TITLE.PM5

PROPERTIES OF PURE SUBSTANCES 95

Dharm

\M-therm/th3-2.p65

Final pressure of steam, p 3 = 760 + 5 = 765 mm

=^765

1000

×1.^3366 (Q 1 m Hg = 1.3366 bar)

~ 1 bar

From steam tables :

At p 1 = p 2 = 12 bar : hf = 798.4 kJ/kg, hfg = 1984.3 kJ/kg

At p 3 = 1 bar : ts = 99.6°C, hf = 417.5 kJ/kg, hfg = 2257.9 kJ/kg

tsup = 110°C (given)

Also h 3 = h 2

()( )hh cT T hfxhf 33 ++fg psssup 33 −=+ 22 fg 2

Taking cps = 2 kJ/kg K, we get

417.5 + 2257.9 + 2[(110 + 273) – (99.6 + 273)] = 798.4 + x 2 × 1984.3

2696.2 = 798.4 + 1984.3 x 2

∴ x 2 =26962 798 4 19843 ..−. = 0.956

Now, quality of steam supplied,

x mmxms

(^1) ws
2 0956 205
= + = 2205
×

• ..
  .
  = 0.87. (Ans.)


• Example 3.28. The following data were obtained in a test on a combined separating
  and throttling calorimeter :
  Pressure of steam sample = 15 bar, pressure of steam at exit = 1 bar, temperature of steam
  at the exit = 150°C, discharge from separating calorimeter = 0.5 kg/min, discharge from throt-
  tling calorimeter = 10 kg/min.
  Determine the dryness fraction of the sample steam.
  Solution. Pressure of steam sample, p 1 = p 2 = 15 bar
  Pressure of steam at the exit, p 3 = 1 bar
  Temperature of steam at the exit, tsup 3 = 150°C
  Discharge from separating calorimeter, mw = 0.5 kg/min
  Discharge from throttling calorimeter, ms = 10 kg/min
  From steam tables :
  At p 1 = p 2 = 15 bar : hf 2 = 844.7 kJ/kg, hfg 2 = 1945.2 kJ/kg
  At p 3 = 1 bar and 150°C : hsup 3 = 2776.4 kJ/kg
  Also, h 2 = h 3
  hxh hf 223 += 2 fg sup
  844.7 + x 2 × 1945.2 = 2776.4
  ∴ x 2 =2776 4 84471945 2..−. = 0.993
  Now, quality of steam supplied,
  x
  xm
  mm
  s
  (^1) sw
  2 0993 10
  10 05
  = + =. +×

. = 0.946. (Ans.)