Question

The Power Input at Gear P is 1.7802 kW, rotating at 222.222 RPM. Assume the the Module m is the same for all gears. a. What is the Input Torque at gear P? b. Assuming 100% efficiency at all gear pairs, what is the Output Power, Torque and RPM at gear Q? c. If the efficiencies of the gear pairs are: ηP-B = 90%, ηC-Q = 95%, what is the Output Power and Torque at gear Q? d. At 100% efficiency, what are the gear tooth tangential and Normal forces at the P-B and C-Q contacts: FT-B , FN-B , FT-C & FN-C ?

Answer 1

Answer:

From the example the answer are (a) 76.499 N-M (b) 179.43N-M (c) 159.88 N-M (d) 867.93N

Explanation:

Solution

From the example given we solve the following problems.

Recall

The power input at given P (Pp) = 1.7802 kW

The speed of the gear P (Np) = 222.222 RPM

Let the module  m be the same for all the gears

(a)Torque input gear at P/Pp =

Tp  *  2πNp/60 =

1.7802 * 10^3 = Tp * 2π * 222.222/60

Tp = 76.499 N-M

(b) Let assume that the there is 100 % efficiency at all pair of gears,

Pp = Pb  =Pc =Pq

= 1.7802 * 10 ^3 = Tb * 2πNb/60 = Tc * 2πNc/60=  Tq * 2πNc/60

now,

dp = 18cm,

db= 40cm

dc= 20cm

dq =44cm

where d = mƵ, Ƶ = Number of teeth

Then

Ƶp /Ƶq = Nq/Np

dp/m/dq/m =Nq/222.222

Nq = 90. 908RPM

The output power which is Pq =1.7802kW

The output torque Tq = 179.43N-M

(c) = Pq = Tq *  2πNq/60 = 1.522071 * 10^3

= Tq *2πNq / 60 = 159.88 N-M

Tq = 159.88 N-M

(d) Now, at efficiency of 100%

(Ftb) * (db/2) = Tb

Ft-b = Ft-p

Ft-p * dp/2 =Tp

Ftp =849.989 N

Ft-b =849.989 N

Ft-b = Ft-n cos 20%

Ft-n = 904.54 N

so,

Ft-c = Ft-q

(Ft -q) * dq/2 = Tq

Ft-q = 815.59 N

where,

Ft-c = 815.59 N

then,

Ft-c = Ft-n cos 20

Ft-n = 867. 93N