Answer :
The horsepower to be taken off the driven shaft is A. 25 hp.
- Given Values:
Diameter of driving pulley, [tex]D_d = 84 \text{ in.}[/tex]
Diameter of driven pulley, [tex]D_f = 48 \text{ in.}[/tex]
Speed of driving pulley, [tex]N_d = 150 \text{ rpm}[/tex]
Tension on the tight side, [tex]T_1 = 300 \text{ lb}[/tex]
Tension on the slack side, [tex]T_2 = 50 \text{ lb}[/tex]
- Belt speed [tex]V[/tex] can be calculated using the formula:
[tex]V = \frac{\pi \times D_d \times N_d}{12 \times 60}[/tex]
- Where:
[tex]\pi[/tex] is approximately 3.1416,
12 is used to convert inches to feet,
60 is used to convert minutes to seconds.
- So, substituting the values,
[tex]V = \frac{3.1416 \times 84 \times 150}{12 \times 60}[/tex]
[tex]= 54.97 \text{ ft/s}[/tex]
- The effective belt tension [tex]T[/tex] is the difference between the tensions on the tight side and slack side of the belt:
[tex]T = T_1 - T_2[/tex]
[tex]= 300 - 50 = 250 \text{ lb}[/tex]
- Horsepower (HP) can be calculated using the formula:
[tex]HP = \frac{T \times V}{550}[/tex]
Substituting the values,
[tex]HP = \frac{250 \times 54.97}{550}[/tex]
[tex]= 24.98[/tex]
Rounding off to the nearest whole number we get,
[tex]\approx 25 \text{ hp}[/tex]
