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Questions (13-15)
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In this lesson, question (13-15) have been explained.

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

Unacademy user
mam aap bhut jldi jldi test lete hain... :/ ..... :) but nice and thnq ji
Heena Malhotra
9 months ago
Wo, isliye kyuki m almost pura syllabus krwa chuki hu :) Welcome :)
Anushka Kashyap
9 months ago
hnji mam... i know :)
Heena Malhotra
9 months ago
On sollution 13, Q1 is going out of the TER. So, it should be with - ve sign before Q1/T1
when use q=Cv ln (t2/t1) & cp ln t2/t1 -r ln P2/P1 ?
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.
  1. COURSE: Detailed Explanation on Entropy Questions Lesson : Questions -(13-15)

  2. About me Graduated from NIT Nagpur in 2008 * Cleared Engineering Services Examination (ESE-UPSC) Exam Got the offer letter from most of the Maharatna and Navratna Companies , Cleared GATE Exam * Rate, Review, Recommend, Share Follow me on Unacademy at: aggarwal

  3. 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

  4. 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)

  5. 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.

  6. 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

  7. 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

  8. 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