Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Chemical Engineering » Thermal Engineering MCQ’s – Impulse Turbines – 1

# Thermal Engineering MCQ’s – Impulse Turbines – 1

Which of the following statements regarding impulse turbine is correct?

a) Only velocity of flow performs work on the blade

b) Only velocity of Whirl performs work on the blade

c) Axial component of velocity is called velocity of whirl

d) Relative velocity of steam to moving blade at the inlet is always less than the relative velocity at outlet

**
View Answer**

Answer: bExplanation: The steam is impinged on the blades at the nozzle angle. Only the tangential i.e. velocity of whirl performs work on the blades. The axial component i.e. velocity of flow causes the steam to flow through the turbine and has no contribution in work. The relative velocity of stem at exit is always less than (or equal if there is no friction loss at the blade) the relative velocity at the entrance.

Which of the following expressions represent the correct formula for calculating the blade efficiency?

a) (Cw1+Cw2)∗CblC21

b) 2(Cw1+Cw2)∗Cbl)C22

c) 2(Cw1+Cw2)∗Cbl)C21

d) 2(Cw1+Cw2)C1

**
View Answer**

Answer: c

Explanation: Blade efficiency or diagram efficiency is the ratio of work done on the blade to energy supplied to the blade.

Mathematically,

ηblade = ṁ(Cw1+Cw2)∗CblṁC222=2(Cw1+Cw2)∗Cbl)C21

Symbols –

Cbl – Linear velocity of moving blade

C1 – Absolute velocity of steam entering the moving blade

Cw1 – Velocity of whirl at the entrance of the moving blade

C2 – Absolute velocity of steam leaving the moving blade

Cr2 –Relative velocity of steam to moving blade at exit

Cw2 – Velocity of whirl at the exit of the moving blade

ṁ – Rate of flow of steam.

The ratio of relative velocity of steam to the moving blade at exit to the relative velocity of steam to the moving at entrance is called _____

a) reduction co-efficient

b) blade velocity co-efficient

c) friction loss

d) factor of reduction

**
View Answer**

Answer: bExplanation: Blade velocity co-efficient (K) is the ratio of relative velocity of steam to the moving blade at the exit to relative velocity of moving blade at the entrance. As the steam passes over the blades of the turbine the velocity is reduced due to the friction present on the surface of the blades.

Blade speed ratio is the _____

a) ratio of blade speed to steam speed at entrance of the moving blade

b) ratio of steam speed at entrance of the moving blade to blade speed

c) ratio of blade speed to steam speed at exit of the moving blade

d) ratio of steam speed at exit of the moving blade to blade speed

**
View Answer**

Answer: a

Explanation: Blade speed ratio (ρ) is the ratio of blade speed to the steam speed at the entrance of the moving blade.

Mathematically,

ρ = CblC1

For maximum efficiency blade speed ratio should be half of the cosine of the nozzle angle.

What is the maximum possible value of blade efficiency?

a) sinα2

b) cosα2

c) sinβ2

d) cosβ2

**
View Answer**

Answer: bExplanation: The maximum value of blade efficiency is obtained when the blade speed ratio (ρ) is half the cosine of the nozzle angle. The maximum possible value of blade efficiency is cosα2 considering the blades to be symmetrical and no friction in fluid passage.

What is the optimum blade speed ratio for maximum blade efficiency in a multi-stage turbine? (n denotes the number of moving blade rows in series)

a) cosα2∗n

b) cosβ2∗n

c) 2∗cosαn

d) 2∗cosβn

**
View Answer**

Answer: aExplanation: The optimum blade speed ratio (ρ) for maximum blade efficiency is given by – cosα2∗n Also, in this case the ratio of total work to the work done in the last row is 2n. The maximum efficiency is cosα2.

