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anzhelika [568]
3 years ago
10

Teaching how to characterize and implement high speed power devices for tomorrow's engineers

Engineering
1 answer:
Simora [160]3 years ago
6 0

Answer: Teaching how to characterize and implement high speed power devices for tomorrow's engineers

Explanation:

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Steam flows steadily through a turbine at a rate of 420 kg/min. The enthalpy of the steam decreases by 600 kJ/kg as it flows ste
Ghella [55]

Answer:

the rate of heat loss from the steam turbine  is Q = 200 kW

Explanation:

From the first law of thermodynamics applied to open systems

Q-W₀ = F*(ΔH + ΔK + ΔV)

where

Q= heat loss

W₀= power generated by the turbine

F= mass flow

ΔH = enthalpy change

ΔK = kinetic energy change

ΔV = potencial energy change

If we neglect the changes in potential and kinetic energy compared with the change in enthalpy , then

Q-W₀ = F*ΔH

Q =  F*ΔH+ W₀

replacing values

Q =  F*ΔH+ W₀ = 420 kg/min * (-600 kJ/kg) * 1 min/60 s * 1 MW/1000 kW + 4 MW = -0.2 MW = -200 kW (negative sign comes from outflow of energy)

4 0
3 years ago
2. What of the following demonstrates a
Dmitry_Shevchenko [17]
B. Always cut away from the body
8 0
3 years ago
Ashrae standard 15 -2013 require that each machinery room activate an alarm and mechanical ventillation?
Kobotan [32]

ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).

<h3>What is ASHRAE Standard 15 - 1994?</h3>
  • The key standards guiding refrigerant identification and usage developed by ASHRAE have been revised to comply with government regulations and achieve improved performance.
  • Standards 15 and 34 provide critical guidance to manufacturers, design engineers, and operators who must stay up to date on new air conditioning and refrigeration requirements.
  • Standard 34 describes a shorthand method of naming refrigerants and assigns safety classifications based on toxicity and flammability data, whereas Standard 15 establishes procedures for operating equipment and systems when those refrigerants are used.
  • Before refrigerant concentrations exceed the TLV-TWA, each machinery room must activate an alarm and mechanical ventilation, according to ASHRAE Standard 15 - 1994 (Threshold Limit Value - Time Weighted Average).

Therefore, ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed the TLV-TWA (Threshold Limit Value - Time Weighted Average).

Know more about ASHRAE Standard here:

brainly.com/question/14483054

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The correct question is given below:
ASHRAE Standard 15 - 1994 requires that each machinery room must activate an alarm and mechanical ventilation before refrigerant concentrations exceed:

5 0
2 years ago
Which other factors do you need to consider when preparing to move gondolas?
Delvig [45]

The factors that  you need to consider when preparing to move gondolas are:

  • Height
  • width
  • length
  • color.

<h3>How do you move a gondola?</h3>

In choosing gondola one need to look at some  factor in decisions such as height.

Know that it is good to  consider the construction material and also the center wall configuration for its building also.

The factors that  you need to consider when preparing to move gondolas are:

  • Height
  • width
  • length
  • color.

Learn more about gondola from

brainly.com/question/10652274

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5 0
2 years ago
Two gage marks are placed exactly 320 mm apart on a 12-mm diameter solid metal rod. When an axial tension force of 15 kN is appl
katrin2010 [14]

Answer:

E = 80.381 GPa

Explanation:

given,

Distance between two gauges, L = 320 mm

diameter,d = 12 mm

Load, P = 15 x 10³ N

distance between gauge marks = 320.5280 mm

Now, elongation of rod is equal to

δ = 320.5280 - 320 = 0.5280 mm

Using elongation formula

\delta = \dfrac{PL}{AE}

E = \dfrac{PL}{A\delta}

E = \dfrac{15\times 10^3\times 320}{\dfrac{\pi}{4}\times d^2\times 0.5280}

E = \dfrac{15\times 10^3\times 320}{\dfrac{\pi}{4}\times 12^2\times 0.5280}

E = 80381.28 MPa

E = 80.381 x 10³ MPa

E = 80.381 GPa

Hence, Modulus of elasticity of the metal is equal to E = 80.381 GPa

4 0
4 years ago
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