In a De Laval turbine steam issues from the nozzle with a velocity of 1200 m/s. The
nozzle angle is 20°, the mean blade velocity is 400 m/s, and the inlet and outlet angles
of blades are equal. The mass of steam flowing through the turbine per hour is 1000 kg.
Calculate
(i) Blade angles.
(ii) Relative velocity of steam entering the blades.
(111) Tangential force on the blades.
(iv) Power developed
(v) Blade efficiency.
Take blade velocity co-efficient as 0.8.
Answers
Answer:
1. 30°
2. 840m/s
3. 308.632 N
4. 123.853 KW
5. 72.9%
Explanation:
The explanation is provided in below pictures.
Blade angles -- 30°
Relative velocity of steam entering the blades -- 830 m/s
Tangential force on the blades -- 363.8 N
Power developed -- 145.5 kW
Blade efficiency -- 72.8%
Explanation:
- Select a suitable scale and draw line LM to represent C,(= 400 m/s).
- At point L make an angle of 20° (a) and cut length LS to represent velocity C,(= 1200 m/s).
- Join MS. Produce M to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.
By measurement : t = 30°, C(r1) = MS = 830 m/s
t = p = 30° (given)
Now, C,, = KC, = 0.8 x 830 = 664 m/s
- At point M make an angle of 30° (6) and cut the length MN to represent C,, (= 664 m/s). Join LN. Produce L to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.
Blade angles 6, 6 :
As the blades are symmetrical (given)
t = p = 30°. (Ans.)
- (ii) Relative velocity of steam entering the blades, C,, 3
C,, = MS = 830 m/s. (Ans.)
- (iii) Tangential force on the blades :
Tangential force = M,(Cy, +Cw,) =
1000 (1310)/60.60 = 363.8 N. (Ans.)
- (iv) Power developed, P :
1000 1310 x 400
P = ti(Cu, +Cu,) Cy = Sox<60 * jon
kW = 145.5 kW. (Ans.)
- (v) Blade efficiency, n,, :
Wwi(Cy, +Cuy) 2x 400 x 1310 (Ans)
ws Ce =~ q9008 = 72.8% (Ans.