Calculate the amount of rock, in tonnes per hour, which can be transported by a belt conveyor having the following characteristics: maximum belt tension in the belt = 40 kN, the contact angle = 199°, the power to overcome friction = 5 kW, belt speed = 1,8 m/s, vertical height = 37 m, and the coefficient of friction is 0,25

Answer 1

To determine the capacity of the belt conveyor to transport rock, we need to understand the system's power mechanics and the load dynamics. Given the characteristics provided, we'll make use of the belt conveyor power formula and the material mass balance equation.

First, let's establish the power available to move the material:
- Total power (P_total) is usually the sum of power needed to overcome friction (P_friction) and the power needed to lift the material (P_lift):
\[ P_{total} = P_{friction} + P_{lift} \]

We are given:
- \( P_{friction} = 5 \) kW
- Belt speed, \( V = 1.8 \) m/s
- Vertical height to be overcome, \( h = 37 \) m

We need to determine the available power \( P_{lift} \) for lifting:
\[ P_{lift} = P_{total} - P_{friction} \]

However, \( P_{total} \) is what typically drives the entire system, but in this problem, we primarily work with overcoming friction specifically mentioned.

Total power applied by the driving mechanism in \( kW \):
\[ P_{total} = 5 \text{ kW} + P_{lift} \]

Next, calculate the belt tension:
- Maximum belt tension \( T \) is given as 40 kN:
\[ T = 40 \text{ kN (which is 40000 N)} \]

- The coefficient of friction \( \mu = 0.25 \)
- The contact angle \( \theta = 199° \) (which needs to be in radians for calculation):
\[ \theta = 199 \times \frac{\pi}{180} \approx 3.47 \text{ radians} \]

To determine the lifting power, we use the relationship for the lifting part:

### Henry's conveyor calculation - Lifting component

1. Calculate the lifting force:
\[ F_lift = \text{mass flow rate (ṁ)} \cdot g \cdot h \]
where the mass flow rate \(\dot{m}\) is what we need to find, \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity), and \( h = 37 \text{ m} \).

We change over power for lifting (P_lift):
\[ P_{lift} = \dot{m} \cdot g \cdot h \text{ (in Watts)} \]

\[ P_{lift} = \dot{m} \cdot 9.81 \cdot 37 \text{ W} \equiv P_{lift} \text{ W} \]

Since 1 kW = 1000 W, thus:
\[ \displaystyle \frac{P_{total} - 5}{9.81 \cdot 37} \cdot 1000 = \dot{m} \]
solving,

\[ \displaystyle \frac{5}{9.81 \cdot 37} \cdot 1000 \approx 13.85 \text{ kilograms per second} \]

To convert kilograms per second to tonnes per hour:
\[ 1 \text{ tonne} = 1000 \text{ kilograms}\]

Thus:
\[ \dot{m} = 13.85 \cdot 3600 / 1000 = 49.86 \text{ tonnes per hour} \]

Therefore, the belt conveyor can transport **49.86 tonnes per hour** of rock, considering the power required to overcome both frictional forces and the vertical lift of the material.