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Ahat [919]
3 years ago
12

True or false 19. Closed systems rely on feedback from outside of the system to operate.

Engineering
1 answer:
Furkat [3]3 years ago
6 0

Answer: True

Explanation: Closed loop relies on feedback from PNS to make modifications in the movement, open loop allows action in the absence of feedback, 2. ... Closed loop can change the initial commands, open loop can not change the initial commands.

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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the
Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})

Using the Boyle equation we have

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})

Where,

C_p = Specific heat at constant pressure

T_1= Initial temperature of gas

T_2= Final temprature of gas

R = Universal gas constant

v_1= Initial specific Volume of gas

v_2= Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

T_1 = T_2, v_2=v_1

The equation would turn out as

\Delta S = C_p ln1-ln1

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

\Delta S = \frac{Q}{T}

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

\Delta S = \frac{230kJ}{(30+273)K}

\Delta S = 0.76 kJ/K

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0
3 years ago
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
A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False
tiny-mole [99]

The answer is : True

4 0
3 years ago
Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.
levacccp [35]

Answer:

\dot W_{out} = 3863.98\,kW

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

h_{in} = 3353.1\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mixture)

h_{out} = 890.1\,\frac{kJ}{kg}

The power produced by the turbine is:

\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})

\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )

\dot W_{out} = 3863.98\,kW

8 0
3 years ago
Natalie solved the problem below incorrectly. Identify his error and justify your reasoning. −19+3x−11+2x=2 5x−19−11=2 5x−30=2 5
arsen [322]
5x-30=2
5x=2+30 (not -28) when the -30 is brought over to the RHS, 30 should be added to 2 instead of subtracted

hence, 5x=32
x = 6.4
5 0
3 years ago
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