enthalpy is:

$h = h_{f}+x\cdot (h_{g}-h_{f})$

$h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)$

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