An oil gusher shoots crude oil 29.0 m into the air through a 0.100 m diameter pipe. Neglecting air resistance but not the resistance of the pipe, and assuming laminar flow, calculate the pressure at the entrance of the 55.0 m long vertical pipe. Take the density of the oil to be 900 kg/m3 and its viscosity to be 1.00 (N/m2)·s. Note that you must take into account the pressure due to the 55.0 m column of oil in the pipe.
To calculate the pressure at the entrance of the vertical pipe, we need to consider the pressure due to the weight of the column of oil in the pipe, as well as the pressure due to the velocity of the oil shooting into the air.
First, let's calculate the pressure due to the weight of the column of oil in the pipe using the formula for pressure:
\[ P_{\text{weight}} = \text{density} \times \text{gravitational acceleration} \times \text{height} \]
where pressure \( P_{\text{weight}} \) is the pressure due to the weight, density is the density of the oil, gravitational acceleration is the acceleration due to gravity (approximately 9.8 m/s^2), and height is the height of the column of oil in the pipe (55.0 m).
Substituting the given values, we get:
\[ P_{\text{weight}} = 900 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 55.0 \, \text{m} \]
Next, let's calculate the pressure due to the velocity of the oil shooting into the air. For this, we use Bernoulli's equation which relates the pressure, density, and velocity at two different points in a fluid flow.
\[ P_{\text{velocity}} = \frac{1}{2} \times \text{density} \times \text{velocity}^2 \]
where \( P_{\text{velocity}} \) is the pressure due to the velocity, density is the density of the oil, and velocity is the velocity of the oil shooting into the air. Since the oil shoots 29.0 m into the air, we will consider the vertical component of its velocity only.
To calculate the velocity, we can use the conservation of energy. The potential energy of the oil at the top is converted into kinetic energy when it reaches the maximum height. Therefore, we can equate the potential energy to the kinetic energy:
\[ \text{gravitational potential energy} = \text{kinetic energy} \]
\[ m \times g \times h = \frac{1}{2} \times m \times v^2 \]
where m is the mass of the oil, g is the gravitational acceleration, h is the height of the oil shooting into the air (29.0 m), and v is the velocity of the oil.
From this equation, we can solve for the velocity v:
\[ v = \sqrt{2 \times g \times h} \]
Substituting the given values, we get:
\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 29.0 \, \text{m}} \]
Now, let's calculate \( P_{\text{velocity}} \):
\[ P_{\text{velocity}} = \frac{1}{2} \times 900 \, \text{kg/m}^3 \times (\sqrt{2 \times 9.8 \, \text{m/s}^2 \times 29.0 \, \text{m}})^2 \]
Finally, to find the pressure at the entrance of the vertical pipe, we sum up the pressures due to the weight and the velocity:
\[ P_{\text{entrance}} = P_{\text{weight}} + P_{\text{velocity}} \]