Question 5.21: In an installation 5.2 kg/s of steam at 30 bar and 350°C is ...

Refer Fig. 23.

From Mollier chart :
h_{1}      = 3115 kJ/kg ;                        h_{3}      = 2675 kJ/kg
v_{1}      = 0.09 m³/kg ;                    v_{3}      = 0.46 m³/kg.

(i) Dimensions of the nozzle :
For supersaturated steam, the index of expansion is assumed same as for superheated steam, i.e., n = 1.3.
Thus isentropic enthalpy drop,

( h_{1}     –    h_{2 })   =  \frac{n}{n  –  1}    \frac{p_{1}v_{1}    ×    10^{5}}{10^{3}}  [1   –   (  \frac{p_{2} }{  p_{1}}) ^{ \frac{ n  –  1}{n} }]   kJ/kg

=  \frac{1.3}{1.3  –   1}    ×    \frac{30    ×    0.9    ×  10^{5}}{10^{3}}  [1   –   (\frac{4}{30} ) ^{\frac{1.3  –   1}{1.3} }  ]

= 1170 (1 – 0.6282) = 435 kJ/kg

∴                     Velocity at point 2,

Also                                v_{2 }   =   v_{1}   (\frac{p_{1}}{p_{2}})^{\frac{1}{n} }   =   0.09  ×   (\frac{30}{4} ) ^{\frac{ 1}{0.3} }  =  0.4239   m³ / kg

∴                      Mass flow rate,                  \dot{m}    =    \frac{A_{2 }    ×   C_{2 }}{v_{2 }}

5.2  =      \frac{A_{2 }    ×   932.7}{0.4239}

∴                                                    A_{2 }  =    \frac{5.2    ×  0.4239}{932.7}    =  0.002363 m²

Since aspect ratio is 3 : 1, if we assume breadth as x the length will be 3x and area of six nozzles will be

A_{2 }  = 6  ×  3x   ×   x

0.002363 = 18x²

∴                                     x =  (\frac{0.002363}{18} )^{1 / 2}       =  0.0114 m   or   11.4 mm. 

and                        length                          = 3 × 11.4 = 34.2 mm.
(ii) Degree of undercooling and supersaturation :
Temperature at point 2 is found as follows :

∴                                        T_{2}  =  T_{1}    ×    (\frac{p_{2} }{  p_{1}}) ^{ \frac{ n  –  1}{n} }   =  (273   +    350)    ×     (\frac{4}{30} ) ^{\frac{ 1.3  –  1}{1.3} }
= 623 × 0.628 = 391.2 K or 118.2°C
From steam tables saturation temperature at 4 bar
= 143.6°C
∴             Degree of undercooling = 143.6 – 118.2 = 25.4°C.

Saturation pressure corresponding to 118.2°C    \simeq    1.9 bar

∴    Degree of super saturation    =    \frac{4}{1.9}    =  2.1.

(iii) Loss in available heat drop :
Isentropic enthalpy drop for expansion under thermal equilibrium conditions as read out from Mollier chart

h_{1}      –     h_{3}  = 3115 – 2675 = 440 kJ/kg
∴                 Loss of available heat drop = 440 – 435 = 5 kJ/kg.

(iv) Increase in entropy =    \frac{5}{(143.6  +  273)}   =  0.012 kJ/kg K. 

(v) Ratio of mass flow rate :
Exit velocity from the nozzle with expansion in thermal equilibrium is given by

C_{3}  =    44.72  \sqrt{(h_{1}      –     h_{3} )}  =    44.72  \sqrt{440}   =  938   m / s

Also specific volume at 4 bar at state point 3 from Mollier chart,
v_{3}   = 0.46 m³/kg
∴                       Mass flow rate for metastable flow

∴                                        \frac{Mass    flow     rate     for    metastable    flow }{Mass    flow    rate    for    thermal    equilibrium       flow}

=  \frac{Area    of    flow  ×    C_{2}}{ v_{2}}    ×    \frac{v_{3} }{Area    of    flow  ×    C_{3}}

=  \frac{v_{3} C_{2}}{ v_{2} C_{3} }    =  \frac{0.46   ×    932.7}{0.4239   ×    938}  =  1.07.