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RUDIKE [14]
3 years ago
7

Which sentence is correct about the exergy of an empty (pressure around zero Pascal) tank with a volume of V, located in an envi

ronment with a pressure and temperature of P0 and T0?
a. The exergy of the tank is positive and equal to PoV.
b. Since the system receives work from the environment the exergy of the tank is negative and equal to -PoV
c. The exergy of the tank can be anything between zero to PoV.
d. Since the pressure of the tank is almost zero the tank has no exergy.
Engineering
1 answer:
sineoko [7]3 years ago
6 0

Answer:

The correct option is;

c.  the exergy of the tank can be anything between zero to P₀·V

Explanation:

The given parameters are;

The volume of the tank = V

The pressure in the tank = 0 Pascal

The pressure of the surrounding = P₀

The temperature of the surrounding = T₀

Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change

The exergy balance equation  is given as follows;

X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}

Where;

X₂ - X₁ is the difference between the two exergies

Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system

Therefore, the exergy of the tank can be anything between zero to P₀·V.

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Answer: N has to be lesser than or equal to 1666.

Explanation:

Cost of parts N in FPGA = $15N

Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

15N < 3N + 20000 lets say this is equation 1

(cost of FPGA lesser than that of gate array)

 Also. 15N < 1N + 100000  lets say this is equation 2  

(cost of FPGA lesser than that of standardcell)

Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

So N <= 1666 (Since N has to be an integer)

N has to be lesser than or equal to 1666.

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3 years ago
You can safely place a jack on a floor pan to keep a vehicle steady.
Elis [28]

Answer: Yes

Explanation:

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bagirrra123 [75]

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What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p
Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is =55\times 25

                                  =1375psf

1 ft = 12 inch

pressure at bottom =\frac{1375}{12^2}

                                = 9.55 psi

so, answer will be option (d) which is 9.55 psi

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3 years ago
Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which
Rudik [331]

ANSWERS:

-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor x_{1} =1

The temperature T_{1} =0^0 F

Isothermal process  T=C

a)

-V_{2} =2V_{1} ( double)

b)

-V_{2} =.5V_{2} (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb

then the state exists in the supper heated region.

a) from standard data

-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF

at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg

at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg

assume linear interpolation

\frac{P_{x}-P_{2(b)}  }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)}  }{v_{x}-v_{y}  }

P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y}  }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2

b)

-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}

from standard data

-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}

then the state exist in the wet zone

-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )

x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%

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3 years ago
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