# ch12 - Problem 12.1 Given Air flow through a filter Find[2...

This preview shows pages 1–7. Sign up to view the full content.

Problem 12.1 [2] Given: Air flow through a filter Find: Change in p, T and ρ Solution: Basic equations: h 2 h 1 c p T 2 T 1 () = p ρ R T = Assumptions: 1) Ideal gas 2) Throttling process In a throttling process enthalpy is constant. Hence h 2 h 1 0 = so T 2 T 1 0 = or T constant = The filter acts as a resistance through which there is a pressure drop (otherwise there would be no flow. Hence p 2 p 1 < From the ideal gas equation p 1 p 2 ρ 1 T 1 ρ 2 T 2 = so ρ 2 ρ 1 T 1 T 2 p 2 p 1 = ρ 1 p 2 p 1 = Hence ρ 2 ρ 1 < The governing equation for entropy is Δ sc p ln T 2 T 1 Rln p 2 p 1 = Hence Δ sR ln p 2 p 1 = and p 2 p 1 1 < so Δ s0 > Entropy increases because throttling is an irreversible adiabatic process

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 12.2 [2]
Problem 12.3 [2] Given: Data on an air compressor Find: Whether or not the vendor claim is feasible Solution: Basic equation: Δ sc p ln T 2 T 1 Rln p 2 p 1 = The data provided, or available in the Appendices, is: p 1 101 kPa = T 1 20 273 + () K = p 2 650 101 + ( ) kPa = T 2 285 273 + K = c p 1004 J kg K = R 287 J kg K = Then Δ p ln T 2 T 1 p 2 p 1 = Δ s 71.0 J kg K = Entropy s Temperature T The second law of thermodynamics states that, for an adiabatic process Δ s0 or for all real processes Δ > Hence the process is feasible!

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 12.4 [2] Given: Adiabatic air compressor Find: Lowest delivery temperature; Sketch the process on a Ts diagram Solution: Basic equation: Δ sc p ln T 2 T 1 Rln p 2 p 1 = T 2 T 1 p 1 p 2 1k k = The lowest temperature implies an ideal (reversible) process; it is also adiabatic, so Δ s = 0, and The data provided, or available in the Appendices, is: p 1 14.7 psi = p 2 100 14.7 + ( ) psi = T 1 68 460 + () R = k 1.4 = Hence T 2 T 1 p 1 p 2 k = T 2 950R = T 2 490°F = Entropy s Temperature T The process is
Problem 12.5 [2] Given: Test chamber with two chambers Find: Pressure and temperature after expansion Solution: Basic equation: p ρ R T = Δ uqw = (First law - closed system) Δ uc v Δ T = Assumptions: 1) Ideal gas 2) Adiabatic 3) No work For no work and adiabatic the first law becomes Δ u0 = or for an Ideal gas Δ T0 = T 2 T 1 = We also have M ρ Vol = const = and Vol 2 2 Vol 1 = so ρ 2 1 2 ρ 1 = From the ideal gas equation p 2 p 1 ρ 2 ρ 1 T 2 T 1 = 1 2 = so p 2 1 2 p 1 = Hence T 2 20 °F = p 2 200 kPa 2 = p 2 100 kPa = Note that Δ sc p ln T 2 T 1 Rln p 2 p 1 = R ln 1 2 = 0.693 R = so entropy increases (irreversible adiabatic)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 12.6 [2] Given: Supercharger Find: Pressure, temperature and flow rate at exit; power drawn Solution: Basic equation: p ρ R air T = Δ sc p ln T 2 T 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 88

ch12 - Problem 12.1 Given Air flow through a filter Find[2...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online