Question 12.CS.1: High-Speed Turbine Geared to Drive a Generator A turbine rot......

See Figures 12.6 and 12.8 and Tables 11.5 through 11.9.

a. The geometric quantities for the gearset are obtained by using Equations (12.1) through 12.5. Therefore,

p_n=p \cos \psi, \quad p_a=p \cot \psi=\frac{p_n}{\sin \psi}           (12.1)

P p=\pi, \quad P_n p_n=\pi, \quad P_n=\frac{P}{\cos \psi}, \quad P=\frac{N}{d}      (12.2)

m=\frac{1}{P} \quad m_n=\frac{1}{P_n}         (12.2′)

\tan \phi_n=\tan \phi \cos \psi         (12.3)

d=\frac{N p}{\pi}=\frac{N p_n}{\pi \cos \psi}=\frac{N}{P_n \cos \psi}         (12.4)

c=\frac{d_1+d_2}{2}=\frac{p}{2 \pi}\left(N_1+N_2\right)=\frac{N_1+N_2}{2 P_n \cos \psi}         (12.5)

\phi=\tan ^{-1} \frac{\tan \phi_n}{\cos \psi}=\tan ^{-1} \frac{\tan 20^{\circ}}{\cos 30^{\circ}}=22.8^{\circ}

P=P_n \cos \psi=10 \cos 30^{\circ}=8.66

d_p=\frac{N_p}{P}=\frac{35}{8.66}=4.04 \text { in. }

N_g=N_p\left(\frac{n_p}{n_g}\right)=35\left(\frac{8000}{1000}\right)=280

d_g=\frac{N_p}{P}=\frac{280}{8.66}=32.3  in.

It follows that

\begin{aligned} c & =\frac{1}{2}\left(d_p+d_g\right) \\ & =\frac{1}{2}(4.04+32.3)=18.17  in . \end{aligned}

Comment: The condition that the center distance is not to exceed 18.5 is satisfied.

The pitch-line velocity equals

V=\frac{\pi d n}{12}=\frac{\pi(4.04) 8000}{12}=8461  fpm

The power that the gear must transmit is about 335/0.95=353 hp. The transmitted load is then

F_t=\frac{33,000 hp }{V}=\frac{33,000(353)}{8461}=1.38  kips

As a result, the radial, axial, and normal components of tooth force are, using Equation (12.10),

\begin{aligned} & F_r=F_t \tan \phi \\ & F_a=F_t \tan \psi \\ & F_n=\frac{F_t}{\cos \phi_n \cos \psi} \end{aligned}           (12.10)

b. From Equation (12.7b), we have

N^{\prime}=\frac{N}{\cos ^3 \psi}         (12.7b)

N^{\prime}=\frac{N}{\cos ^3 \psi}=\frac{35}{\cos ^3 30^{\circ}}=53.9

Then, for 53.9 teeth and \phi=22.8^{\circ} by interpolation from Table 11.2, Y =0.452. Using Table 11.3, \sigma_o \approx 18 ksi. Applying Equation ((11.33), modified) with K_f =1,

F_b=\frac{\sigma_o b}{K_f} \frac{Y}{P}         (11.33)

F_b=\sigma_o b \frac{Y}{P_n}=18(8) \frac{0.452}{10}=6.51  kips

Corresponding to \phi=22.8^{\circ} , interpolating in Table 11.9, K \approx 68 psi. Through the use of Equation (11.40),

Q=\frac{2 N_g}{N_p+N_g}        (11.40)

Q=\frac{2 N_g}{N_p+N_g}=\frac{2(280)}{35+280}=\frac{112}{63}

The limit load for wear, by Equation ((11.38), modified), is

F_w=d_p b Q K        (11.38)

\begin{aligned} F_w & =\frac{d_p b Q K}{\cos ^2 \psi} \\ & =\frac{4.04(8)(68)(112)}{60\left(\cos ^2 30^{\circ}\right)}=5.21  kips \end{aligned}