Which of the following statements regarding velocity compounded impulse turbine is FASLE?

a) Maximum stage efficiency decreases with the with the increase in number of moving blades’ rows

b) It has low efficiency and high steam consumption

c) Optimum value of blade speed ratio for maximum efficiency increases with the with the increase in number of moving blades’ rows

d) It requires comparatively small number of stages

**
View Answer**

Answer: cExplanation: As the number of moving blades’ rows increases the optimum value of blade speed ratio for maximum efficiency decreases. A velocity compounded impulse turbine requires comparatively less number of stages, due to relatively large heat drop. Usually, more than two rows are not preferred.

Steam issues from the nozzle with a velocity of 1000 m/s in a De Laval turbine. The nozzle angle is 20°, mean blade velocity is 350 m/s. The inlet and outlet angles of the blades are equal. The mass of the steam flowing through the turbine per second is 0.3 kg. Calculate the tangential force on the blades. Take blade velocity coefficient as 0.8.

a) 451 N

b) 256 N

c) 318 N

d) 196 N

**
View Answer**

Answer: c

Explanation: Cbl = 350 m/s, C1 = 1000 m/s, α = 20°, K = 0.8, ṁ = 0.3 kg/sec, θ = Ф

Steps of Construction –

Select a suitable scale and draw a line LM to represent C_{bl} (= 350 m/s).

At point M make angle of 20° (α) and cut length MS to represent the velocity 1000 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 30.4°, C_{r1} = 681.7 m/s

θ = Ф = 30.4°

C_{r2} = KC_{r1} = 0.8(681.7) = 545.36 m/s

At point L, make an angle of 30.4° (Ф) and cut the length LN to represent C_{r2}(= 545.36 m/s). Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C_{w1} = 939.69 m/s and C_{w2} = 120.38 m/s

Tangential force on the blades = ṁ (C_{w1} + C_{w2}) = 0.3(939.69 + 120.38) = 318.021 N ≈ 318 N

Velocity of the steam leaving the nozzle is 1200 m/s. The nozzle angle is 20° and the mean blade velocity is 500 m/s. The inlet and outlet angles of blades are equal. The mass of the steam flowing through the turbine per second is 0.4 kg. Considering the blades to be frictionless determine the Power Developed.

a) 251.30 kW

b) 150.21 kW

c) 123.65 kW

d) 321.56 kW

**
View Answer**

Answer: a

Explanation: Cbl = 500 m/s, C1 = 1200 m/s, α = 20°, ṁ = 0.4 kg/sec, θ = Ф

Steps of Construction –

Select a suitable scale and draw a line LM to represent C_{bl} (=500 m/s).

At point M make angle of 20° (α) and cut length MS to represent the velocity 1200 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 33°, C_{r1} = 749.91 m/s

θ = Ф = 33°

C_{r2} = C_{r1} = 749.91 m/s

At point L, make an angle of 33° (Ф) and cut the length LN to represent C_{r2}(= 749.91 m/s). Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C_{w1} = 1127.6 m/s and C_{w2} = 128.9 m/s

Power Developed = ṁ (C_{w1} + C_{w2}) C_{bl} = 0.4(1127.6 + 128.9)500 = 251300 W or 251.3 kW

Steam is issued from the nozzle at a velocity of 1000 m/s. The mean blade velocity is 450 m/s and the nozzle angle is 25°. The inlet and outlet angles of blades are equal. The mass of the steam flowing through the turbine per hour is 1000 kg. Calculate the blade efficiency if the blade velocity coefficient is 0.85.

a) 88%

b) 95%

c) 50%

d) 76%

**
View Answer**

Answer: d

Explanation: Cbl = 450 m/s, C1 = 1000 m/s, α = 25°, K = 0.85, θ = Ф, ṁ = 1000 kg/hr or 0.278 kg/sec

Steps of Construction –

Select a suitable scale and draw a line LM to represent C_{bl} (=450 m/s).

