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3 years ago

What does it mean to say that PEER is a data-driven, consumer-centric, and comprehensive system?

Engineering

2 answers:

Reika [66]3 years ago

3 0

Answer:

have you heard of gnoogle?

Explanation:have you heard of goongle?

vodomira [7]3 years ago

3 0

Answer:

PEER establishes clear performance metrics, it serves consumer needs, and it covers all aspects of energy system performance.

Explanation:

on edg. just did it

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Determine the following for a south facing surface at 30� slope in

lesantik [10]

Answer:

z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m², Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM

Explanation:

A. Zenith Angle:

As we know that,

Zenith angle=z=90⁰-α=L(latitude)=29.68⁰

Another way to do it is to find α first,

At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as

sinα=cosL

α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°

B. Angle of incidence:

angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°

i=cos⁻¹(sin92.32°)=0.32°

C. Beam Radiation:

First we need to calculate extra terrestrial radiations

Iext.=1353[1+0.034cos(360n/365)]

where n=264

=1345 W/m²

Now,

Beam Radiation=CIext⁻ⁿ

where n=0.1/sin60.32°

Beam Radiation = 1097.2 W/m²

D. Diffude Radiation:

difuse radiation = Id = 0.0921ₙcos²(β/2)

where β=30°

Id = 94.2 W/m²

E. Reflected Radiations:

Ir=pIn(sinα+0.092)sin²(β/2)

= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)

= 14.1W/m²

F. Total Radiation:

total radiation = beam radiation + diffuse radiation + reflected raddiation

= 1205.4 W/m²

G. Local Time:

LST= ST-ET-(lₓ-l(local))4min/₀

= 12:00-7.9min-(75°-82.27°)4min/₀

=12:21PM

Local time

LDT=LST+=12:21+1:00=1:21PM

5 0

3 years ago

The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy

DanielleElmas [232]

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

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3 years ago

Matthew bought 13 used video games that were on sale at a store. He paid $84.37 for the games. If each video game cost the same

Gre4nikov [31]

$6.49. hoped this helped:)

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3 years ago

Read 2 more answers

provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki

Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0

3 years ago

A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve t

lesantik [10]

Answer:

a) 0.50613

b) 22.639 kJ

Explanation:

From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C

first step : calculate the volume of R-13a ( values gotten from table A-11 )

V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3

next : calculate final specific volume ( v2 )

v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg

<u>a) Calculate the mass of refrigerant that entered the tank </u>

v2 = Vf + x2 * Vfg

v2 = Vf + [ x2 * ( Vg - Vf ) ] ----- ( 1 )

where: Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg ( insert values into equation 1 above )

x2 = ( 0.01652 - 0.0008261 ) / 0.031834

= 0.50613 ( mass of refrigerant that entered tank )

<u>b) Calculate the amount of heat transfer </u>

Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )

uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg

therefore U2 = 164.737 Kj/kg

The mass balance ( me ) = m1 - m2 --- ( 3 )

energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he

therefore Qin = 41.184 - 422.2 + 403.655 = 22.639 kJ

3 0

3 years ago