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State and derive steady flow energy equation and apply it to a boiler, condenser, nozzle and turbine.

Mumbai university > Mechanical Engineering > Sem 3 > Thermodynamics

Marks: 8M

Year: Dec 2015,

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## The Subject Name - May 2018

### Computer Engineering (Semester 8)

TOTAL MARKS: 80many practical problems, the rate at which the fluid flows through a machine or piece of apparatus is constant. This type of flow is called steady flow.

Assumptions:

The following assumptionsare made in the system analysis:

(i) The mass flow through the system remains constant.

(ii) Fluid is uniform in composition.

(iii) The only interaction between the system and surroundings are work and heat.

(iv) The state of fluid at any point remains constant with time.

(v) In the analysis only potential, kinetic and flow energies are considered.

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The steady flow equation can be expressed as follows: u1+C212+Z1g+p1v1+Q=u2+C22)2+Z2g+p2v2+Wu1+C122+Z1g+p1v1+Q=u2+C22)2+Z2g+p2v2+W………(1)

u1+p1v1+C212+Z1g+Q=u2+p2v2+C222+Z2g+Wu1+p1v1+C122+Z1g+Q=u2+p2v2+C222+Z2g+W h1+C212+Z1g+Q=h2+C222+Z2g+Wh1+C122+Z1g+Q=h2+C222+Z2g+W [∵h=u+pv]

Above equation is steady flow energy equation.

For boiler: A boiler transfers heat to the incoming water and generates the steam.

For this system, ∆Z=0 and ΔC222∆C222=0

W= 0 since neither any work is developed nor absorbed.

Applying energy equation to the system

h1+Q=h2h1+Q=h2 Condenser: The condenser is used to condense the steam in case of steam power plant and condense the refrigerant vapour in the refrigeration system using water or air as cooling medium.

For this system:

∆PE=0,∆KE=0(as their values are very small compared with enthalpies)

W=0(since niether any work s developed nor absorbed)

Using energy equation to steam flow

h1−Q=h2h1−Q=h2 Where Q = Heat lost by 1 Kg of steam passing through the condenser.

Nozzle: In case of a nozzle as the enthalpy of the fluid decreases and pressure drops simultaneously the flow of fluid is accelerated. This is generally used to convert the part of the energy of steam into kinetic energy of steam supplied to the turbine.

For this system, ∆PE=0,W=0,Q=0

Applying the energy equation to the system, h1+C212=h2+C222h1+C122=h2+C222$Turbine: In a steam or gas turbine steam or gas is passed through the turbine and part of its energy is converted into work in the turbine. This output of the turbine runs a generator to produce electricity. The steam or gas leaves the turbine at lower pressure or temperature. Applying energy equation to the system, Here, Z1=Z2Z1=Z2 h1+C312−Q=h2+C222+W \ltbr\gt TOTAL TIME: 3 HOURS \lt/span\gt \ltspan class='paper-points'\gt (1) Question 1 is compulsory.\ltbr\gt (2) Attempt any \ltb\gtthree\lt/b\gt from the remaining questions.\ltbr\gt (3) Assume data if required.\ltbr\gt (4) Figures to the right indicate full marks. \lt/span\gt \lt/span\gt \ltspan class='paper-question'\gt \ltspan class='paper-ques-desc'\gt \ltb\gt1(a)\lt/b\gt TYPE QUESTION NO.1 HERE - Example: write equation like this$a^2$(5 marks) 1(b) TYPE QUESTION NO.2 HERE (10 marks) ======= 1 Answer 1 1.6kviews In many practical problems, the rate at which the fluid flows through a machine or piece of apparatus is constant. This type of flow is called steady flow. Assumptions: The following assumptionsare made in the system analysis: (i) The mass flow through the system remains constant. (ii) Fluid is uniform in composition. (iii) The only interaction between the system and surroundings are work and heat. (iv) The state of fluid at any point remains constant with time. (v) In the analysis only potential, kinetic and flow energies are considered. The steady flow equation can be expressed as follows:$u_1+\frac{C_1^2}{2}+Z_1g+p_1 v_1+Q=u_2+\frac{C_2^2)}{2}+Z_2 g+p_2 v_2+W$………(1)$u_1+p_1 v_1+\frac{C_1^2}{2}+Z_1 g+Q=u_2+p_2 v_2+\frac{C_2^2}{2}+Z_2 g+Wh_1+\frac{C_1^2}{2}+Z_1 g+Q=h_2+\frac{C_2^2}{2}+Z_2 g+W$[∵h=u+pv] Above equation is steady flow energy equation. For boiler: A boiler transfers heat to the incoming water and generates the steam. For this system, ∆Z=0 and$∆\frac{C_2^2}{2}$=0 W= 0 since neither any work is developed nor absorbed. Applying energy equation to the system$h_1+Q=h_2$Condenser: The condenser is used to condense the steam in case of steam power plant and condense the refrigerant vapour in the refrigeration system using water or air as cooling medium. For this system: ∆PE=0,∆KE=0(as their values are very small compared with enthalpies) W=0(since niether any work s developed nor absorbed) Using energy equation to steam flow$h_1-Q=h_2$Where Q = Heat lost by 1 Kg of steam passing through the condenser. Nozzle: In case of a nozzle as the enthalpy of the fluid decreases and pressure drops simultaneously the flow of fluid is accelerated. This is generally used to convert the part of the energy of steam into kinetic energy of steam supplied to the turbine. For this system, ∆PE=0,W=0,Q=0 Applying the energy equation to the system,$h_1+\frac{C_1^2}{2}=h_2+\frac{C_2^2}{2}

Turbine: In a steam or gas turbine steam or gas is passed through the turbine and part of its energy is converted into work in the turbine. This output of the turbine runs a generator to produce electricity. The steam or gas leaves the turbine at lower pressure or temperature.

Applying energy equation to the system,

Here, $Z_1 =Z_2$

$h_1+\frac{C_1^3}{2}-Q=h_2+\frac{C_2^2}{2}+W$

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