Air at 800 kPa and 400^∘ C enters a steady-flow nozzle with a low velocity and leaves at 100 kPa

Air at 800 kPa and 400^∘ C enters a steady-flow nozzle with...

Key Concepts

Nozzle Flow

Nozzle flow involves fluid acceleration through a device that converts pressure energy and enthalpy into kinetic energy. The design and analysis of nozzles commonly assume adiabatic and often isentropic processes, making them a key component in various applications such as propulsion systems and turbines where the speed of the exiting fluid is critical.

Isentropic Process

An isentropic process is an idealization of a reversible and adiabatic process, where entropy remains constant. In the context of nozzle flow, assuming an isentropic (and therefore reversible and adiabatic) process allows for the use of ideal gas relations to predict changes in pressure, temperature, and velocity, thereby simplifying the determination of the maximum exit velocity.

Ideal Gas Behavior

The ideal gas assumption provides a simplified model for evaluating the thermodynamic properties of gases. By applying ideal gas relationships, calculations of pressure, temperature, and enthalpy become more straightforward, which is especially useful in the analysis of adiabatic, isentropic nozzle flows where the conversion of thermal energy into kinetic energy is of primary interest.

Steady-Flow Energy Equation

This equation applies the first law of thermodynamics to a control volume with a continuous mass flow, such as a nozzle. It relates the change in enthalpy of the fluid to changes in kinetic energy, potential energy, work, and heat transfer. In an adiabatic nozzle, where no heat or work interactions occur apart from the pressure work, the decrease in enthalpy directly translates into an increase in the fluid's kinetic energy, allowing calculation of the exit velocity.

Adiabatic Process

An adiabatic process is one in which there is no heat exchange between the system and its surroundings. For a nozzle flow, assuming an adiabatic process means that all changes in energy are due to work interactions, not heat transfer. This simplification helps isolate the conversion of internal energy (enthalpy) into kinetic energy.

Transcript

00:01 The given data of this question is we here consider the air that is flowing with the pressure p1, which is given in the question that is 100 kylpascal at temperature t1, which is also given that is 400 degree centigrade.

00:29 And the air is entered steadily flow nozzle with the low velocity and leaves the nozzle at the pressure p2 which is given that is 100 kiosk and then air is accelerated is an zanthropic nozzle when air is going through azantropic nozzle the air is accelerated then what we have to find at that point we have to find out the maximum velocity at the exit which means v2 we have to find the maximum velocity so at the exit so for this our solution is the properties of air at room temperature as there is no specific temperature is given to find the property so at room temperature the air are cp which is equals to one 0 .005 kilojowl per kg into kelvin and k is 1 .4 and these values we get from the table a to a so these values we must know now first of all we have to find out the exit temperature and exit temperature is determined from ideal gas as in tropic relation so that relation is t2 which is equals to t1 p2 by p1 and raise to par k minus 1 divided by k now putting the values which are given as t1 is 400 degree centigrade we have the temperature in kelvin so we add with 273 kelvin capital k kelvin and pressure p 2 that is 100 kiosk while the pressure p 1 that is 800 k and the value of k we have find from the table that is 1 .4 minus 1 divided by 1 .4 so this equals to 371 .5k so this is our t to the exit temperature the final temperature now we take nozzle as the system which is a control volume so the energy balance for this steady flow system can be expressed in the rate form so the rate form is this the rate of net energy transfer by heat and work all this this is basically the rate of net energy transfer by heat work and mass which is equals to the rate of change in internal kinetic potential energies etc basically the rate of change of the system energy so this equals to zero as because of steady flow so at this part is equal to zero so are the rate of energy in which is equals to rate of energy out this equals to this so this becomes zero equals to h2 minus h1 plus v two square minus zero will divide by two and zero is equals to cp t2 minus t plus v2 squared divided by 2 therefore v2 as we have to find out the exit velocity v2 so we take this v2 and so our equation becomes like this under root whole under root 2 cp t1 minus t2 now we put the values in it so whole under root 2.

06:32 Cp is the constant that is 1 .005 kilojoule per kg into kelvin and t1 is 630, 6773 minus 371 .5 which we have 5 .t2 holder unit kelvin...

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