Question

Determine the specific entropy (kJ/kg-K) of superheated water using the table below. The pressure and temperature are given as P = 398.19 kPa and T = 260°C, respectively. Note: Give your answer to four decimal places. Properties of Superheated Water Vapor Tv u hs v u hs °C m/kg kJ/kg kJ/kg kJ/kg-K °C m®/kg kJ/kg kJ/kg kJ/kg-K P= 3.0 bar P= 4.0 bar Tsar= 133.55 °C Tsar= 143.63 °C Sat. 0.6058 2543.1 2724.8 6.9911 Sat. 0.4625 2553.1 2738.1 6.8951 150 0.6339 2570.3 2760.5 7.0770 150 0.4708 2564.0 2752.4 6.9291 200 0.7163 2650.2 2865.1 7.3107 200 0.5342 2646.4 2860.1 7.1698 220 0.7485 2681.5 2906.1 7.3955 220 0.5588 2678.3 2901.8 7.2562 240 0.7805 2712.7 2946.8 7.4765 240 0.5831 2709.9 2943.1 7.3384 260 0.8122 2743.8 2987.5 7.5543 260 0.6071 2741.4 2984.2 7.4170 280 0.8438 2775.0 3028.1 7.6291 280 0.6310 2772.9 3025.3 7.4925 300 0.8753 2806.2 3068.8 7.7014 300 0.6548 2804.4 3066.3 7.5654 320 0.9067 2837.6 3109.6 7.7714 320 0.6785 2835.9 3107.4 7.6358

Answer 1

Solution:

In the given problem,we need to find specific entropy of superheated water at a pressure of 398.19 kpa and temperature of 260 Celsius.

GIVEN DATA : p = 398-19 kpa - 3.9819 bar. T= 260°c TO FIND : Specific entropy of superheated water Solution: P = 3.9814 bar T- 260°C. To find specific entropy 1 (kJ/kg kl for P=3.9819 bar of super heated vapour in and T= 260c. specific entropy Accurately, we need to find at P=3.9819 bar and T = 260e. have but, we ha specific entropy at P= abar and T=266€

80, to final Specific entopy at P= 3.9819 bar, and T=26oc. we need to interpolate . P= 3 bars and" - respective specific entropy with P=Abar and T=2600 respective specific entropy. To find, specific entropy at P= 3.9819 bar and T = 260c. 9 P= 3 bars and T= 2600 - - - - - - - - - Fayl mare wonderstandingan ? A P= 4 bar P 4 bar p= 3,9819 bar -- --- Di= 3 bar --- S:? ? S2 = 7.4170 kJ/kak -... →S -- To ***-.-.-- Si=7.5543 (KJ/KGK) make a cph, from the table , P= 3bar and T = 2600 Si= 1.5543 KK . Kg-k --

And, from the table P=4 bar; T = 2600 - S2= 7.4170 kg . .kg .k but, we need P= 3.9819 bar; T = 260°c. iS=? from graph between P-S, We know, Sa-si = P2-Pi .... (1) s-s, p-pi We knows from the graph Pg = 4bari Di = 3bar; p= 3.9819 bar S2 = 7.4170 kJ ; S1 = 7.5543 kJ ; S = ? kg k g kg.k Substitute in equation (1) 7.4170 - 7.5543 - 4-3 3.9819-3 S- 7 5543

-0.1373 S-7-5543 0.9819 s = 7.41948 kJ Answer K9 pressure of 3.9819 bar S = 7.41948 kJ At a 1 kg.k and at a temperature of T= 260°c.