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2 years ago

Is making design tradeoffs the same as, or different from, the practice of combining the best parts of different solutions

Engineering

2 answers:

mezya [45]2 years ago

5 0

Answer:

Explanation:

Using a process of informed decision-making, the designer or design team compares different solutions to the requirements of the problem and either chooses

sineoko [7]2 years ago

4 0

Answer:The engineering design process usually begins by stating a need or want as a clearly defined challenge in the form of a statement with criteria and constraints. For example, a group of engineers might be given the task of designing, for example, a cell phone with a particular set of features, of a particular size and weight, with a certain minimum battery life, and that is able to be manufactured at a particular cost. Criteria are characteristics of a successful solution, such as the desired function or a particular level of efficiency. Constraints are limitations on the design, such as available funds, resources, or time. Together, the criteria and constraints are referred to as the requirements for a successful solution.

Once the challenge is defined, the next steps are often to investigate relevant scientific and technical information and the way that similar challenges have been solved in the past and then to generate various possible solutions. This generation of potential solutions is the most creative part of the design process and is often aided by sketching and discussion. Using a process of informed decision-making, the designer or design team compares different solutions to the requirements of the problem and either chooses the most promising solution or synthesizes several ideas into an even more promising potential solution. The next step is usually to try out the solution by constructing a model, prototype (first of its kind), or simulation and then testing it to see how well it meets the criteria and falls within the constraints. An additional characteristic of engineering design is that ideas are tested before investing too much time, money, or effort

hope it helped

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Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

Furkat [3]

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

$W = \dot{m}c_p \Delta T$

Where

$C_p =$ Specific heat

$\dot{m}$ = Mass flow rate

$\Delta T$ = Change in Temperature

Our values are given as

$C_p = 1.005kJ/kgK \rightarrow$ Specific heat of air

$T_1 = 50\°C$

$\dot{m} = 2kg/s$

$W = 8kW$

From our equation we have that

$W = \dot{m}c_p \Delta T$

$W = \dot{m}c_p (T_2-T_1)$

Rearrange to find $T_2$

$T_2 = \frac{W}{\dot{m}c_p}+T_1$

Replacing

$T_2 = \frac{8}{2*1.005}+(50+273)$

$T_2 = 326.98K \approx 53.98\°C$

Therefore the exit temperature of air is 53.98°C

6 0

3 years ago

Question Completion Status:

Mashutka [201]

Answer: c. Centre of pressure

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

6 0

3 years ago

W²-5w+14

alexira [117]

Answer:w²-5w+14

4x²+11x+16

6x²+7x-10

x²-16x+56

Explanation:

7 0

3 years ago

Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki

storchak [24]

Answer:

True, Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor.

Option: A

Explanation:

To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. Regeneration technique is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine.

Thermal efficiency of a turbine is increased as the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.

6 0

3 years ago

Read 2 more answers

An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p

Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

24.9x10 m

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

mol

AUs¡ = Ws¡ = -5601.55

3 0

3 years ago