A stage of a turbine with Parson’s blading delivers dry saturated steam at 2.7 bar from the fixed blades at 90 m/s. The mean blade height is 40 mm, and the moving blade exit angle is 20°. The axial velocity of steam is 3/4 of the blade velocity at the mean radius. Steam is supplied to the stage at

Question

A stage of a turbine with Parson’s blading delivers dry saturated steam at 2.7 bar from the fixed blades at 90 m/s. The mean blade height is 40 mm, and the moving blade exit angle is 20°. The axial velocity of steam is 3/4 of the blade velocity at the mean radius. Steam is supplied to the stage at the rate of 9000 kg/h. The effect of the blade tip thickness on the annulus area can be neglected. Calculate :
(i) The wheel speed in r.p.m. ; (ii) The diagram power ;
(iii) The diagram efficiency ; (iv) The enthalpy drop of the steam in this stage.

Accepted Answer

The velocity diagram is shown in Fig. 47 (a) and the blade wheel annulus is represented in Fig. 47 (b).
Pressure = 2.7 bar, x = 1, [latex]C _{1} [/latex] = 90 m/s, h = 40 mm = 0.04 m
α = φ = 20°, [latex]C_{f_1} = C_{f_0} = 3/4 C_{bl} [/latex]
Rate of steam supply = 9000 kg/h.

(i) Wheel speed, N :
[latex] C_{f} = 3/4 C_{bl} = C _{1} \sin 20° = 90 \sin 20° = 30.78 m/s [/latex]
∴ [latex] C_{bl} [/latex] = 30.78 × 4/3 = 41.04 m/s

The mass flow of steam is given by : [latex]\dot{m}_{s} = \frac{C_{f} A}{v} [/latex]

(where A is the annulus area, and v is the specific volume of the steam).

In this case, [latex]v = v_{g} [/latex] at 2.7 bar = 0.6686 m³/kg

∴ [latex]\dot{m}_{s} = \frac{9000}{3600} = \frac{30.78}{0.6686}[/latex] or A = [latex] \frac{9000 × 0.6686}{3600 × 30.78} [/latex] = 0.054 m²

Now, annulus area, A = πDh
(where D is the mean diameter, and h is the mean blade height)

∴ 0.054 = πD × 0.04 or D = [latex] \frac{0.054}{π × 0.04} [/latex] = 0.43 m

Also, [latex] C_{bl} = \frac{πDN}{60}[/latex] or 41.04 = [latex] \frac{π × 0.43 × N}{60} [/latex]

or N = [latex] \frac{ 41.04 × 60}{π × 0.43 } [/latex] = 1823 r.p.m.

(ii) The diagram power :

Diagram power = [latex]\dot{m}_{s} C_{w} C_{bl}[/latex]
Now, [latex]C_{w} = 2C_{1} \cos α – C_{bl}[/latex]
=[latex] 2 × 90 × \cos 20° – 41.04 = 128.1 m/s[/latex]

∴ Diagram power = [latex] \frac{9000 × 128.1 × 41.04 }{3600 × 1000}[/latex] = 13.14 KW.

(iii) The diagram efficiency :

Rate of doing work per kg/s = [latex]C_{w} C_{bl}[/latex] = 128.1 × 41.04 N m/s
Also, energy input to the moving blades per stage

= [latex] \frac{C_{1}²}{2} + \frac{C_{r_0} ² - C_{r_1} ²}{2} = \frac{C_{1}²}{2} + \frac{C_{1}² - C_{r_1} ²}{2} = C_{1}² - \frac{ C_{r_1} ²}{2} (∵ C_{r_0} = C_{1} ) [/latex]

Referring to Fig. 47 (a), we have

[latex]C_{r_1} ² = C_{1}² + C_{bl}² - 2 C_{1} C_{bl} \cos α [/latex]

[latex]= 90² + 41.04² – 2 × 90 × 41.04 × \cos 20°[/latex]
= 8100 + 1684.28 – 6941.69

∴ [latex]C_{r_1} [/latex] = 53.3 m/s

Energy input = 90² – [latex] \frac{53.3²}{2} [/latex] = 6679.5 Nm per kg/s

∴ Diagram efficiency = [latex] \frac{128.1 × 41.04 }{6679.5} [/latex] = 0.787 or 78.7%.

(iv) Enthalpy drop in the stage :
Enthalpy drop in the moving blades

= [latex] \frac{C_{r_0}² - C_{r_1}²}{2} = \frac{90² - 53.3²}{2 × 1000} = 2.63 kJ/kg (∵ C_{r_0} = C_{r_1} )[/latex]

∴ Total enthalpy drop per stage = 2 × 2.63 = 5.26 kJ/kg.

Question 6.32: A stage of a turbine with Parson’s blading delivers dry satu...

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