Air with a mass flow rate of 23 kgs enters a horizontal...

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3 m/s. At the exit, the temperature is 300 K and the velocity is 460 m/s. Using the ideal gas model for air with constant c_p = 1.011 kJ/kg*K, determine (a) the area at the inlet, in m^2 (b) the heat transfer between the nozzle at its surroundings in kW. (c) Specify whether the heat transfer is to or from the air.

Transcript

00:01 Hello, in the question we have given that air with a mass flow rate of 2 .3 kgs per second, so we have given m dot that is equals to 2 .3 kg per second enters a horizontal nozzle operating at a steady state at 450 kelvins.

00:20 So, t1 is equals to 450 kelvins is given and 350 kilopascal.

00:30 So, p1 is given that is 350 kilopascal and velocity, so velocity we will call it as v1 that is equals to 3 meters per second at the exit the temperature is 300 kelvin.

00:48 So, t2 we are considering the temperature t2 at the exit that is given to be 300 kelvins further and the velocity is 460 meters per second.

00:58 So, v2 is equals to 460 meters per second.

01:04 So, using the ideal gas model for air with a constant, so we have given the specific heat at constant pressure that is 1 .011 kilojoule per kg kelvin.

01:18 So, we have to find out the area of inlet then we have to find out the heat transfer and specify whether the heat transfer is to or from air.

01:29 So, now in order to answer this we will use the ideal gas equation.

01:34 So, we will use the ideal gas equation that is pv is equals to mrt.

01:39 So, if we rearrange so we will get p by rt that is equals to m times v but m times v is nothing but density.

01:47 So, density will be equal to so let us call this as rho 1 and p1 divided by rt1.

01:57 So, the density rho 1 will be equal to p1 is 350 then gas constant.

02:10 So, we have given the this specific heat.

02:13 So, for the specific heat the gas constant is 0 .287.

02:22 So, here we should mention so r so i will mention over here r is equals to 0 .287 kilojoule per kg kelvin.

02:38 So, at the specific heat so 350 and here it will be 450.

02:49 So, if we calculate this we will get the answer.

02:52 So, rho 1 will be equal to 2 .71 kg per meter cube.

03:01 So, now we have to find out the area.

03:09 So, the inlet area will be so area we need to calculate.

03:16 So, how can we find out? so, we know that the this mass is equals to so mass flow is given by rho 1 a1 times v1...

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