Question

2. The properties of a certain fluid are related as follows:
1960.7180
pv=0.287 (t + 273)
Where u is the speafic internal energy (kJ/kg), t is in °C, p is pressure (kN/m), and vis specifle
volume (m/leg). For this fluid, find e anda
(Ans. 0.719, 1.005 kJ/kg-K)
This system expands in an adiabatic process following pv= constant. The initial
conditions are 1 MPa. 200°C and final pressure is 0.1 MPa. Determine the work done and change
in intemal energy for the process, if the mass of the fluid is 2 kg.
(Ang dw = 216.66 kJ, dU = -216.66kJ)​

Answer 1

Answer:

The properties of a certain fluid are related as follows. µ=196+0.718t, pv=0.287(t+273) Where µ is the specific internal energy (KJ/kg), t is in 0C, p is pressure (KN/m2) and v is the specific volume (m3/kg)

The value of Cv will be

Answer

0.718 KJ/kg-k