Question

3. Air flows through a long duct at 0.25 kg/s. A short section of the duct is cooled by liquid nitrogen that surrounds the duct. The absolute pressure, temperature, and velocity entering the cooled section are 190 kPa, 400 K, and 250 m/s, respectively. At the outlet, the absolute pressure and temperature are 215 kPa and 350 K. Considering the flow to be compressible in nature, compute the duct cross-sectional area, the change in enthalpy, the change in internal energy, and the change in entropy for this flow

Apparently, you can solve this problem without knowing the heat loss, and I'm not sure how.

Answer 1

For solving this problem we'll assume that at given temperature and pressure air behaves like an ideal gas...

Here don't get confused by consideration of Compressibility. Remember all the gases including Ideal gases are compressible.

Further solution is given in the images below::

Assumption.. main Air is assumed to behave as Ideal gas PV = nRT is So that Valid section't Where R Characteristic gab Constant Sections a universal R (R= 8.314 0.25 legls Igas Mgas, constand givena mair P, 190kPa Pq = 215 kPa TIF 400k Te KO Kama). K Myasz = 350k Molar Mass I Kamol of gas V = 250 mis 1 four Air By Ideal gas Relations: Rair R Mairi 8.3 14 kmink 29 kg pr=MRT imal l RT pa <3 Rair 0.287 KJ Kak Awa densityofgas (pais alroos K5 legk -0,717 K3 Kok Pi 190 kPa i. Density of Air at andet l' hors Kg my 0.2.97 kJ Kak 400K Density of Air 215kPa Se Pa RTE at Section 2 2.14 leg m3 KJ icg 0.267 350 K - Since mass flowe Rate is constant throughout the Pipe Maire maire = 0,25 Kols Di Aivi =0,25 2.655x. Axeso =0,25kg A = 0.006042m² A= 6.042 am² And

Change in Enthalpy AH in COT mcp (TE-Ti) = 0,2 400x is set X ). OU SKJ Kak (350-90 매 יו 42.5685 kW Change in Internal Energy am GoT = 0.25*0,717 (250-400) -8.5685 kW Change in Entropy [(y den 17 h + Roln Derivation of this formula in from Steady flow Energy equations ; {Assuming No KE : مرد Changes} ( for open system) bmhit nhąt o Wopen system a= mche-ha) mfydpea specific volume for smaller changes mda mvdp So = m. [dha Velp q dll dh - Velf USE ... Entropy Change T mm (p dT T & apo DS ت

None from Ideal gas pram RT equations in >TE RT p G 11 V specific Vorrime RU Р from ② ir q 2. Cp dT u [ Spal osa im T 1 P, min far som 7 - Alan [Golem + klime P, m for our case 0.25 [koos en portant 0.257 In 18 to -0.25 0.163 G KT ASA -0.0424 Sair TE
The value of entropy change turns out to be Negative, which is in concurrence of our observations. As we are removing heat from the gas, it's entropy goes down thus explaining the negative sign.

Hope you got the concept well...!!!