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4 years ago

Describe the differences, if any, between fin efficiency and fin resistance.

Engineering

1 answer:

mario62 [17]4 years ago

7 0

Answer:

$\eta =\dfrac{1}{mL}$

$R=\sqrt {\dfrac{KA}{hP}}$

Explanation:

Fin efficiency:

Fin efficiency is the ratio of actual heat transfer through fin to the maximum heat transfer through fin.

For infinite long fin:

Actual heat transfer

$q=\sqrt{hPKA}\Delta$

So the efficiency of fin

$\eta =\dfrac{1}{mL}$

Where

$m =\sqrt{\dfrac{hP}{KA}}$

h is heat transfer coefficient,P is the perimeter,K is the thermal conductivity and A is the cross sectional area.

$q=\sqrt{hPKA}\Delta$ \

From equation we can say that fin resistance

$R=\sqrt {\dfrac{KA}{hP}}$

Thermal resistance offer resistance to flow of heat.

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Explain three (3) modes of heat transfer in air conditioning system.

LenKa [72]

Answer:

1. Conduction

2. Convection

3. Radiation

Explanation:

The 3 modes of heat transfer i an air conditioning system:

1. Conduction:

The transfer of heat by conduction takes place in solid and is when the conduction takes place as a result of direct contact in between the interacting material which transfer the heat energy from particle to particle thus conducting the heat through out the system.

2. Convection:

The other mode for the transfer of heat which takes place especially in fluids - gases and liquids is through the technique of convection in which the transfer of heat takes place by the circular motion of the atoms and molecules of the fluid which carries the heat energy and results in the distribution of the heated fluid in the entire system thus transferring all the heat energy in the entire system.

3. Radiation:

The third mode of heat transfer in the air conditioning system is through radiation. This method transfers the heat by making use of the electro-magnetic radiation in the infra red spectrum where the waves of the spectrum transfers the heat energy with the help of a medium or without any medium at all.

Thus making the radiation method of heat transfer as the only method out of the three methods which does not require the material medium for the transfer of heat energy.

4 0

3 years ago

A particular cloud-to-ground lightning strike lasts 500 ÂµÂµsec and delivers 30 kA across a potential difference of 100 MV. Assu

sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

the residential retail value of the energy delivered by the strike is $ 40.83

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

we know that; I = Q/t

so Q = I × t

we substitute

Q = 30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹² × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

given that; 1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.

7 0

3 years ago

150 lb of force is applied at the end of a shaft that is 2 1/2 ft long. It is applied

Alona [7]

Answer:

Torque: 500

Power: 1700

Explanation:

Since the force is 150 lbs and if the shaft is rotating at 350 rpms, then you will need 1700 for the power in order to keep it running.

5 0

3 years ago

Describe two components of the MEDC mechanized agricultural system that may lead to an increase in water stress

Zina [86]

Mechanized agriculture refers to agricultural machinery capable of facilitating and enhancing agriculture. Mechanized agriculture does not require moving the soils and therefore is hard for water to enter into it.

<h3>What is mechanized agriculture?</h3>

Mechanized agriculture refers to the use of different technologies (e.g., precision agriculture GPs) in order to increase farmworker productivity.

Mechanized agriculture is a common practice in developed countries (MEDCs), whereas this practice is relatively uncommon in developing countries (LEDCs).

The techniques of mechanized agriculture do not require moving the soils and therefore is hard for water to enter into it.

Learn more about mechanized agriculture here:

brainly.com/question/1543791

8 0

3 years ago

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m

inn [45]

Answer:

$2.9\times 10^{-6}$

Explanation:

$Q_s$ = Energy for defect formation = 1.86 eV

T = Temperature = $573^{\circ}\text{C}=573+273.15=846.15\ \text{K}$

k = Boltzmann constant = $8.62\times 10^{-5}\ \text{eV/K}$

The fraction of lattice sites that are Schottky defects is given by

$\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}$

The required ratio is $2.9\times 10^{-6}$ .

6 0

3 years ago