Esercizi Termodinamica applicata

Efficienza politropica e adiabatica isentropica

L_poly. = C_p (T_2' - T₁) = 877,53 (611,6 - 298) = 283,9 kJ/kg

C_p = n/(n-1) R = 1.486/(1.486-1) · 287 = 877,53 J/kg K

Polytropic efficiency: η_pol = L_pol/L_r

Adiabatic isentropic efficiency: η_ad = L_ad.is./L_r = 0,83

L_poly = C_p (T₂ - T₁) = 877,53 (611,6 - 298) = 283,9 KJ/kg

C_p = n/(n-1) R = 1,486/(1,486 - 1) · 287 = 877,53 J/kg K

Polytropic efficiency: η_pol = L_pol/L_r

Adiabatic isentropic efficiency: η_ad = L_ad.is./L_r = 0,83

Open system

A dynamic compressor draws air (ideal gas with R = 287 J/kgK and k = 1.4) at a pressure of 1 bar and temperature of 288K. The compression ratio is β = 10. Evaluate the specific adiabatic isentropic work. Additionally, given that the final temperature in the real process exceeds the corresponding isentropic adiabatic final temperature by 10%, calculate the specific real work, the polytropic exponent of the equivalent transformation, and the polytropic and adiabatic isentropic efficiencies.

Lavoro isentropico

l_ad.is. = c_p (T₂ − T₁) = 1004,5 (556 − 288) = 269,2 kJ/kg

c_p = k/(k-1) R = 1.4/0.4 × 287 = 1004.5 J/kg K

T₂ = T₁ β^(k−1)/k = 288 × 10^0.4/1.4 = 556 K

10%

T'₂ = 1.1 T₂ = 611,6 K

l_r = Δh = c_p (T'₂ − T₁) = 325.1 kJ/kg

Polytropic exponent

n = const => p₁v₁ⁿ = p₂v'₂ⁿ

T'₂ = T₁ β^(n−1)/n => T'₂/T₁ = β^(n−1)/n

ln (T'₂/T₁) = ((n−1)/n) ln β

n = ln β / (ln β + ln (T₁/T'₂)) = 1.486

Scambio termico in un condotto

Air at a temperature of T₁ = 293 K and velocity w₁ = 1 m/s flows through a duct with a constant cross-section. The height difference between the inlet and outlet sections is 10 m. At the outlet, the air temperature is T₂ = 40 °C and the velocity è w₂ = 3 m/s. Determine the amount of heat exchanged per unit mass along the duct, assuming no work exchange. Use the Langen formulas to calculate the specific heat at constant pressure for air (a = 992.082 J/kgK, b = 0.134 J/kgK).

First law of thermodynamics for open systems

q = Δh + cc + C_p

q = h₂ - h₁ + (w₂² - w₁²) / 2 + g Δz

C_p2 T₂ - C_p T₂

Langen formulas: C_p = a + bT

C_p1 = 992,092 J/kgK + 0,134 J/kgK × 293 K = 1031,34 J/kg K

C_p2 = 1034 J/kg K

q = 1034 · 313 - 1031 · 293 + (3² - 1²) / 2 + 9,81 · 10 = 21661 J/kg

Sistema isolato con acqua

A container with a fixed volume and adiabatic walls is divided into two parts by a non-adiabatic partition. One part contains 10 kg of water in the liquid state at a temperature of 20°C, while the other, with a mass of 2.5 kg, contains water in the liquid state at a temperature of 80°C. Determine the final temperature of the system once thermal equilibrium is reached.

Isolated system: No mass exchange nor energy.

Sometimes with adiabatic walls

q = 0

L = 0

Δu = 0

Since the mass (m) of the system is constant and doesn’t experience changes in elevation (z) or velocity (u).

Q + L = Δu = 0

Q₁ = Δu₁; Q₂ = Δu₂

Q₁ = -Q₂ → Δu₁ = Δu₂

Heat exchanged by a liquid related to its temperature change:

Q - m cw ΔT It's the same since they are in thermal equilibrium

m₁cw(T_F - T₁) = - m₂cw(T_F - T₂)

10 kg 20°C; 2.5 kg 90°C

10 cw T_F - 10 cw · 20 = - 2.5 cw T_F - 2.5 cw 90

(10 + 2.5) cw T_F =

T_F = 32°C