Esercizio Termodinamica applicata

A turbocharger compresses air

A turbocharger compresses air (with c_p = 1000 J/(kg K)) in steady-state conditions from 100 kPa and 280 K to 600 kPa and 400 K. Assuming that the air mass flow rate is 0.02 kg/s, that a heat loss of 16 kJ/kg occurs during the process, and that the variations in kinetic and potential energy are negligible, determine the power to be supplied to the compressor. Assume air as an ideal gas.

P₁ = 100 kPa

P₂ = 600 kPa

T₁ = 280 K

T₂ = 400 K

Open system

ṁ = 0.02 kg/s

q = -16 kJ/kg

Δe_c and Δe_p = 0

1^st law of thermodynamics for open systems:

q + L = Δh + Δe^c + Δe^p

L = c_p (T₂ - T₁) - q = 1000 J/kg K (400 K - 280 K) - -16 kJ/kg

L = 136000 J/kg K = 136 kJ/kg

P = ṁ L = 0.02 kg/s · 136 kJ/kg = 2720 W

A turbocharger compresses air

A turbocharger compresses air (with c_p = 1000 J/(kg K)) in steady-state conditions from 100 kPa and 280 K to 600 kPa and 400 K. Assuming that the air mass flow rate is 0.02kg/s, that a heat loss of 16kJ/kg occurs during the process, and that the variations in kinetic and potential energy are negligible, determine the power to be supplied to the compressor. Assume air as an ideal gas.

P₁ = 100kPa

P₂ = 600kPa

T₁ = 280 K

T₂ = 400 K

Open system

ṁ = 0,02kg/s

q = -16kJ/kg

Δe_c y Δe_p = 0

1^st law of thermodynamics for open systems:

q + L = Δh + Δe_c^o + Δe_p^o

L = c_p (T₂ - T₁) - q = 1000 J/kg K (400K - 280K) - -16kJ/kg

[L = 136000 J/kg K = 136 kJ/kg]

[P = ṁL = 0.02 kg/s · 136 kJ/kg = 2720 W]

P = ṁL

P₁ = 100 kPa

P₂ = 600 kPa

T₁ = 280 K

T₂ = 400 K

Open system

ṁ = 0.02 kg/s

q = -16 kJ/kg

Δec y Δep = 0

1^st law of thermodynamics for open systems:

q + L = Δh + Δec_o + Δep_o

L = cp (T₂ - T₁) - q = 1000 J/kg K (400 K - 280 K) - -16 kJ/kg

L = 136000 J/kg K = 136 kJ/kg

P = ṁL = 0.02 kg/s · 136 kJ/kg = 2720 W

An adiabatic gas turbine

An adiabatic gas turbine provides a power output of 5 MW under the inlet and outlet conditions reported in the table.

Considering the gas passing through the turbine as an ideal gas with a specific heat at constant pressure of 1121 J/(kg K), determine:

• The changes in enthalpy, kinetic energy and potential energy: Δh, Δe_c and Δe_p;
• The work done by the gas per unit mass
• The mass flow rate of the gas

a) Δh = c_p (T₂ - T₁) = 1121 J/kg K (600 - 1200) = -672600 J/kg

Δe_c = (w₂² - w₁²)/2 = 14950 J/kg

Δe_p = g Δz = -39,24 J/kg = -0.04 kJ/kg

b) L = Δh + Δe_c + Δe_p

L = -672.6 + 14.95 - 0.04 = -657.69 kJ/kg

c) P = ṁL; ṁ = P/L = 5 MW/657.69 = 7.6 kg/s

Exercise M1.4

Consider a mixing heat exchanger where hot water at 60 °C is mixed with cold water at 10 °C. Determine the ratio of the mass flow rates of hot water and cold water in the case where a steady flow of warm water at 40 °C is desired. Assume that heat losses from the mixing chamber are negligible and that mixing occurs at a pressure of 140 kPa. Also, neglect changes in kinetic and potential energy.

T₁ = 60°C = 333K

Mass conservation equation

∑ṁ_in = ∑ṁ_out ⇒ ṁ₁ + ṁ₂ = ṁ₃

∑ṁ_inh_in = ∑ṁ_outh_out ↔ ṁ₁h₁ + ṁ₂h₂ = (ṁ₁ + ṁ₂)h₃

y = ṁ₁/ṁ₂

yh_₁ + h_₂ = (y + 1)h_₃

y = (h_₃-h_₂) / (h_₁-h_₃) = (c_p*(T_₃-T_₂)) / (c_p*(T_₁-T_₃)) = (40-10)/(60-40) = 30/20 = 1.5