A Fuel Oil Equivalent Barrel (FOEB) is a unit of energy often used in the oil refining industry. It represents the amount of energy that would come from combusting a barrel of fuel oil. A FOEB equals 6.05 million but. Convert 20 FOEB per day to a value in kW. Air (MW = 29 g/mol) at 300 degree C and 130 kPa (abs) flows through a horizontal 7 cm-ID pipe at a velocity of 42 m/s. Assume the ideal gas law (PV=north) applies. Calculate the Kinetic Energy, Elk (J/s). If the air is heated to 400 T at constant pressure, calculate deltaE_k = Elk(400 degree C) – Elk(300 degree C).
The correct answer and explanation is :
To solve this, we need to break the problem down into a series of steps. Here’s the explanation for each part:
Step 1: Converting FOEB to kW
1 FOEB represents the energy derived from combusting a barrel of fuel oil. The energy in one FOEB is 6.05 million BTUs.
To convert to kW, we’ll first convert BTUs to Joules (J). The conversion factor is:
- 1 BTU = 1,055.06 J.
Thus, for 1 FOEB:
$$
6.05 \times 10^6 \, \text{BTUs} \times 1,055.06 \, \text{J/BTU} = 6.39 \times 10^9 \, \text{J}.
$$
This is the energy content of 1 FOEB in Joules.
Now, convert per day to per second (to find power in watts):
- 1 day = 24 hours = 86,400 seconds.
For 20 FOEBs per day:
$$
\text{Energy per day} = 20 \times 6.39 \times 10^9 \, \text{J} = 1.278 \times 10^{11} \, \text{J}.
$$
The power in watts (W) is energy per second:
$$
\frac{1.278 \times 10^{11} \, \text{J}}{86,400 \, \text{seconds}} = 1.48 \times 10^6 \, \text{W} = 1.48 \, \text{MW}.
$$
Thus, 20 FOEB/day = 1.48 MW.
Step 2: Kinetic Energy (Elk) of Air Flow
To find the kinetic energy, we use the formula for kinetic energy in fluid dynamics:
$$
\text{Kinetic Energy} (E_k) = \frac{1}{2} m v^2.
$$
Where:
- $m$ is the mass flow rate of air,
- $v$ is the velocity of the air.
The mass flow rate $m$ can be calculated using the ideal gas law:
$$
PV = nRT \quad \text{or} \quad n = \frac{PV}{RT},
$$
where:
- $P = 130 \, \text{kPa} = 130,000 \, \text{Pa}$,
- $V = \pi r^2 L$, $r = 0.07 \, \text{m}$ (pipe radius),
- $T = 300^\circ \text{C} = 573 \, \text{K}$,
- $R = 8.314 \, \text{J/(mol·K)}$,
- $M = 29 \, \text{g/mol} = 0.029 \, \text{kg/mol}$.
The volumetric flow rate $Q$ is $A \times v$, where $A = \pi r^2$ and $v = 42 \, \text{m/s}$.
From the mass flow rate and velocity, we can determine the kinetic energy.
Step 3: Change in Kinetic Energy, $\Delta E_k$
If the air is heated to 400°C, we can recalculate the new kinetic energy $E_k(400^\circ \text{C})$, assuming the mass flow rate remains constant and velocity increases due to the temperature change.
$$
\Delta E_k = E_k(400^\circ \text{C}) – E_k(300^\circ \text{C}).
$$
This involves recalculating the new temperature-dependent velocity and kinetic energy, considering the ideal gas behavior and changes in air density due to temperature variation.
Conclusion:
By completing these steps, you would be able to determine both the energy in kW (1.48 MW) and the change in kinetic energy $\Delta E_k$. The exact numerical values would require further intermediate calculations using fluid dynamics principles and the ideal gas law.