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konstantin123 [22]
3 years ago
12

Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says t

hat a way to prevent galvanic corrosion is to reuse coated bolts. Who is right
Engineering
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

Technician A

Explanation:

Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.

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Explanation:

i did this last year

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Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

4 0
3 years ago
A 2 m3 rigid tank initially contains air at 100 kPa and 22 degrees C. The tank is connected to a supply line through a valve. Ai
Vaselesa [24]

Answer:

a. 9.58kgs b. 340.32KJ

Explanation:

Volume of tank= 2m³

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Initial Temperature Ti= 22 C= 295K

Line Pressure P₁= 600 KPa

Line Temperature T= 22 C= 295K

Final Pressure P2= 600 KPa

Final Temperature T2= 77 C= 350K

Use Ideal Gas Equation

PV= mRT

P₁V₁= m₁RT₁

m₁= (100 x 2)/(0.287 x 295) = 2.3622kg

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m₁ + mi = m₂ + me

as per above condition me= 0

mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg

Applying energy equation

m₁u₁ + mihi + Q = m₂u₂ + mehe + W

me and W=0

m₁u₁ + mihi + Q = m₂u₂

m₁CvT₁ + miCpTi + Q =  m₂CvT₂

Q =   m₂CvT₂- m₁CvT₁ - miCpTi

Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)

Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)

4 0
4 years ago
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