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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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How is an orthographic drawing similar to or different from an isometric drawing?
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An isometrical drawing is a nearly 3d drawing showing the object's width and depth in a complete image, from each curved plane of the orthhographic view, the viewpoint is at a 45 degree angle. From an observations point of view, isometric differs, since all longitudes are true.
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Explanation:

7 0
3 years ago
The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc
Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

\theta = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑f_{y} = 0, ∑M = 0

If ∑M_{A} = 0

⇒F_{BC}×0.9 - P × 0.6 = 0

⇒F_{BC}×3 - P × 2 = 0

⇒F_{BC} = \frac{2P}{3}

If ∑M_{B} = 0

⇒F_{AD}×0.9 = P × 0.3

⇒F_{AD} ×3 = P

⇒F_{AD} = \frac{P}{3}

Now,

Area, A = \frac{\pi }{4} X (0.012)^{2} = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , \delta = \frac{P l}{A E}

Now,

\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 9.1626 × 10⁻⁹ P

\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 1.83253 × 10⁻⁸ P

Given that,

\theta = 0.015°

⇒\theta = 2.618 × 10⁻⁴ rad

So,

\theta =  \frac{\delta_{BC} - \delta_{AD}}{0.9}

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0
3 years ago
A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a)
Bess [88]

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

v=\sqrt{2gh}

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

v=\sqrt{2\times 9.81\times 90}=42.02m/s

now we know the relation between discharge and velocity as dictated by contuinity equation is

Q=V\times Area

Applying values in the above equation and solving for area we get

Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}

The circular area is related to diameter as

Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m

Thus the diameter of the nozzle is 0.246 meters

7 0
3 years ago
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