Four kilograms of steam in a piston/cylinder device at 400 kPa and 175 °C underg

Question

Four kilograms of steam in a piston/cylinder device at 400 kPa and 175 °C undergoes isothermal and mechanically reversible process to a final pressure such that the steam is completely condensed (i.e., became a saturated liquid). Determine Q and W for this process: (b) Using generalized correlations and Equations (6.70 – 6.74). Comment on the accuracy of your answer. [Answer: Q = -9,461 kJ, W = 1494.9 kJ] Four kilograms of steam in a piston/cylinder device at 400 kPa and 175 °C undergoes isothermal and mechanically reversible process to a final pressure such that the steam is completely condensed (i.e., became a saturated liquid). Determine Q and W for this process: (b) Using generalized correlations and Equations (6.70 – 6.74). Comment on the accuracy of your answer. [Answer: Q = -9,461 kJ, W = 1494.9 kJ] (b) Using generalized correlations and Equations (6.70 – 6.74). Comment on the accuracy of your answer. [Answer: Q = -9,461 kJ, W = 1494.9 kJ] (b) Using generalized correlations and Equations (6.70 – 6.74). Comment on the accuracy of your answer. [Answer: Q = -9,461 kJ, W = 1494.9 kJ]

Explanation / Answer

Overall process can be divided into 2 process:

one is isothermal process to reach the pressure at vapor pressure of 175 DegC

2nd process is condensation of vapor to liquid stream

Now for 1st process, it is isothermal, final temperature will be same as 175 DegC

Steam is fully condensed at this temperature, so pressure in system = vapor pressure of water at 175 DegC = 892.5 kPa

Now Work done for isothermal process = W1 = nRTln(P1/P2) =

(4000/18)*8.314*(273+175)*ln(400/892.5) = -664.3J

Now for 2nd process W2 = P*total mass * change in specific volume

change in specific volume = specific value of saturated liquid-specific vol of saturated vapor

= 0.0011-0.2166 = -0.2155 m3/Kg ( reffered steam table)

W2 = 892.5*1000*4*(-0.2155) = 769335 J = -769.335 kJ

Now W = W1 + W2 = -664.3 + -769.335 = -1433.6 kJ

work done on the system = -W = 1433.6 kJ (please note answer slightly differs than given in question which might be bacause of difference in steam table data of mine and the one used in question)

Now Q1 = Delta U1 + W1 = W1 = -664.3 (because for isothermal process Delta U1 = 0)

Now Q2 = Mass *Delta Hcondensation + W2 at 892.5 Pa = -2031.55*4 = -8126-769.335= -8895.5

Q= Q1+ Q2 = --664.3-8895.5 = -9559.8 (please note answer slightly differs than given in question which might be bacause of difference in steam table data of mine and the one used in question)

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.