Air at 27 °C and 100 kPa enters in a steady flow to a nozzle at a velocity of 100 m/s. If the inlet area of the nozzle is 0.5 m2, what is the mass flow rate through the system?
58 kg/s
This problem requires calculating the mass flow rate of air entering a nozzle under specific conditions. The mass flow rate is a crucial parameter in fluid dynamics, representing the mass of fluid passing through a given cross-section per unit time.
We are provided with the following information about the air entering the nozzle:
First, we need to convert the given temperature and pressure into standard SI units (Kelvin for temperature and Pascals for pressure) to use them in thermodynamic equations.
We can calculate the density of air at the inlet using the ideal gas law, which relates pressure, density, specific gas constant, and temperature.
The ideal gas law is given by:
$$ P = \rho R T $$
Where:
For air, the specific gas constant ($R_{air}$) is approximately 287 J/(kg·K).
Rearranging the formula to solve for density ($\rho_1$):
$$ \rho_1 = \frac{P_1}{R_{air} T_1} $$
Now, substitute the values:
$$ \rho_1 = \frac{100 \times 10^3 \text{ Pa}}{287 \text{ J/(kg·K)} \times 300.15 \text{ K}} $$
$$ \rho_1 \approx \frac{100000}{86143.05} \text{ kg/m}^3 $$
$$ \rho_1 \approx 1.1608 \text{ kg/m}^3 $$
The mass flow rate ($\dot{m}$) for a steady flow system is calculated by multiplying the density, the cross-sectional area, and the velocity of the fluid.
The formula for mass flow rate is:
$$ \dot{m} = \rho \times A \times V $$
Using the inlet conditions:
$$ \dot{m} = \rho_1 \times A_1 \times V_1 $$
Substitute the calculated density and given area and velocity:
$$ \dot{m} = 1.1608 \text{ kg/m}^3 \times 0.5 \text{ m}^2 \times 100 \text{ m/s} $$
$$ \dot{m} = 1.1608 \times 50 \text{ kg/s} $$
$$ \dot{m} \approx 58.04 \text{ kg/s} $$
The calculated mass flow rate through the nozzle is approximately 58.04 kg/s. Comparing this value to the given options, it closely matches 58 kg/s.
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