Question

In a stage of an impulse turbine provided with a single row wheel, the mean diameter of blade ring is 0.08 m and the speed of rotation is 3000 rpm. The steam issues from the nozzle with a velocity of 300 m/s and the nozzle angle is 20 degree. The blades are equiangular and blade coefficient is 0.85. determine the power developed in the blades when the axial thrust on the blade is 150 N. Also determine the diagram efficiency.

Answer 1

Answer:

In a stage of an impulse turbine provided with a single row wheel, the mean diameter of the blade ring is 800 mm and the speed of rotation is 3000 rpm. The steam issues from the nozzles with a velocity of 300 m/s and the nozzle angle is 20°. The rotor blades are equiangular and the blade friction factor is 0.86. What is the power developed in the blading when the axial thrust on the blades is 140 newtons?

Answer 2

The power developed in the blades is 373.20 kw

Explanation:

Given:

The diameter of a blade ring D = 0.08 m

Speed of rotation N= 3000 rpm

Velocity V= 300  m/s

$V_{b}=\frac{\pi DN}{60}$

⇒$\frac{(3.14).(0.08).(3000)}{60}$

⇒$12.56 m/s$

$v_{1} = 300 m/s$

$\frac{Vr_{2} }{Vr_{1} } =0.85$

$Vf_{1} =V_{1} sin20$

⇒$300sin20$

⇒$102.6$ m/s

$Vw_{1} =v_{1} cos20$

⇒$300cos20$

⇒$281.9 m/s$

$tan$θ= $\frac{Vf_{1} }{Vw_{1} - V_{b} }$

⇒$\frac{102.6}{281.9 -12.56}$

⇒$\frac{102.6}{269.3}$

⇒$0.380$

$tan$θ = $0.380$

θ =$tan^{-1} 0.380$

θ = $20.8^{o}$

sinθ = $\frac{Vfx_{1} }{Vr_{1} }$

$Vr_{1} =\frac{Vf_{1} }{sin}$θ

⇒$\frac{102.6}{sin20.8}$

⇒$\frac{102.6}{0.928}$

⇒$110.56$

$Vr_{1} =110.56$

$\frac{Vr_{2} }{Vr_{1} } =0.85$

$Vr_{2} =(0.85)(110.56)$

$Vr_{2} =93.97 m/s$

$Vf_{2} = 93.97 sin 20.8$

$Vf_{2} = (93.77)(0.9287)$

$Vf_{2} = 87.08 m/s$

$Vw_{2} = 93.97cos20 - V_{b}$

$Vw_{2} =93.97(0.4080) - 12.56$

$Vw_{2} =38.33 -12.56$

$Vw_{2} =25.7 m/s$

$P = m_{s}[Vw_{1} +Vw_{2} ]V_{b}$

Axial thrust $F_{T} =m_{s}[Vf_{1} -Vf_{2} ]$

$150 = m_{s}[102.6-87.08]$

$150=m_{s}[15.52]$

$m_{s} = 9.66 kg/s$

$P = m_{s}[Vw_{1} +Vw_{2} ]V_{b}$

$p=9.66[281.9+25.7]12.56$

$p=9.66(307.6)(12.56)$

$p = 373.20 kw$

Final answer:

The power developed in the blades with the axial thrust  is 373.20 kw

#SPJ3