Question

(1 point) A rigid, insulated 0.43 m tank initially contains saturated water liquid at 362 kPa. The tank is connected to a steam pipe line carrying steam at 1.53 MPa and 305 °C through a valve. When the valve is opened, steam enters the tank from the pipe. When the valve is closed, the tank is filled with a saturated vapor with a pressure of 0.8 MPa. How much mass was added to the tank during this process?

Answer 1

Volume of tank = 0.43m³

Pressure in the tank = 362 kPa

From following steam table:

Temperature at 362 kPa or 3.62 bar:

Saturated Water and Steam (Pressure) Tables (vp (vo) (h ) 121.8 123.3 124.7 126.1 127.4 0.001 062 0.001 064 0.001 065 0.001 066 0.001 068 0.845 86 0.809 80 0.776 77 0.746 41 0.718 40 511.3 517.6 523.7 529.6 535.3 2 197.2 2 193.0 2 188.9 2 184.9 2 181.1 2 708.5 2 710.6 2 712.6 2 714.5 2 716.4 1.547 1.563 1.578 1.593 1.607 5.564 5.532 5.502 5.473 5.445 7.111 7.095 7.080 7.066 7.052 1.621 1.634 128.7 130.0 131.2 132.4 133.5 0.001 069 0.001 070 0.001 071 0.001 072 0.001 074 0.692 47 0.668 40 0.646 00 0.625 09 0.605 53 540.9 546.2 551.4 556.5 561.5 2 177.3 2 173.7 2 170.1 2 166.6 2 163.2 2 718.2 2 719.9 2 721.5 2 723.1 2 724.7 1.647 5.418 5.392 5.367 5.342 5.319 7.039 7.026 7.014 7.002 6.991 1.660 1.672 134.7 135.8 136.8 1.683 1.695 1.706 5.297 5.274 5.253 5.232 5.212 6.980 6.969 6.959 6.949 6.939 137.9 1.717 138.9 1.727 0.001 075 0.001 076 0.001 077 0.001 078 0.001 079 0.007 080- 0.001081, 0.001 082 0.001 083 0.001 084 0.587 18 0.569 95 0.553 73 0.538 43 0.523 97 .510 29 0.497 33 0.485 02 0.473 33 0.462 20 566.2 2 159.9 2 726.1 570.9 2 156.7 2 727.6 575.5 2 153.5 2 729.0 579.9 2 150.4 2 730.3 584. 3 2 147.3 2 731.6 588.5 2 144.4 2 732.9 592-712141427341 596.7 2 138.6 2 735.3 600. 8 2 135.7 2 736.5 604.7 2 132.9 2 737.6 139.9 140.8 141.8 142.7 143.6 1.738 1.748 1.758 1.767 1.776 5.192 5.173 5.154 5.136 5.118 6.930 6.921 6.912 6.903 6.894

3.7-3.6 3.62-3.6 140.8-139.9 -139.9

r= 140.08°C

Hence temperature of water in the tank = 140.08⁰C

Mass of water in the tank at above conditions:

PV x 18 Mass = RT

Mass = 362000 x 0.43 x 18 8.314 x (140.08 + 273.15)

Mass = 815.54(Kg)

Mass of water in tank initially = 815.54 Kg

Now,

Suppose mass of steam added = 'x' kg

Final mass in the tank = M_final

> Mfinal = (1 +815.54) Kg

From following table:

Saturated Water and Steam (Pressure) Tables (V) - (vo) (h) (he) (Sp) () 153.3 154.8 156.2 157.5 158.8 0.001 095 0.001 096 0.001 098 0.001 099 0.001 101 0.001 102 0.001 104 0.001 105 0.001 107 0.001 108 1.875 1.890 1.904 1.918 1.931 0.361 06 0.348 44 0.336 69 0.325 72 0.315 46 0.305 84 0.296 80 0.288 29 0.280 26 0.272 68 646.5 652.8 658.8 664.7 670.4 676.1 681.5 686.8 692.0 697.1 (h) 2 749.2 2 750.9 2 752.5 2 754.0 2 755.5 2 756.9 2 758.2 2 759.5 2 760.8 2 762.0 2 102.7 2 098.1 2 093.7 2 089.3 2 085.1 2 080.8 2076.7 2072.7 2068.8 2064.9 (85) 4.931 4.903 4.877 4.851 4.827 4.803 4.780 4.757 4.735 4.713 6.806 6.793 6.781 6.769 6.758 6.747 6.736 6.725 6.715 6.705 160.1 161.4 162.6 163.8 165.0 1.944 1.956 1.968 1.980 1992 166.1 167.2 168.3 169.4 170.4 7.6 7.8 0.001 110 0.001 111 0.001 112 0.001 114 0.001 113 0.265 50 0.258 70 0.252 24 0.246 10 .240 20 702.0 706.9 711.7 716.3 720.9 2061.2 2 057.4 2 053.7 2 050.1 2.046. 3 2763.2 2 764.3 2 765.4 2.766.4 767.4 2003 2014 2 025 2035 2.046 4.693 4.672 4.652 4.633 6.696 6.686 6.677 6.668 7.660 2

Final temperature in the tank = 170.4⁰C

Therefore, at final stage we can write:

Mfinal = x + 815.54 = 800000 x 0.43 x 18 8.314 x 443,55

*I=863.57(Kg)

Therefore, mass added in the tank = 863.57 Kg