Question 5.10: Cooling of Refrigerant-134a by Water Refrigerant-134a is to ...

Refrigerant-134a is cooled by water in a condenser. The mass flow rate of the cooling water and the rate of heat transfer from the refrigerant to the water are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus \Delta m_{ CV }=0 \text { and } \Delta E_{ CV }=0 . 2 The kinetic and potential energies are negligible, k e \cong p e \cong 0 . 3 Heat losses from the system are negligible and thus \dot{Q} \cong 0 . 4 There is no work interaction.
Analysis We take the entire heat exchanger as the system (Fig. 5–37). This is a control volume since mass crosses the system boundary during the process. In general, there are several possibilities for selecting the control volume for multiple-stream steady-flow devices, and the proper choice depends on the situation at hand. We observe that there are two fluid streams (and thus two inlets and two exits) but no mixing.
(a) Under the stated assumptions and observations, the mass and energy balances for this steady-flow system can be expressed in the rate form as follows:

Mass balance:    \dot{m}_{ in }=\dot{m}_{ out }

for each fluid stream since there is no mixing. Thus,

\begin{aligned} &\dot{m}_{1}=\dot{m}_{2}=\dot{m}_{w} \\ &\dot{m}_{3}=\dot{m}_{4}=\dot{m}_{R} \end{aligned}

Energy balance:    \underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{d E_{\text {system }} /dt^{\nearrow ^{0(steady)}} }_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc., energies } \end{array}}=0

\dot{E}_{\text {in }}=\dot{E}_{\text {out }}

\dot{m}_{1} h_{1}+\dot{m}_{3} h_{3}=\dot{m}_{2} h_{2}+\dot{m}_{4} h_{4} \quad(\text { since } \dot{Q} \cong 0, \dot{W}=0, \text { ke } \cong \text { pe } \cong 0)

Combining the mass and energy balances and rearranging give

\dot{m}_{w}\left(h_{1}-h_{2}\right)=\dot{m}_{R}\left(h_{4}-h_{3}\right)

Now we need to determine the enthalpies at all four states. Water exists as a compressed liquid at both the inlet and the exit since the temperatures at both locations are below the saturation temperature of water at 300 kPa (133.52°C). Approximating the compressed liquid as a saturated liquid at the given temperatures, we have

\begin{aligned} &h_{1} \cong h_{f @ 15^{\circ} C }=62.982 kJ / kg \\ &h_{2} \cong h_{f @ 25^{\circ} C }=104.83 kJ / kg \end{aligned} \quad(\text { Table } A -4)

The refrigerant enters the condenser as a superheated vapor and leaves as a compressed liquid at 35°C. From refrigerant-134a tables,

\begin{aligned} &\left.\begin{array}{l} P_{3}=1 MPa \\ T_{3}=70^{\circ} C \end{array}\right\} \quad h_{3}=303.85 kJ / kg \quad(\text { Table } A -13) \\ &\left.\begin{array}{l} P_{4}=1 MPa \\ T_{4}=35^{\circ} C \end{array}\right\} \quad h_{4} \cong h_{f @ 35^{\circ} C }=100.87 kJ / kg \quad(\text { Table } A -11) \end{aligned}

Substituting, we find

\begin{aligned} \dot{m}_{w}(62.982-104.83) kJ / kg &=(6 kg / min )[(100.87-303.85) kJ / kg ] \\ \dot{m}_{w} &=29.1 kg / min \end{aligned}

(b) To determine the heat transfer from the refrigerant to the water, we have to choose a control volume whose boundary lies on the path of heat transfer. We can choose the volume occupied by either fluid as our control volume. For no particular reason, we choose the volume occupied by the water. All the assumptions stated earlier apply, except that the heat transfer is no longer zero. Then assuming heat to be transferred to water, the energy balance for this single-stream steady-flow system reduces to

\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{d E_{\text {system }} / dt^{\nearrow ^{0(steady)}}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc., energies } \end{array}}=0

\begin{aligned} \dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \\ \dot{Q}_{w, \text { in }}+\dot{m}_{w} h_{1} &=\dot{m}_{w} h_{2} \end{aligned}

Rearranging and substituting,

\begin{aligned} \dot{Q}_{w, \text { in }}=\dot{m}_{w}\left(h_{2}-h_{1}\right) &=(29.1 kg / min )[(104.83-62.982) kJ / kg ] \\ &=1218 kJ / min \end{aligned}

Discussion Had we chosen the volume occupied by the refrigerant as the control volume (Fig. 5–38), we would have obtained the same result for \dot{Q}_{R, \text { out }}  since the heat gained by the water is equal to the heat lost by the refrigerant.