A 200 t train experiences wind resistance equivalent to $$ 62.5 \mathrm{~N} / \mathrm{t} $$. The operating efficiency of the driving motors is 0.87 and the cost of electrical energy is $$ 0.8 \mathrm{Rand} / \mathrm{kW} \mathrm{h} $$. What is the cost of the energy required to make the train travel 1 km ? If the train is supplied at a constant voltage of 1.5 kV and travels with a velocity of $$ 80 \mathrm{~km} / \mathrm{h} $$, what is the supply current? [R3.2; 214A]

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**Step 1. Calculate the total force due to wind resistance**

The wind resistance is given as  
$$
62.5\ \frac{\mathrm{N}}{\mathrm{t}}
$$
For a $$ 200\ \mathrm{t} $$ train, the total resistance force is  
$$
F = 62.5\ \frac{\mathrm{N}}{\mathrm{t}} \times 200\ \mathrm{t} = 12500\ \mathrm{N}.
$$

**Step 2. Determine the work done to travel 1 km**

Work done is the product of force and distance:
$$
W = F \times d,
$$
where $$ d = 1\ \mathrm{km} = 1000\ \mathrm{m} $$. Thus,
$$
W = 12500\ \mathrm{N} \times 1000\ \mathrm{m} = 12.5 \times 10^6\ \mathrm{J}.
$$

**Step 3. Adjust the work for the motor efficiency**

The operating efficiency of the motors is $$ \eta = 0.87 $$. This means that the energy drawn from the electrical supply must be higher than the mechanical work output. The actual energy required is:
$$
W_{\text{supplied}} = \frac{W}{\eta} = \frac{12.5 \times 10^6\ \mathrm{J}}{0.87} \approx 14.37 \times 10^6\ \mathrm{J}.
$$

**Step 4. Convert the energy to kilowatt-hours and calculate its cost**

Recall that  
$$
1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}.
$$
Thus, the energy in kilowatt-hours is:
$$
E_{\mathrm{kWh}} = \frac{14.37 \times 10^6\ \mathrm{J}}{3.6 \times 10^6\ \mathrm{J/kWh}} \approx 3.99\ \mathrm{kWh}.
$$
The cost of electrical energy is given as $$ 0.8\ \mathrm{Rand/kWh} $$. Therefore, the cost is:
$$
\text{Cost} = 3.99\ \mathrm{kWh} \times 0.8\ \frac{\mathrm{Rand}}{\mathrm{kWh}} \approx 3.19\ \mathrm{Rand}.
$$

**Step 5. Determine the supply current when the train travels at $$ 80\ \mathrm{km/h} $$**

1. **Calculate the train's speed in meters per second:**  
   $$
   v = 80\ \frac{\mathrm{km}}{\mathrm{h}} = \frac{80}{3.6}\ \frac{\mathrm{m}}{\mathrm{s}} \approx 22.22\ \frac{\mathrm{m}}{\mathrm{s}}.
   $$

2. **Calculate the mechanical power required to overcome resistance at this speed:**  
   $$
   P_{\text{traction}} = F \times v = 12500\ \mathrm{N} \times 22.22\ \frac{\mathrm{m}}{\mathrm{s}} \approx 277777.5\ \mathrm{W}.
   $$

3. **Account for the driving motor efficiency:**  
   Since the motors operate at $$ 87\% $$ efficiency, the electrical power drawn is:
   $$
   P_{\text{electrical}} = \frac{P_{\text{traction}}}{0.87} \approx \frac{277777.5\ \mathrm{W}}{0.87} \approx 319080\ \mathrm{W}.
   $$

4. **Calculate the supply current using $$ I = \frac{P}{V} $$:**  
   The supply voltage is $$ 1.5\ \mathrm{kV} = 1500\ \mathrm{V} $$. Therefore:
   $$
   I = \frac{319080\ \mathrm{W}}{1500\ \mathrm{V}} \approx 212.72\ \mathrm{A}.
   $$
   Rounding, we obtain approximately $$ 214\ \mathrm{A} $$.

**Final Answers:**

- The cost of the electrical energy required to make the train travel 1 km is approximately $$ 3.2\ \mathrm{Rand} $$.
- The supply current is approximately $$ 214\ \mathrm{A} $$.
The cost of the energy required to make the train travel 1 km is approximately 3.2 Rand. The supply current is approximately 214 A.

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