At point M make angle of 25° (α) and cut length MS to represent the velocity 1000 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 43°, C_{r1}=621.95 m/s

θ = Ф = 43°

C_{r2} = KC_{r1} = 0.85(621.95) = 528.66 m/s

At point L, make an angle of 43° (Ф) and cut the length LN to represent C_{r2}(= 528.66 m/s). Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C_{w1} = 906.31 m/s and C_{w2} = 66.36 m/s (Opposite direction)

Blade efficiency = \(\frac{2(C_{w1}+ C_{w2})C_{bl}{C_1^2} = \frac{2(906.31-66.36)450}{1000^2}\) = 0.76 or 76%.

If the blade and nozzle efficiencies are 75% and 85% respectively, determine the stage efficiency.

a) 75.65%

b) 45.32%

c) 63.75%

d) 55.32%

**
View Answer**

Answer: cExplanation: ηblade = 0.75, ηnozzle = 0.85 ηstage = ηblade*ηnozzle = 0.75*0.85= 0.6375 or 63.75%.

If for a certain simple impulse turbine, the blade inlet and outlet angles are 35° and 30° respectively and the relative velocity of steam to moving blade at entrance and exit are 350 m/s and 320 m/s respectively, determine the axial thrust on the bearings. The turbine has a steam flow rate of 20 kg/s.

a) 850 N

b) 815 N

c) 750 N

d) 456 N

**
View Answer**

Answer: b

Explanation: Cr1 = 350 m/s, θ = 35, Cr2 = 320 m/s, Ф = 30°, ṁ = 20 kg/s

Cf1 = cr1*sinθ = 350*sin35 = 200.75 m/s

Cf2 = cr2*sinФ = 320*sin30 = 160 m/s

Axial Thrust = ṁ (Cf1 – Cf2) = 20 (200.75 – 160) = 815 N.

A simple impulse turbine has a mean blade speed of 250 m/s. The nozzle angle is 20° and the steam velocity from the nozzle is 600 m/s. The turbine uses 3000 kg/h of steam. The blade outlet angle is 25°. Determine the amount of energy converted to heat by blade friction if the blade friction factor is 0.86.

a) 65.32 kJ

b) 30.50 kJ

c) 45.32 kJ

d) 78.32 kJ

**
View Answer**

Answer: b

Explanation: Cbl = 250 m/s, α = 20°, C1 = 600 m/s, ṁ=3000 kg/h or 0.833 kg/s, K = 0.86, Ф = 25°

Steps of Construction –

Select a suitable scale and draw a line LM corresponding to 250 m/s (C_{bl}).

At M, at an angle of 20° draw a line MS having a length of corresponding to 600 m/s (C_{1}). Join L and S. The inlet triangle is completed.

By measurement –

C_{r1} = 374.96 m/s

C_{r2} = K(C_{r2}) = 0.86(374.96) = 322.46 m/s

To complete the outlet triangle, draw a line of length corresponding to 322.46 m/s (C_{r2}) i.e. LN at an angle of 25° at L. Join MN.

Amount of energy converted to heat by blade friction = ṁ (C_{r1}^{2}-C_{r2}^{2}) = 0.833 (374.96^{2}-322.46^{2}) = 30499.92 J or 30.5 kJ.

Steam is issued from the nozzles at 600 m/s in a simple impulse turbine. The nozzle angle is 25° and the mean blade velocity is 350 m/s. The blade velocity coefficient is 0.88. Determine the blade speed ratio.

a) 0.325

b) 0.583

c) 0.654

d) 0.785

**
View Answer**

Answer: bExplanation: C1 = 600 m/s, α = 20°, Cbl = 350 m/s, K = 0.88 Blade speed ratio (ρ) = CblC1 = 350/600 = 0.583.

Nozzle angle of a simple impulse turbine is 25°. Determine the blade speed ratio if the turbine works at maximum efficiency. The Turbine blades are to be assumed frictionless.

a) 0.4532

b) 0.2136

c) 0.5465

d) 0.6987

**
View Answer**

Answer: a

Explanation: α = 25°

Since the turbine efficiency is maximum,

Blade speed ratio (ρ) = cosα2=cos252 = 0.4532.