Question

Air exits a turbine at 200 kPa and 1508C with a volumetric flow rate of 7000 liters/s. Modeling air as an ideal gas, determine the mass flow rate, in kg/s.

Answer 1

Answer:

mass flow rate = 11.464 kg/sec

Explanation:

given data

P = 200 kPa = 2 × $10^{5}$  Pa

temperature = 150°C  = 423 k

volumetric flow rate = 7000 liters/s

solution

first we get here molar flow rate that is express as

molar flow rate  = $\frac{Pv}{RT}$   .................1

put here value

molar flow rate  = $\frac{2 \times 10^5 \times 7}{8.314\times 423}$  

molar flow rate  = 398.06 mol/sec

and

now we get mass flow rate

mass flow rate = molar flow rate × average  mol weight ..............2

mass flow rate = 398.06 × 28.8

mass flow rate = 11464.3 g/sec

mass flow rate = 11.464 kg/sec