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3 years ago

Why should engineers avoid obvious patterns?

Engineering

2 answers:

lozanna [386]3 years ago

5 0

Answer:

Patterns may cloud the vision of the design process and send the engineer off track.

Explanation:

Engineers use a cyclical process and must stay on track during this process. A key point to an engineer is being able to have the ability to use creative and critical thinking.

seraphim [82]3 years ago

5 0

Answer:

Patterns tend to impose a prescribed way of thinking that will limit possible solutions.

Explanation:

While patterns help scientists and mathematicians solve problems, engineers should avoid obvious patterns. These patterns tend to lock in a prescribed way of thinking that will limit possible solutions. While making observations is helpful, don't force them into categories. Watch for observations that are unexpected, and compare and contrast all information before deciding on a final solution.

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A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

Tech A says that thrust angle refers to the direction the front wheels are pointing. Tech B says that scrub radius refers to the

kenny6666 [7]

Both of the technicians are correct.

<u>Explanation:</u>

The thrust angle is defined as the angle formed by the imaginary line drawn perpendicular to the rear axis center line. It is used in alignment of the four wheels.

It is also used to determine the direction the front wheels are pointing. While the scrub radius is the intersecting point of the vertical center line of front tires with the imaginary line drawn from the steering knuckles.

Thus both the technicians are saying correct. The thrust angle and scrub radius are used to determine the alignment of wheels, if there is any misalignment in wheels then it needs to make it correct to prevent accidents.

4 0

3 years ago

A high molecular weight hydrocarbon gas A is fed continuously into a heated mixed flow reactor (0.1liter) where it is thermally

dimulka [17.4K]

Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

5 0

3 years ago

The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d

Colt1911 [192]

Answer:

Explanation:

The given equation is :

$\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f$

5 0

2 years ago

Implement the following Matlab code:

vagabundo [1.1K]

28384 *x soít cos estematema

3 0

3 years ago