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3 years ago

5

Identify renewable energy sources you will propose. Explain the key elements to your solution and the basic technical principles

involved.

1 answer:

MArishka [77]3 years ago

3 0

Answer:

A renewable electricity generation technology harnesses a naturally existing energy. But they have other features that a few fringe customers value.

Explanation:

You might be interested in

Engineer Smith, who is licensed in several jurisdictions, recently had his license acted upon by a licensing authority in one of

Nimfa-mama [501]

Answer:

Assuming the infraction in the other jurisdiction is an infraction in Florida, having approximately the same penalty imposed by the FBPE as was imposed in the other jurisdiction

Explanation:

The idea of 8 hours continuation in the options is irrelevant to this case. Rather, Engineer Smith should expect about the same penalty imposition.

3 0

3 years ago

An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a

DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π × 0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0

3 years ago

What is a heat engine? State its efficiency (a sketch would be appropriate).

dimulka [17.4K]

Answer:

$\eta =\dfrac{Q_1-Q_2}{Q_1}$

Explanation:

Heat engine

Heat engine is a device which produce work while consuming some amount of energy .This engine works in cycle.In other words we can say that Heat engine is a device which covert thermal or chemical energy in to mechanical energy and after that this mechanical energy can be use for producing mechanical work.

Heat engine takes heat from high temperature and rejects heat to lower temperature and producing work.

We know that efficiency given as

$\eta =\dfrac{out\ put}{in\ put}$

From diagram we can say that

$Q_1=Q_2+W$

So efficiency

$\eta =\dfrac{w}{Q_1}$

$\eta =\dfrac{Q_1-Q_2}{Q_1}$

3 0

3 years ago

How do scientists go about their work? do you agree or disagree with the statement ?

likoan [24]

It should be noted that scientists do their work by making hypotheses, testing them through and formulating conclusions based on the evidence

<h3>Who is a scientist?</h3>

It should be noted that a scientist is a professional that conducts and gathers research to further knowledge in a particular area.

This is a person who has expert knowledge of one or more of the natural or physical sciences.

Therefore, it should be noted that scientists do their work by making hypotheses, testing them through and formulating conclusions based on the evidence

Learn more about scientist on:

brainly.com/question/17216882

#SPJ1

3 0

1 year ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

3 years ago