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Kruka [31]
4 years ago
7

9.4 Lead sheet is pop-riveted with special bronze fasteners. The total exposed area of the sheet is 100 times that of the fasten

ers. The environment for the assembly is seawater. Which metal will corrode? If half as many fasteners are used, what influence would this have on the total rate of corrosion? How could corrosion be reduced?
Engineering
2 answers:
Alinara [238K]4 years ago
7 0

Answer:

(a) Lead will corrode

(b) fasteners does not have any influence on the rate of corrosion.

(c) The use of protective coating like paint, powder

Explanation:

(a) Which metal will corrode: Since bronze is resistant to oxidation, corrosion cannot take place. Lead will corrode due to its reaction with the seawater.

(b) Influence on the total rate of corrosion; No matter the fasteners used, corrosion will still take place as far as the surface of the metal is exposed. therefore fasteners does not have any influence on the rate of corrosion.

(c) there are so many ways to reduce corrosion, some of which includes;

-- The use of protective coating like paint, powder

-- By keeping the surface of the metal dry

-- Not exposing the metal to moisture

---By electroplating it with another metal

joja [24]4 years ago
4 0

Answer:

Questions are answered in the explanation section.

Explanation:

a) Of both metals, lead will corrode, since it reacts with seawater to form oxides, chlorides, etc., while bronze is more resistant to oxidation.

b) The corrosion rate will remain constant, because the concentration of seawater does not change over time, therefore, the corrosion rate is independent of the area that is exposed to seawater.

c) Corrosion can be reduced by using protective coatings, so that the metal is isolated from sea water, for example, paints, acrylic powders, etc.

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An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumpt
scZoUnD [109]

Answer:

s =0.21\ kg/Kw.hr

Explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

\eta =\dfrac{BP}{\dot{m_f}\times CV}

Where BP is the brake power

\eta =\dfrac{BP}{\dot{m_f}\times CV}

0.379 =\dfrac{BP}{\dot{m_f}\times 44000}

\dfrac{BP}{\dot{m_f}}=16676

Brake specific fuel consumption (s)

s =\dfrac{\dot{m_f}}{BP}

s =\dfrac{3600\times \dot{m_f}}{BP}

s =\dfrac{3600}{16,676}\ kg/Kw.hr

s =0.21\ kg/Kw.hr

7 0
3 years ago
A plane surface 25 cm wide has its temperature maintained at 80°C. Atmospheric air at 25°C ows parallel to the surface
Gre4nikov [31]

Answer:

See the detailed answer in attached file.

Explanation :

Download docx
3 0
3 years ago
Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel f
Morgarella [4.7K]

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

T_{ \infty} = 27^0C= 273+27 = 300K

T_s= 127^0C = (127+273)= 400K

P = 1atm

Film temperature

T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K

dynamic viscosity =

\mu =20.9096\times 10^{-6} m^2/sec

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}

=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043

we have,

Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k

we have,

heat transfer rate from top plate

\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w

7 0
3 years ago
On a ship the price gift is 24 euros .What is the difference in the price on a day when the exchange rate is £1=2378
bija089 [108]

Answer:

The answer is "\$29.7072".

Explanation:

Please find the complete question in the attachment file.

As the original prices are 24 euros, \$ 1.2378 must be multiplied by 24.  

\to 1.2378 \times 24 = \$29.7072

Therefore the new price value will be \$ 29.7072

4 0
3 years ago
PLEASE HELP
Anna11 [10]
PLEASE HELP
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A. Satellites that enter Earth’s atmosphere land mostly in the ocean.


B. NASA does a good job informing people of when satellites will enter the atmosphere.


C. Satellites that enter Earth’s atmosphere land mostly in unpopulated areas.


D. Much of a satellite is destroyed during the process of entering Earth’s atmosphere.
8 0
3 years ago
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