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enot [183]
3 years ago
14

A 360 kg/min stream of steam enters a turbine at 40 bar pressure and 100 degrees of superheat. The steam exits the turbine as a

100% saturated vapor at a pressure of 5 bar. Write and simplify the appropriate energy balance and then determine the energy generated by the steam as it passes through the turbine in kW.
Engineering
1 answer:
Brrunno [24]3 years ago
7 0
Answer:
skskkdkdkfkgkgkgkkgkgkgigooigigi lol
Explanation:
Oof
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Define Plastic vs elastic deformation.
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Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

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3 years ago
A Service Schedule is...
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A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0
3 years ago
3.8 LAB - Select lesson schedule with multiple joins
dem82 [27]

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

4 0
2 years ago
A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta
strojnjashka [21]

Answer:

\dot W_{out} = 399.47\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

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

The work done by the turbine is:

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

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

P = 10\,MPa

T = 520\,^{\textdegree}C

h = 3425.9\,\frac{kJ}{kg}

Outlet (Superheated Steam)

P = 1\,MPa

T = 280\,^{\textdegree}C

h = 3008.2\,\frac{kJ}{kg}

The work output is:

\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW

\dot W_{out} = 399.47\,kW

5 0
3 years ago
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