Evaluate the specific enthalpy - thermodynamics, Mechanical Engineering

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Evaluate the specific enthalpy - thermodynamics:

Steam is being generated in a boiler at a pressure of 15.25 bar. Calculate the specific enthalpy when

(i)  Steam is dry saturated

 (ii)  Steam is wet saturated having 0.92 as the dryness fraction, and

  (iii)  Steam is already superheated, the temperature of steam being 270°C.

Sol.: Note. Specific enthalpy means enthalpy per unit mass. From steam table which is given below, we get the following data:

 

Absolute pressure (P)

bar

Saturation temperature (t) ºC

Specific enthalpy kJ/kg

Water (hf)

Latent heat (hfg)

15

15.55

198.3

200.0

844.6

852.4

1947

1941

Now  15.55 - 15.25 = 0.30 bar

15.55-15 = 0.55bar

For difference of pressure of 0.55 bar, difference in t (or ts) = 200 - 198.0 = 2.0°C For difference of pressure of 0.30 bar, difference of

t = (2/0.55) × 0.30 =1.091°C. Corresponding to 15.25 bar, perfect value of

t = 200 - 1.091 = 198.909°C

For difference of pressure of 0.55 bar, difference of hf (which is heat of the liquid) = 852.4 - 844.6 = 7.8 kJ/kg For difference of pressure of 0.30 bar, difference of hf  =   (7.8/0.55) × 0.30 = 4.255 kJ/kg. Corresponds to 15.25 bar, exact value of

hf  = 852.4 - 4.255 = 848.145 kJ/kg.

Again, for difference of pressure of 0.55 bar, difference of hfg (which is latent heat of evaporation)

= 1947 - 1941 = 6 kJ/kg.

For difference of 0.30 bar, difference of

hfg  = (6/0.55) × 0.30 = 3.273 kJ/kg. Corresponding to 15.25 bar, perfect value of

hfg = 1941 + 3.273 = 1944.273 kJ/kg

[Greater the pressure of steam generation less is latent heat of evaporation.]

The data computed above are written in a tabular form as below:

 

Absolute pressure (P)

bar

Saturation temperature (t) ºC

Specific enthalpy kJ/kg

Water (hf)

Latent heat (hfg)

15.25

198.909

848.145

1944.273

 

(i) When steam is dry saturated, its enthalpy can be given by

HDry = hf  + hfg  kJ/kg

= 848.145 + 1944.273 = 2792.418 kJ/kg                                           .......ANS

(ii) When steam is wet saturated, its enthalpy can be given by

Hwet = hf  + xhfg   kJ/kg

= 848.145 + 0.92 × 1944.273 = 2636.876 kJ/kg                            .......ANS

(iii) When steam is superheated, its enthalpy can be given by

Hsup = hf  + hfg  + Cp(tsup  - ts) kJ/kg


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