Question 9.1: A refrigerated space is maintained at −20°C, and cooling wat...

(a) Allowing 5°C temperature differences, the evaporator temperature is −25°C = 248.15 K, and the condenser temperature is 26°C = 299.15 K. Thus, by Eq. (9.3) for a Carnot refrigerator,

\omega=\frac{T_C}{T_H-T_C}       (9.3)

\omega=\frac{248.15}{299.15-248.15}=4.87

(b) With HFC-134a as the refrigerant, enthalpies for states 2 and 4 of Fig. 9.2 are read directly from Table 9.1.

The entry at −25°C indicates that HFC-134a vaporizes in the evaporator at a pressure of 1.064 bar. Its properties as a saturated vapor at these conditions are:

H_2=383.45 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}                         \quad S_2=1.746 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}

The entry at 26°C in Table 9.1 shows that HFC-134a condenses at 6.854 bar; its enthalpy as a saturated liquid at these conditions is:

H_4=235.97 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

If the compression step is reversible and adiabatic (isentropic) from saturated vapor at state 2 to superheated vapor at state 3′,

S_3^{\prime}=S_2=1.746 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}

This entropy value and the condenser pressure of 6.854 bar are sufficient to specify the thermodynamic state at point 3′. One could find the other properties at this state using Fig. F.2, following a curve of constant entropy from the saturation curve to the condenser pressure. However, more precise results can be obtained using an electronic resource such as the NIST WebBook. Varying the temperature at a fixed pressure of 6.854 bar shows that the entropy is 1.746  kJ⋅kg^{−1}⋅K^{−1} at T = 308.1  K . The corresponding enthalpy is:

H_3^{\prime}=421.97 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

and the enthalpy change is:

(\Delta H)_S=H_3^{\prime}-H_2=421.97-383.45=38.52 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

By Eq. (7.17) for a compressor efficiency of 0.80, the actual enthalpy change for step 2 → 3 is:

\eta \equiv \frac{(\Delta H)_S}{\Delta H}         (7.17)

H_3-H_2=\frac{(\Delta H)_S}{\eta}=\frac{38.52}{0.80}=48.15  \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Because the throttling process of step 1 → 4 is isenthalpic, H_1 = H_4. The coefficient of performance as given by Eq. (9.4) therefore becomes:

\omega=\frac{H_2-H_1}{H_3-H_2}          (9.4)

\omega=\frac{H_2-H_4}{H_3-H_2}=\frac{383.45-235.97}{48.15}=3.06

and the HFC-134a circulation rate as given by Eq. (9.5) is:

\dot{m}=\frac{\dot{Q}_C}{H_2-H_1}            (9.5)

\dot{m}=\frac{\dot{Q}_C}{H_2-H_4}=\frac{120,000}{383.45-235.97}=814 \mathrm{~kg} \cdot \mathrm{h}^{-1}