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3 years ago

The thermodynamic property used to determine the amount of work produced by a power turbine in a heat engine is:_______

Engineering

1 answer:

Sveta_85 [38]3 years ago

3 0

Answer:

Enthalpy

Explanation:

Thermodynamic properties could be Intensive or extensive properties. Intensive properties are those that don't depend on the quantity of matter. Examples are pressure & temperature. Meanwhile, extensive properties are those whose values depend on the mass of the system. Examples are Energy, enthalpy and volume.

Now, in turbines in thermodynamics, the work done is produced by the flow required to turn the turbine and shaft. Recall that from the law of conservation of energy, the work in the turbine per mass airflow would be equal to the change in specific enthalpy of the flow from the entrance to the exit point of the turbine.

Thus, the property required to determine the work is Enthalpy.

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One kilogram of water fills a 150 L rigid container at an initial pressure of 2MPa. The container is cooled to 40 oC. Find the i

Tatiana [17]

Answer:

The initial temperature is 649 K (376 °C).

The final pressure is 0.965 MPa

Explanation:

From the ideal gas equation

PV = nRT

P is the initial pressure of water = 2 MPa = 2×10^6 Pa

V is intial volume = 150 L = 150/1000 = 0.15 m^3

n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol

R is gas constant = 8.314 m^3.Pa/mol.K

T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C

From pressure law,

P1/T1 = P2/T2

P2 (final pressure) = P1T2/T1

T2 (final temperature) = 40 °C = 40 + 273 = 313 K

P1 (initial pressure) = 2 MPa

T1 (initial temperature) = 649 K

P2 = 2 × 313/649 = 0.965 MPa

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4 years ago

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insens350 [35]

Answer:

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3 years ago

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frutty [35]

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Explanation:

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With advancing technologies, the uses of GPS can be extended to improve the productivity of the workforce by identifying location services and field operation insights.

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3 years ago

Is an ideal way for a high school student to see what an engineer does on a typical day but does not provide a hands-on experien

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Answer:

Look at some engineering colleges and set up or join a public zoom meeting. For example, some colleges have sign ups for zoom calls on set dates and you‘re able to ask questions.

Explanation:

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3 years ago

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The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin

ddd [48]

Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.

The surface of the cross-section of the stapes was determined:

A_ab= 10 mm^2

A-cd= 15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab $\leq$ σ_allow

σ_cd = F_cd/A_cd $\leq$ σ_allow

In the next step we will determine the static size: Picture b).

We apply the conditions of equilibrium:

∑F_x=0

∑F_y=0

∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

==> F_ab = w*k*N

Now we determine the load w

Sector AB:

σ_ab = F_ab/A_ab $\leq$ σ_allow=300 KPa

= w/10*10^-6 $\leq$ σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

Sector CD:

σ_cd = F_cd/A_cd $\leq$ σ_allow=300 KPa

= 2*w/15*10^-6 $\leq$ σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

==> w=2.25 * 10^-3 kN/m

The solution is:

w=2.25 N/m

note:

find the attached graph

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