Steam at 3 MPa and 400^∘ C enters an adiabatic nozzle steadily with a velocity of 40 m / s and l

Steam at 3 MPa and 400^∘ C enters an adiabatic nozzle...

Key Concepts

Energy Conversion Between Enthalpy and Kinetic Energy

In nozzles, a primary phenomenon is the conversion of fluid enthalpy into kinetic energy as the fluid accelerates. By applying the energy balance, the decrease in enthalpy (which is directly related to temperature for compressible fluids) is used to power an increase in velocity, and this relationship is central to determining changes in thermodynamic properties such as temperature at the exit.

Steady Flow Energy Equation

This principle states that for a control volume in steady flow, the sum of energy forms entering the system must equal the sum of those leaving, allowing conversion between enthalpy, kinetic, and potential energy. In nozzle analysis, where potential energy changes are often negligible, the energy equation is used to relate the change in fluid enthalpy to the change in kinetic energy as the fluid accelerates or decelerates.

Adiabatic Process

An adiabatic process involves no heat transfer between the system and its surroundings. In the context of a nozzle, this means that any change in the fluid's energy is due solely to work interactions or kinetic energy changes, rather than heat exchange, simplifying the energy analysis and reinforcing the reliance on internal energy and enthalpy changes.

Conservation of Mass (Continuity Equation)

The continuity equation ensures that the mass flow rate into a control volume equals the mass flow rate out under steady flow conditions. This is essential in nozzle analysis because it relates the fluid's density, velocity, and cross?sectional area at different points within the nozzle, thereby allowing calculations of area ratios when the velocity and density change.

Transcript

00:01 Okay, so we know the energy transfer into the system minus the energy transfer out of the system should be equal to the change of energy in the system, which is delta e system here.

00:10 So since there's no heat loss, in other words, heat exchange with the surroundings and system, so therefore the change of heat, i'm sorry, the change of the energy in the system should be equal zero.

00:21 So i have ein, is equal e .r, which is equal to m .d.

00:24 Times hin plus 1 1 have v1 squared, equal to m.

00:27 M doth 10 hr plus 1 1 1 half v2 square as you can tell mddd can cancel out so therefore have h nis plus 1 1 1⁄2 square is equal to hr plus 1 half b2 square h in is the anthopis ellic inlet and v1 is the inlet velocity h r is the anthopis elli axis and v2 is the axis of velocity so if we do some arrangement here we'll have the enthalpy ellic axis is equal to h in plus 1 half times v1 square minus v2 so we know if you take a look at the table for the superheated steam.

01:06 So therefore at 3 megapascal and 400 degrees celsius, we'll have the anthopies is equal to 3 ,231 .7 kilojou per kilogram.

01:17 We know the v1 square, which is the inlet velocity square, is equal to 1 ,600 meter square per second square.

01:26 If we convert to kilo, per kilogram, it's about 1 .6 kilo.

01:29 Jub per kilogram.

01:31 And v2 square is equal to 300 meter per second square, which is equal to 90 ,000 meters square per second square.

01:38 If we convert the kilojou per kilogram, it's about 90 kilojou per kilogram.

01:43 So now let's plug in the values back into the equation.

01:46 So we have the enthalpy's that the axis is equal to 3 ,231 .7 kilo kilo kilo kilo per kilogram plus 1 .6 kilojou per kilogram minus 90 kilojou per kilogram and this will give us the anthropis at the axis is about 3 ,187 .5 kilojou per kilogram.

02:47 You know what? let me make it tighter here.

02:54 So 3187.

02:57 0 .5 kilojou per kilogram.

03:02 Okay? so if you take a look at a table, at 2 .5 mega pascal, and when the enthalpy is equal to 3 ,187 .5 kilojou per kilogram, you'll find out that the corresponding temperature, which is the xxate temperature, is about 376 0 .6 degrees celsius.

03:39 So therefore, this is the axis temperature.

03:43 For the next question, while we know, the mass flow rates is constant through the inlets and the axis part...

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