COURSE: Detailed Explanation on Entropy Questions Lesson : Questions -(13-15)

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Question 13 A reversible engine, as shown in Figure during a cycle of operations draws 5 MJ from the 400 K reservoir and does 840 kJ of work. Find the amount and direction of heat interaction with other reservoirs. 200 K 300 K 400 K Q3 Q2

Solution 13 Let Q2 and Qa both incoming i.e. out from the system Q Q5000 200 300 400 200 K 300 K 400 K Q3 w 840 kJ Or 5000+0+0-0 or 6Q+4 Q2t3 x 5000 0 Qs t Q+5000 840 0 Heat balance or 4Qa t 4 Q16640 0 (iii) - gives Qs- +820 kJ Q-4980 kJ 2 Qa+1640 (Here-ve sign means heat flow opposite to our assumption)

Question 14 For a fluid for which pu/T is a constant quantity equal to R, show that the change in specific entropy between two states A and B is given by A fluid for which R is a constant and equal to 0.287 kJ/kg K, flows steadily through an adiabatic machine, entering and leaving through two adiabatic pipes. In one of these pipes the pressure and temperature are 5 bar and 450 K and in the other pipe the pressure and temperature are 1 bar and 300 K respectively. Determine which pressure and temperature refer to the inlet pipe For the given temperature range, cp is given by Cp = a ln T + b where T is the numerical value of the absolute temperature and a-0.026 kJ/kg K, b 0.86 kJ/kg K.

Solution 14 Intrigation both side with respect A to B C dT R T V or SSA V_R T P or CpdT Vdp or C,dT R T P 300 {(In 300)"-(In 450)" } + bln_-0.287 In | = or Sg SA 450 or sp- SA 0.05094 kJ/kg -K A is the inlet of the pipe

Question 15 Two vessels, A and B, each of volume 3 m3 may be connected by a tube of negligible volume. Vessel a contains air at 0.7 MPa, 95 o C, while vessel B contains air at 0.35 MPa, 205 C. Find the change of entropy when A is connected to B by working from the first principles and assuming the mixing to be complete and adiabatic

Solution 15 Let the find temperature be T) Mass of (m)- 700 x 3 0.287 x 368 kg -19.88335 kg Or T 398.6 K AC If final pressure (pt) -1.005 kJ/kg K 0.7 MPa 700 kPa 368 350 kPa 478 K cy 0.718 kJ/kgK R 0.287 kJ/kg.K 27.5372 x 0.287 x 398.6 6 kPa525 kPa Mass of gas (m,) RT. 350x37.653842 kg (AS)A mplnRIn3.3277 B. 0.287 x 478 In R n-2.28795 kJ/K For adiabatic mixing of gas Internal Energy must be same (AS)univ-(AS)a + (AS)I + 0 = 0.9498 kJ/K -19.88335 x 0.718 x 368 kJ 5253.66 kJ -7.653842 x 0.718 x 478 kJ 2626.83 kJ

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Harshit Aggarwal

Cleared UPSC ESE twice with Rank 63 and 90 in mechanical engg. Got 99 percentile in GATE. Cracked ONGC, BHEL,ISRO, SAIL, GAIL successfully

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Jeffrey Selvaraju

4 months ago

Maam this is a very good course , Thank you so much for this . It is helpful.

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Prithwish Chakraborty

25 days ago

On sollution 13, Q1 is going out of the TER. So, it should be with - ve sign before Q1/T1

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Mohammed MUBASHIR A

9 months ago

when use q=Cv ln (t2/t1)
&
cp ln t2/t1 -r ln P2/P1
?

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Prithwish Chakraborty

25 days ago

If you see for GATE qtn, antwhere given "pressure changing from.... Or volume changing from....." apply second form. = its a short cut to identify.
Conceptually, TER/TER or TER/BODY ptoblems must be dealt with first form and system involving ideal gas scenario must be dealt with second form.