Question

Air is at 100 kPa and 150°C, and undergoes a constant pressure process. The final temperature of the air is 1000°C.

a) Compute the specific volume ratio.

b) Compute the boundary work (kJ/kg)

c) Compute the change in specific internal energy by:
1. Using the tables.
2. Integrating the polynomial (cp)
3. Assuming constant specific heat.

d) Repeat c for specific enthalpy

e) Compute the heat

Answer 1

Initial pressure = P1 = 100 Kpa

Initial tempe = 150 C = 423 K

Final temperaure = T2 = 1000 C = 1273 K

It is said that the process ocuur at constant pressure.

So, final pressure = P2 = 100 Kpa.

a) Specific volume ratio= ?

From Ideal gas relation, we know that Pv = RT

Where P = Pressure

T = Temperature

v = specific volume.

R = Gas constant = 0.287 KJ/kg-K.

For first process,

100*v₁ = 0.287*423

Specific volume at state 1 = v₁ = 1.214

For second process,

100*v₂ = 0.287*1273

Specific volume at state 2 = v₂ = 3.65

Specific volume ratio = v₁ / v₂ = 1.214/3.65 = 0.332

b) Boundary work= ?

Boundary work = W = P*dv

Boundary work = W = 100*(v₂ - v₁) = 100*( 3.65 - 1.214) = 243 KJ / kg.

c) Change inspecific internal energy = U = C_v*dT

Change in internal energy = 0.718 * (1000-150) = 610.3 KJ/kg

d) Change in specific enthalphy = H = C_p*dT = 1.013 * (1000-150) = 861 KJ/kg.

e) Heat = Q = ?

We know from the first law of thermodynamics, Q = dU+W

Heat Q = 610.3 + 243 = 853.3 KJ