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miss Akunina [59]
4 years ago
13

A flat-plate solar collector is 2 m long and 1 m wide and is inclined 60o to the horizontal. The cover plate is separated from t

he absorber plate by an air gap of 2 cm thick. If the temperatures of the cover plate and the absorber plate are 305 K and 335 K, respectively. Calculate the convective heat loss. Take the pressure at 1 atm.

Engineering
1 answer:
Deffense [45]4 years ago
5 0

Answer:

164.4 W

Explanation:

We can define Heat loss as a measure of the total transfer of heat through the fabric of a building from inside to the outside, either from conduction, convection, radiation, or any combination of the these.

Please kindly check attachment for the solution of the heat loss as asked in the question.

The attached file have the solved problem.

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State the number of terms for each of the following algebraic expression 2x+1
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Answer:

Expressions are made up of terms.

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Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

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Led test lights are used to test circuits that include controllers and computers. True or false
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True

Explanation:

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3 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

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3 years ago
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