Thermal Engineering MCQ’s – Reaction Turbines – 2

Answer: c
Explanation: Given, N = 1400 r.p.m., D = 1 m, hd = 8.530 kJ/Kg, Ф = 20°, ρ = 0.8
Cbl = πDN60=π∗1∗140060 = 73.30 m/s
we know that,
ρ = CblC1 0.8 = 73.30C1 C1 = 91.63 m/s
For Parson’s reaction turbine,
α = Ф = 20°
Steps of Construction:
Select a suitable scale and draw a line LM to represent Cbl (= 73.30 m/s).
At point M, at an angle of 20° (α), cut a length MS to represent the velocity C1 (= 91.63 m/s). Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.
For a Parson’s turbine:
Cr2 = C1 = 91.63 m/s
and Ф = 20°
At L, at an angle of 20° (Ф), cut a length LN to represent Cr2 (= 91.63 m/s). Join MN.
Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.
By measurement:
Cw1 + Cw2 = 98.91 m/s
ηstage=\frac{C_{bl}(C_{w1}+C_{w2})}{h_d} = \frac{73.30(91.63)}{8530} = 0.7874 or 78.74%.