A turbojet aircraft flies with a velocity of 1100 km / h at an altitude where the air temperatu

1. First, we need to find the isentropic temperature at the compressor exit. We can use the isen

A turbojet aircraft flies with a velocity of 1100 km / h at...

A turbojet aircraft flies with a velocity of $1100 \mathrm{~km} / \mathrm{h}$ at an altitude where the air temperature and pressure are $-35^{\circ} \mathrm{C}$ and $40 \mathrm{kPa}$. Air leaves the diffuser at $50 \mathrm{kPa}$ with a velocity of $15 \mathrm{~m} / \mathrm{s},$ and combustion gases enter the turbine at $450 \mathrm{kPa}$ and $950^{\circ} \mathrm{C}$. The turbine produces $800 \mathrm{~kW}$ of power, all of which is used to drive the compressor. Assuming an isentropic efficiency of 83 percent for the compressor, turbine, and nozzle, and using variable specific heats, determine $(a)$ the pressure of combustion gases at the turbine exit, $(b)$ the mass flow rate of air through the compressor, ( $c$ ) the velocity of the gases at the nozzle exit, and $(d)$ the propulsive power and the propulsive efficiency for this engine. This problem is solved using appropriate software.

Transcript

00:01 Hello students, it is given that v is equal to 1100 km per hour and it can be written as 305 .66 m per second.

00:12 So, the relation at stage 1 and the enthalpy and the relative pressure at stage 4 are determined from a17 i .e.

00:20 For the given temperature h1 is equal to 238 .02 kj per kg and h4 is equal to 1304 .85 kj per kg and that of the p is equal to 257 .34.

00:45 Now, the enthalpy at stage 2 is determined from the energy balance in process 1 to 2.

00:51 For that, the equation can be written as h2 is equal to h1 plus v1 square minus v2 square divided by 2.

00:58 Let's substitute the values into this equation.

01:00 Hence, we will get h2 is equal to 238 .02 plus v1 value is 305 .66 the whole square minus 15 the whole square divided by 2.

01:13 The whole should be multiplied by 10 raised to minus 3 to convert to kj per kg.

01:17 Hence, we will get h2 is equal to 284 .62 kj per kg.

01:25 Now, the relative pressure at stage 3 is determined by, the equation will be pr3 will be equal to p3 by p2 into pr2.

01:40 Let's substitute the values is equal to 450 divided by 50 into 1 .1512.

01:47 Hence, we will get 10 .36.

01:50 Now, the isentropic enthalpy at this state is determined from this value with data from a17 using interpolation that is h of 3 is equal to 533 .84 kj per kg.

02:07 So, the actual enthalpy at stage 3 is determined from the isentropic compressor efficiency relation that is the relation h3 is equal to h2 plus h3 minus h2 divided by nc.

02:21 Here, let's mark it as h3s.

02:24 So, substituting the values into this, we can write 284 .62 plus 533 .84 minus 284 .62 divided by 0 .83.

02:39 So, here we will get the value as 584 .89 kj per kg.

02:48 Now, let's move on to the part a of the question that is the isentropic enthalpy at stage 5 can be determined by that is h5s will be equal to h4 minus w dot divided by m into eta t.

03:07 Therefore, this will be equal to 1304 .85 minus 800 divided by 2 .66 into 0 .83.

03:18 Hence, we will get the value as 942 .5 kj per kg.

03:23 Now, let's find the value that the relative pressure at state 5 that is we know that pr5 is equal to 78 .17...

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