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Air at a temperature of 150C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 8000C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 6500C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle where it expands until the temperature falls to 5000C. If the air flow rate is 2 kg/s, calculate the following.


(i) Rate of heat transfer to the air in the heat exchanger,


(ii) Power output from the turbine assuming no heat loss, and


(iii) Velocity at exit from the nozzle, assuming no heat loss.



Take the enthalpy of air as h =cpt, where cp is the specific heat equal to 1. 005kJ/kg K and t is the temperature.

Answers

Answer 1

The power output from turbine assuming no heat loss and the velocity at exit from nozzle where Enthalphy h CpT i.e Cp specific heat = 1.005KJ/KgK.

What is meant by velocity?

Velocity is the direction speed of an item in motion as a measure of the rate at which its location is changing as seen from a certain point of view and as measured by a specific unit of time.A particle or object's movement with respect to time is expressed vectorially as velocity.As an alternative, the velocity magnitude can be expressed in centimetres per second (cm/s).Simply said, velocity is the rate of motion of an object in a specific direction. Using an automobile driving north on a highway as an example, or a rocket after it has launched, as an example of speed.

The rate of flow of heat transfer to air exchanger i.e

[tex]w(h1+v^{2} 1/2+Z1g)+Q_{1-2} =w(h2+v2^{2} /2+Z1g)W_{1-2}[/tex]

[tex]Wh_{1} +Q_{1-2} =Wh_{2}[/tex]

[tex]Q_{1-2} =w(h_{1} -h_{2} )[/tex]

[tex]Q_{1-2} =w_{cp} (t_{2} -t_{3} )[/tex]

[tex]Q_{1-2}[/tex] = 2xl.005(8000-150)

[tex]Q_{1-2}[/tex] = 1577.8kgj/s

Velocity at exit from the nozzle, assuming no heat loss.

Given t₁ = 150°C, t₂ = 8000°c

V₁ = 30m/s, V₂ = 30m/s

t₃ = 6500°C , V₃ = 60m/s
[tex]W(v_{2} )^{2} /2+h_{2} )=Wh_{3} + (v3)^{2} /2+Wt[/tex]

[tex](v2^{2}) -(v3^{2}) /2=(h2-h3)=Wt/w[/tex]

[tex](30^{2} -60^{2} )x10^{-3} /2+(1.005)(8000-6500)=Wt/w[/tex]

Wt = 1.506.15xw

Wt = 1506.15x2 = 3012.3kW

the enthalpy of air as h =cpt, where cp is the specific heat equal to 1. 005kJ/kg K and t is the temperature is 3012.3kW

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Answer:

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