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Brilliant_brown [7]
3 years ago
8

List three reasons for surfacing metals.

Engineering
2 answers:
Dmitrij [34]3 years ago
8 0
Chemical,fatigue,and metal transfer or adhesion.
Solnce55 [7]3 years ago
4 0

Answer:

JUST BECUASE

WHY NOT

LOOKS BETTER

Explanation:

You might be interested in
Show that -40 F is approximately equal to -40 C.
V125BC [204]

Answer: Please see below the explanation.

Explanation:

The Celsius (ºC) and Farenheit (ºF), are different temperature scales, but both have identifiable values for two important physical phenomena, water freezing point and water boiling point.

For º C, these are follows:

Freezing Point= 0º C            Boiling Point: 100ºC

For º F, we have the following:

Freezing Point= 32º F            Boiling Point: 212 ºF

Assuming that all steps (in each scale) between two following values are equal, we have 100 º C between both points in ºC, and 190º in ºF, so degrees in Farenheit are "shorter" than in Celsius, being the relationship, approximately the following:

1 º C = 9/5 ºF.

Taking into account that, by definition, 0ºC = 32º F, we can use the following equations in order to convert ºC to ºF, and viceversa:

ºC = (ºF-32) 5/9    

ºF = 32 + 9/5ºC

Now, let's calculate -40ºF, in ºC:

ºC= (-40-32)* 5/9 = -40º C

Let's do the same, converting -40ºC to ºF:

ºF = 32 + (-40)*9/5 = 32 + (-72) = -40ºF.

8 0
3 years ago
What does product integration refer to in an advanced manufacturing setting?
creativ13 [48]

Answer:

C!!

Explanation:

Combining all manufacturing processes to provide higher efficiency and fulfilling the requeriments.

6 0
4 years ago
Read 2 more answers
A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min
hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min (1)

The Laplace transform of this function is:

\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)    (3)

\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}    (4)

4 0
4 years ago
¿Qué son alimentos funcionales?
Mnenie [13.5K]
DescripciónAlimentos funcionales son aquellos alimentos que son elaborados no solo por sus características nutricionales sino también para cumplir una función específica como puede ser el mejorar la salud y reducir el riesgo de contraer enfermedades
4 0
3 years ago
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
gladu [14]

Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

4 0
4 years ago
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