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3 years ago

9

The thermal efficiency of two reversible power cycles operating between the same thermal reservoirs will a)- depend on the mecha

nisms being used b)- be equal regardless of the mechanisms being used c)- be less than the efficiency of an irreversible power cycle

1 answer:

mestny [16]3 years ago

4 0

C ,, i’m pretty sure .

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Use the map to answer the question.

djverab [1.8K]

Answer:

Captain falcon

Explanation:

7 0

2 years ago

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by $\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}$

So availability of system $=\frac{900}{900+100}=\frac{900}{1000}=0.9$

So the availability of system will be 0.9

7 0

3 years ago

A piston–cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine t

drek231 [11]

Answer:

See explanation

Explanation:

Given:

Initial pressure,

p

1

=

15

psia

Initial temperature,

T

1

=

80

∘

F

Final temperature,

T

2

=

200

∘

F

Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

R

=

0.04513

Btu/lbm.R

C

v

=

0.158

Btu/lbm.R

Find the work done during the isobaric process.

w

1

−

2

=

p

(

v

2

−

v

1

)

=

R

(

T

2

−

T

1

)

=

0.04513

(

200

−

80

)

w

1

−

2

=

5.4156

Btu/lbm

Find the change in internal energy during process.

Δ

u

1

−

2

=

C

v

(

T

2

−

T

1

)

=

0.158

(

200

−

80

)

=

18.96

Btu/lbm

Find the heat transfer during the process using the first law of thermodynamics.

q

1

−

2

=

w

1

−

2

+

Δ

u

1

−

2

=

5.4156

+

18.96

q

1

−

2

=

24.38

Btu/lbm

7 0

3 years ago

A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu

Mumz [18]

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

$Density = \frac{mass}{volume}$

Therefore, mass is:

$mass = denisty *volume$

Density of air = $2 kg/m^3$ ; Volume of the tank = $3.5 m^3$

$=> Mass = 3.5 * 2 = 7 kg$

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = $6.5 kg/m^3$

The new mass will be:

$Mass = 6.5 * 3.5 = 22.75 kg$

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

6 0

3 years ago

A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic

Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0

3 years ago