Question 6.18: In a single stage steam turbine saturated steam at 10 bar ab...

Supply steam pressure (to nozzles) = 10 bar abs.
Nozzle angle,                               α = 20°
Mean blade speed,                    C _{bl}  = 400 m/s
Steam pressure leaving the nozzle= 1 bar abs.
Number of nozzles used                = 5
Area of throat at each nozzle        = 0.6 cm²
Nozzle efficiency,                                η _{nozzle}  = 88%
Blade friction co-efficient,                       K  =     \frac{ C _{r_0}}{ C _{r_1}}   = 0.87.

The velocity of steam at the outlet of nozzle is found representing the expansion through nozzle on h-s chart as shown in Fig. 27.

From h-s chart,
h_{1}    –    h_{3}    \simeq  402 kJ/kg

η_{nozzle}   =  \frac{ h_{1}    –    h_{3}  ^{′}}{ h_{1}    –    h_{3} }    =  0.88

∴                                                h_{1}    –    h_{3}  ^{′}   = 0.88 × 402 = 353.76 kJ/kg

Also                                          \frac{C_{3}  ^{′}²}{2}     =  h_{1}    –    h_{3}  ^{′}

or                                         C_{3}  ^{′}   =   \sqrt{2  ( h_{1}    –    h_{3}  ^{′} )}   =   \sqrt{2      ×      353.76     ×     1000}    =  841.14 m/s

(i) Blade angles :
Construct the velocity triangles as per data given as shown in Fig. 27.
By measurement, θ (= Φ) = 35.5°.

(ii) Maximum power developed, P :
For finding out the maximum power developed by the turbine let us first find out the maximum mass of steam passing through the nozzle.
The required condition for the maximum mass flow through the nozzle is given by

\frac{p_{2}}{p_{1}}  =    (\frac{2}{n   +   1})^{\frac{n}{n   –    1}}
where,                  p_{1} = Pressure of steam at inlet of the nozzle,
p_{2} = Pressure of steam at the throat of the nozzle, and
n (index of expansion) = 1.135 as steam is saturated.

∴                                              \frac{p_{2}}{p_{1}}  =    (\frac{2}{1.135   +   1})^{\frac{1.135}{0.135 }}   = 0.58

∴                                            p_{2}  = 10 × 0.58 = 5.8 bar

From h-s chart (Fig. 27)

h_{1}    –    h_{2}    \simeq    105 kJ/kg

and                                     h_{1}    –    h_{2}   ^{′}   = 0.88 × 105 = 92.5 kJ/kg
v_{2}   ^{′}    (specific    volume    at    point     2  ^{′}  )  \simeq   0.32 m³/kg
The maximum velocity of steam at the throat of the nozzle is given by

C  =   \sqrt{2  ( h_{1}    –    h_{2}  ^{′} )}   =   \sqrt{2      ×      92.4     ×     1000}     =   429.88   m/s

Using the continuity equation at the throat of the nozzle, we can write
m . v_{2}   ^{′}  = A × C where A is the area of the nozzle.
∴                                         m × 0.32 = 0.6 × 10^{–4} × 429.88

∴                                        m =  \frac{0.6    ×    10^{–4}  ×    429.88}{0.32}   =  0.0806 kg/s.

Total mass of steam passing through 5 nozzles per second is given by
m_{t}    = 0.0806 × 5 = 0.403 kg/s

∴     Power developed by the turbine =  \frac{m_{t}    ×     C_{w}      C_{bl}}{1000}   kW

From velocity diagram, C_{w}  = 750 m/s (by measurement)

∴    Power developed =  \frac{0.403    ×    750    ×     400}{1000}    = 120.9 kW.