Esercizi Termodinamica applicata 2

Thermal power extracted from the refrigerated environment

Thermal power extracted from the refrigerated environment: \( \dot{Q_L} = \dot{m} \, (h_1 - h_4) = 7.13 \, \text{kW} \)

\( P = \dot{m} \, (h_2 - h_1) = 1.80 \, \text{kW} = L \)

\( \dot{Q_H} = \dot{m} \, (h_2 - h_3) = 9.93 \, \text{kW} \, \text{or} \, \dot{Q_H} = P + \dot{Q_L} = 8.93 \, \text{kW} \)

COP (Coefficient of performance) \( COP = \frac{\dot{Q_L}}{P} \) = 3.96

Evaporator heat exchange: \( q_L \)

\( q_L = h_1 - h_4 \)

Condenser heat: \( q_H \)

\( q_H = h_2 - h_3 \)

Compressor work: \( L = h_2 - h_1 \)

Thermal power and COP

Thermal power extracted from the refrigerated environment: Q_L = ṁ (h₁ - h₄) = 7.13 kW

P = ṁ (h₂ - h₁) = 1.80 kW = L

Q_H = ṁ (h₂ - h₃) = 8.93 kW or Q̇_H = P + Q̇_L = 8.93 kW

COP (Coefficient of Performance)

COP = Q_L/P = 3.96

Evaporator heat exchange: q_L = h₁ - h₄

Condenser heat: q_H = h₂ - h₃

Compressor work: L = h₂ - h₁

Closed piston-cylinder system

Consider a closed piston-cylinder system without friction containing a mass of 0.5 kg of air (ideal gas, c_v = 0.7 kJ/kgK, R = 287 J/kgK) at a temperature T₁ = 350 K and pressure p₁ = 200 kPa. Calculate the initial volume occupied by the mass. Subsequently, the mass is compressed isobarically to a volume of V₂ = 0.20 m³. Determine the final temperature, the specific work, the change in internal energy of the system, and the heat exchanged with the surroundings.

V₁? T₂? L? Δu? q?

p₁V₁ = mRT₁ → V₁ = mRT₁/p₁ = 0.5 · 287 · 350/200 = 0.251 m³

Isobaric transformation

Isobaric transformation: p₁ = p₂; mRT₁/V₁ = mRT₂/V₂; T₁/V₁ = T₂/V₂

T₂ = T₁ · V₂/V₁ = 350/0.25 · 0.20 = 280 K

Work and internal energy

Work for closed system: (Isobaric)

L = - ∫₁² p dV = - p ∫₁² dV = [-pV]₁² = -p (V₂ - V₁) = -200 k (0.20 - 0.25) = 10 kJ

Work per unit mass

L = L/m = 20 kJ/kg

Change in specific internal energy

Δu = c_v(T₂ - T₁) = -49 kJ/kg

Heat exchanged:

q + L = Δu

q = Δu - L = -69 kJ/kg

Exercise M3.1: Single compression refrigeration cycle – ideal conditions

A refrigeration machine uses R-134a as the working fluid and operates with a simple vapor compression cycle, between pressures of 0.14 MPa and 0.80 MPa. The refrigerant flow rate is 0.05 kg/s. Referring to the ideal operating cycle, determine:

• The thermal power extracted from the refrigerated environment.
• The mechanical power required by the compressor.
• The thermal power rejected to the environment (high-temperature source).
• The COP of the machine.

State 1

State 1: h₁ = 236.04 KJ/Kg (Enthalpy)

s₁ = 0.9322 KJ/Kg K (Entropy)

State 3

State 3: h₃ = 93.42 KJ/Kg = h₄ → Expansion valve is isenthalpic

State 2

State 2: s₁ = s₂ = 0.9322 KJ/Kg K

Interpolation:

(h₂ - 264.15)/(273.66 - 264.15) = (s₂ - 0.9066)/(0.9374 - 0.9066)

h₂ = 232.05 KJ/Kg