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2 years ago

12

After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st

age of technological design does this best describe?
A. implement the solution
B. evaluate the solution
C. design a solution
D. identify a problem

1 answer:

Mazyrski [523]2 years ago

4 0

B evaluate the solution. Sorry if I'm wrong.

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For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.

Elza [17]

Assumptions:

Steady state.
Air as working fluid.
Ideal gas.
Reversible process.
Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.

Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

$r =\frac{V_1}{V_2}$

Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

$r = \frac{V_4}{V_3} = \frac{V_1}{V_2}$

Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

$\eta = 1-(\frac{1}{r^{\gamma - 1} } )$

where:

$\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio$

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

$\gamma = 1.4$

Answer:

See image attached.

5 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, <em>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</em>.

Option: A

<u>Explanation: </u>

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

<em>Thermal efficiency</em> of a turbine is increased as <em>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. </em>

6 0

2 years ago

Read 2 more answers

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

$TC=m\cdot c$

8 0

2 years ago

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

Mrac [35]

Answer:

$z_{2} = 91.640\,m$

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

$\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}$

The velocity of water delivered by the fire hose is:

$v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}$

$v_{1} = 0.267\,\frac{m}{s}$

The maximum height is cleared in the Bernoulli's equation:

$z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}$

$z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}$

$z_{2} = 91.640\,m$

7 0

2 years ago

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

Vilka [71]

Answer:

<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

6 0

2 years ago