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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.

Respuesta :

Answer:

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

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