Thermodynamics - Lecture - 11 - Unit - 3 - Steam Properties and Problems

Problem: 1 

A vessel of volume 0.04 m³contains a mixture of saturated water and steam at a temperature of 250°C The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and internal energy.

Given data:

V=0.04m³

T= 250°C

m₁= 9 kg

To find:

p, m, v, h, S, and U

 

Solution:

From Steam Tables corresponding to 250°C,

v_f=v₁ = 0.001251 m/kg

V_g = V_s = 0.050037 m/kg

p= 39.776 bar

 

Total volume occupied by the liquid,

V₁= m₁ v₁ = 9 x 0.001251 = 0.0113 m³

Total volume of the vessel,

V = Volume of liquid + Volume of steam V₁+V_s

V_s=0.0287 m³

 

Mass of steam, m_s = V_s / V_s  =0.0287 / 0.050037

=0.574 kg

Mass of mixture of liquid and steam,

m=m₁+m_s = 9+ 0.574 = 9.574 kg

 

Total specific volume of the mixture,

v = V / m  = 0.04 /9.574 = 0.00418 m³/kg

 

We know that, v=v_f+x V_fg

V_fg = v_g - v_f

0.00418 = 0.001251 + x (0.050037 -0.001251)

x=0.06

 

From Steam Tables corresponding to 250°C,

h_f=1085.8 kJ/kg; h_fg=1714.6 kJ/kg

S_f= 2.794 kJ/kg K; s_fg = 3.277 kJ/kg K

Enthalpy of mixture,

h=h_f+x h_fg=1085.8 +0.06 x 1714.6 = 1188.67 kJkg

Entropy of mixture,

s = S_f+x S_fg =2.794 +0.06 x 3.277=2.99 kJ/kg K

Internal energy, u=h-pv

= 1188.67 – 39.776 x 10² x 0.00418 = 1172 kJ/kg

 

Problem 2

3 kg of steam at 18bar occupy a volume of 0.2550m³. During a constant volume the heat rejected is 1320KJ. Determine final internal energy and find initial dryness and work done.

Given data:

V₁ =  0.255m

P₁ = 18 bar

m = 3kg

Heat rejected, Q =- 1320kJ

Solution:

Specific volume, v₁ = V₁ / m = 0.255 / 3 = 0.085 m³ / kg

From Steam Tables, corresponding to 18 bar,

h_f1 = = 884.5 kJ/kg;

h_fg = 1910.3 kJ/kg

s_f1 = 2.398kJ/kgK;  s_fg = 3.977 kJ/kgK

V_g10.11033 m/kg

0.085 = x1*0.11033

x₁ = 0.77

h₁ = h_f1 + xh_fg1 = 884.5 +0.77 x 1910.3 = 2355.43 kJ/kg

s1=s_f1+xs_fg1

= 2.398 +0.77 x 3.977 = 5.46 kJ/kg

For constant volume process,

V₁ = v₂ and W=0

By first law of thermodynamics,

Q=U

Heat transfer, Q = U₂ –U₁

= m[ U₂ – (h₁ – p₁v₁)]

1320 = 3 [U₂ – (2355.43 – 18 x 100x0.085)

= 2642.43 KJ/Kg

 

 

Problem: 3

Steam initially at 0.3MPa, 250°C is cooled at constant volume. At what temperature will the steam become saturated vapour? What is the steam quality at 80°C.  Also find what is the heat transferred per kg of steam in cooling from 250°C to 80°C.

Given data:

P₁= 0.3MPa = 3 bar

T₁ = 250°C

T₂ =80°C

To find:

T'₂, X₂ and Q.

Solution:

From Steam Tables (in super heated region),

At 3 bar and 250°C

V₁= 0.7964 m³/kg

h₁ = 2967.9kJ/kg

For constant volume process, v₁ = v’₂ = v₂

But, v’₂ = v_g2 = 0.7964 m³/kg

Corresponding to v_g2 = 0.7964 m³/kg, from saturated table, the saturation temperature can be noted down.

 

The saturated vapour temperature, T'₂ = 122.93°C

 

We know that v₂ = v₂ + x₂ v_fg2

where v_i = V₂

0.7964 = 0.001029 + x₂ * (3.4091 - 0.001029)

X₂ = 0.2334

 

From saturated table at 80°C

h_f2 = 334.9 kJ/kg

h_fg2 = = 2308.9 kJ/kg

h₂ = 334.9+0.2334 x 2308.9 = 873.8 kJ/kg

According to first law of thermodynamics,

Work, W = Heat, Q + Internal energy, U

Heat transfer, Q=- U here, work, W=0 for constant volume process

= m (u₁ - u₂)

But, h= u + pv

Heat transfer, Q = (h₂ - p₂V₂) - (h_ı-P_IV_I)

= (h₂-h_i) - (p₂V₂-P₁V_I)

= (h₂-h_i) - V_i (P₂-P₁) [: V₁ = v₂ for constant volume process]

= (873.8–2967.9) - 0.7964 (47.36-300) here P_sat at 80°C = P₂

=-1892.897kJ/kg

 

Problem 4

Ten kg of water of 45°C is heated at a constant pressure of 10bar until it becomes superheated vapour at 300°C. Find the changes in volume, enthalpy, internal energy and entropy.

 

Given data:

m= 10 kg

P₁= P₂ = 10 bar

T₂ = 300°C

 

To find

 ΔV.  Δh,  ΔS and  ΔU

 

Solution:

From Steam Tables, corresponding to 45°C,

h₁, h_f1 = 188.4 kJ/kg

s₁=s_f1 = 0.638 kJ/kg K

 

v₁ =  V_f1= 0.001010 m./kg;

From Steam Tables, corresponding to 10 bar and 300°C,

Cycle

s₂ = 7.125 kJ/kg K;

v₂ = 0.258 m³/kg;

h2 = 3052.1 kJ/kg;

Change in volume,   Δ v = m (v₂ - v₁)

10 (0.258 – 0.001010) = 2.5699 m Ans.

Change in volume,  Δ h=m (h₂ – h₁)

= 10 (3052.1 - 188.4) = 28637 kJ Ans.

Change in entropy, Δ S = m (s₂ - s₁)

= 10 (7.125 -0.638) = 64.87 kJ/K

Ans.

Change in internal energy,

Δ U= m (U₂ - U₁)

= m [(h₂ – h₁) - (p₂v₂ – P₁v₁)

= m[(h₂ – h₁) – P₁ (v₂ - v₁)        [p₁=p₂]

= 10[(3052.1-88.4) - 1000(0.258 -0.001010)]

= 26067.1 kJ