Question

A jet turbine rotates at a velocity of 7,500 rpm. Calculate the stress acting on the turbine blades if the turbine disc radius is 70 cm and the cross-sectional area is 15 cm2. Take the length to be 10 cm and the alloy density to be 8.5 g/cm3.

Answer 1

Answer:

       stress   = 366515913.6 Pa

Explanation:

given data:

density of alloy = 8.5 g/cm^3 = 8500 kg/m^3

length turbine blade = 10 cm =  0.1 m

cross sectional area = 15 cm^2 = 15*10^-4 m^2

disc radius = 70 cm = 0.7 m

angular velocity = 7500 rpm = 7500/60 rotation per sec

we know that

stress = force/ area

force = m*a

where a_{c} is centripetal acceleration =

$a_{c} =r*\omega ^{2}= r*(2*\pi*\omega)^{2}$

         =$0.70*(2*\pi*\frac{7500}{60})^{2}$

         = 431795.19 m/s^2

mass = $\rho* V$

Volume = area* length = 15*10^{-5}  m^3

$mass = m = \rho*V = 8500*15*10^{-5} kg$

force = m*a_{c}

         $=8500*15*10^{-5}*0.70*(2*\pi*\frac{7500}{60})^{2}$

force = 549773.87 N

stress = force/ area

          = $\frac{549773.87}{15*10^{-5}}$

       stress   = 366515913.6 Pa