Question

Determine the density of 1 kg of steam initially at a pressure of 10 bar having a dryness fraction of 0.78. If 500 kJ of heat is added at constant pressure. Also determine the condition and internal energy for the final state of steam. Take specific heat of steam as 2.1 kJ/kgK.

Answer 1

Density is  1/0.78 x 0.194608 6.5879 kg/$m^{3}$.

Internal Energy is  2626.714 kJ/kg.

Given:

• The 1 kg of the steam is initially at a pressure of 10 bar having a dryness fraction of 0.78.

• The 500kJ of heat is added at the constant pressure.

• specific heat of steam as 2.1 kJ/kgK.

Data:

Case (i) : Wet steam, in = 1kg, P = 10 bar (abs) x = 0.78,  Pwet =?

Solution:

From Steam Table 1

P           tu      vA,m31kg          hf            hfg           hg

10bar  452.84K  0.194608  762.2 kJ/kg 2013.8 kJ/kg 2776.1 kJ/kg

Density of wet steam, Pwet = 1/x . ∪g.

Pwet = 1/0.78 x 0.194608 6.5879 kg/$m^{3}$.

Case (ii) : Additional heat 500 kJ, Cp = 2.1 kJ/kgK

Final condition of steam = ? Internal Energy = ?

Enthalpy of wet steam,

hwet = hf  + x • hfg

hwet = 762.2 + (0.78 x 2013.8) = 2332.964 kJ/kg    

     

After adding 500 kJ of heat, hfinal = 2332.964 + 500 = 2832.964 kJ/kg

Referring to the Steam tables hfinal > hg  Hence the final condition of steam is superheated.

i.e., h final = hu  But  hu = hg + Cp(tu—t8)

2832.964 = 2776.1 + 2.1 (t.— 452.84) tu= 479.92 K

The internal energy of superheated steam, Uu = hu- Eu

where, Eu= 100 Pvu. and Vu = tuvg / tg.

Vu  = (479.92 x 0.194608) / 452.84 = 0.20625 $m^{3}$/kg

Eu  = 100 x 10 x 0.20625

     = 206.25 kJ/kg

Uu = hu — Eu

     = 2832.964 — 206.25

Uu = 2626.714 kJ/kg

Hence the  density is   Pwet = 1/0.78 x 0.194608 6.5879 kg/$m^{3}$.

Internal Energy is    Uu 2626.714 kJ/kg.

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