Argon gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are 570oR and 150 ft/s, respectively. At the exit, the temperature is 480oR and the pressure is 40 lbf/in2. The area of the exit is 0.0085 ft2. Use the ideal gas model with k = 1.67, and neglect potential energy effects. Determine the velocity at the exit, in ft/s, and the mass flow rate, in lb/s.

a) Ve = 762.7 ft/s

b) m (flow) = 1.69 lb/s

Explanation:

Given:-

- The fluid : Argon

R = 0.04971 Btu / lbR = 38.684 ft lbf / lbR

- The inlet conditions:

Ti = 570 R

Vi = 150 ft/s

- The exit conditions:

Te = 480 R

Pe = 40 lbf/in^2

- The exit area, Ae = 0.0085 ft^2

- The ideal gas model constant, k = 1.67

Find:-

Determine the velocity at the exit, in ft/s, and the mass flow rate, in lb/s.

Solution:-

- Determine the specific heat constant (cp) for argon gas:

cp = kR / (k-1)

cp = 1.67*0.04971 / 0.67

cp = 0.124 Btu / lbR

- The idealized Heat-energy equation for argon gas is of the form, with no heat transfer, zero work done and changes in elevation are negligible:

0 = 2 * (h2 - h1) + Ve^2 - Vi^2

Ve^2 = 2 * (h1 - h2) + Vi^2

Ve^2 = 2*cp * (Ti - Te) + Vi^2

Ve = (2*0.124*32.2*778 * (570 - 480) + 150^2) ^0.5

Ve = 762.7 ft/s

- According to ideal gas Law the specific volume at exit (ve) would be:

ve = R*T2 / P2

ve = (38.684) * 570 / 40*144

ve = 3.82810 ft^3 / lb

- The mass flow rate can now be determined:

m (flow) = Ae*Ve / ve

m (flow) = (0.0085*762.7) / 3.82810

m (flow) = 1.69 lb/s