If the sum of whirl components of velocity at the inlet and outlet of a DE level turbine is found to be 1200 m/s and the mass flow rate of steam is 15 kg/minute, then the tangential force on the blade is: 

Explanation:

Concept:

De Laval Turbine:

• It is also known as a simple rotor impulse turbine.
• It consists of a single set of rotors and a nozzle which is attached to the wheel/disc.
• The nozzle converts the pressure energy into kinetic energy, on the other hand, the rotor uses this kinetic energy and converts it into rotational mechanical power.
• The velocity diagram is shown below:

Where, V₁ & V₂ = absolute velocity at inlet & outlet

V_r1 & V _r2 = relative velocity at inlet & outlet

V_w1 & V_w2 = whirl component of absolute velocity at inlet & outlet

V_f1 & V_f2 = axial component of absolute velocity at inlet & outlet

U = mean blade velocity

\(\alpha_{1} \) = nozzle angle, \(\alpha_{2}\) = absolute angle of steam at the outlet

\(\beta_{1}\) & \(\beta_{2}\) = blade angle at inlet & outlet.

m = steam flow rate through blades (kg/s)

From Newton's 2^nd law of motion:

Tangential force, F_t = mass \(\times\) tangential acceleration = mass flow rate \(\times\) change in velocity

\(F_{t}=\dot m(V_{w1}\pm V_{w2})\)

Note: 

• +ve sign is used when V_w2 & U are in opposite direction
• -ve sign is used when V_w2 & U are in same direction.

Calculation:

Given:

V_w1 + V_w2 = 1200 m/s

̇ṁ = 15 kg/min

Then, \(F_t=\dot m(V_{w1}+V_{w2})\) = \(\frac{15}{60}\times1200\)

\(F_t=300\space N\)

Thus, option (3) is correct answer.