Because the permissible load in wear is less than that allowable in bending, it is used as the dynamic load F_d . So from Equation ((11.24c), modified),

F_d=\frac{78+\sqrt{V}}{78} F_t \quad(\text { for } V>4000  fpm )        (11.24c)

\begin{aligned} F_d & =\frac{78+\sqrt{V}}{78} F_t \\ 5.21 & =\frac{78+\sqrt{8461}}{78} F_t=2.18 F_t \end{aligned}

or

F_t=2.39  kips

c. Application of Equation (11.36) leads to

\sigma_{\text {all }}=\frac{S_t K_L}{K_T K_R}        (11.36)

\sigma_{\text {all }}=\frac{S_t K_L}{K_T K_R}

where

S_t = 20.5 ksi (interpolating, Table 11.6 for a Bhn of 150)

K_L = 1.0 (indefinite life, Table 11.7)

K_T = 1.0 (from Section 11.9)

K_R = 1.25 (by Table 11.8)

The preceding equation is therefore

\sigma_{\text {all }}=\frac{20,500(1.0)}{(1.0)(1.25)}=16.4  ksi

The tangential force, by Equation (11.35) with \sigma =\sigma _{all} , is

\begin{aligned} & \sigma=F_t K_o K_{ \upsilon } \frac{P}{b} \frac{K_s K_m}{J} \quad \text { (US customary units) } \\ & \sigma=F_t K_o K_{ \upsilon } \frac{1.0}{b m} \frac{K_s K_m}{J} \text { (SI units) } \end{aligned}         (11.35)

F_t=\frac{\sigma_{\text {all }}}{K_o K_{ \upsilon }} \frac{b}{P} \frac{J}{K_s K_m}    (a)

Here, we have

K_o =1.5 (Table 11.4)

K_ {\nu} =2.18 (from curve B of Figure 11.15)

K_s =1.1 (from Section 12.5)

K_m =1.5 (by Table 11.5)

J =0.47 (for N_p =35 and \psi= 30, Figure 12.6(a))

J − multiplier=1.02 (for N_g=280 and \psi= 30, Figure 12.6(b))

J =1.02×0.47=0.48

Equation (a) is then

F_t=\frac{16,400(8)(0.48)}{(1.5)(2.18)(8.66)(1.0)(1.5)}=1.48  kips

compared to the approximate result 2.39 kips for part b.

Comments: The tangential load capacity of the gearset, 1.48 kips, is larger than the force to be transmitted, 1.38 kips (allowance is made for 95% generator efficiency); the gears are safe. Since a wide face width is used, the design should be checked for combined bending and torsion at the pinion [1].

Remarks: A turbine is a rotary mechanical device that extracts energy from a fluid flow and converts it into work. This work created by a turbine-generator assembly can be used for generating electrical power. A turbine is a turbomachine with at least one moving part termed a rotor assembly, that is, a shaft with blades attached. Moving fluid acts on the blades so that they move and transmit rotational energy to the rotor. Basic types of turbines are water, steam, gas, and wind turbines. The identical principles apply to all turbines; however, their specific designs differ sufficiently to merit separate descriptions.

Turbines are often part of a large machine. Almost all electric power on earth is generated with a turbine of some type. Gas, steam, and water turbines have a casing around the blades that contains and controls the working fluid. A steam turbine is used for the generation of electricity in thermal power plants, such as plants using coal, fuel oil, or nuclear fuel. Gas turbines are sometimes referred to as turbine engines. Such engines usually include an inlet, fan, compressor, combustor, and nozzle in addition to one or more turbines. Water turbines convert the potential energy of water on an upstream level into kinetic energy.

A wind turbine, illustrated in the following photo, is designed to convert the wind energy that exists at a location to electricity. We observe that, in this turbine there are three components: the blades converting wind energy to low speed rotational energy; the housing (including a drive shaft, gear train, high speed shaft, couplings, and generator); the structural support consisting of tower, rotating thrust bearing, and rotor yaw mechanism. Aerodynamic modeling is used to find the optimum tower height, control systems, number of blades, and blade shape.