Question

Refrigerant 134 a enters a compressor with a mass flow rate of \(5 \mathrm{kg} / \mathrm{s}\) and a negligible velocity. The refrigerant enters the compressor as a saturated vapor at \(10^{\circ} \mathrm{C}\) and leaves the compressor at \(1400 \mathrm{kPa}\) with an enthalpy of \(281.39 \mathrm{kJ} / \mathrm{kg}\) and a velocity of \(50 \mathrm{m} / \mathrm{s}\). The rate of work done on the refrigerant is measured to be \(132.4 \mathrm{kW}\). If the elevation change between the compressor inlet and exit is negligible, determine the rate of heat transfer associated with this process, in \(\mathrm{kW}\).

Short answer

The rate of heat transfer associated with this process is 297 kW.

Step-by-step solution

Write down the First Law of Thermodynamics for Steady Flow Processes

The first law for the steady flow process is given by: \(\dot{Q} - \dot{W} = \dot{m}(h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1))\), where \(\dot{Q}\) is the rate of heat transfer, \(\dot{W}\) is the work done, \(\dot{m}\) is the mass flow rate, \(h_1\) and \(h_2\) are the initial and final enthalpies, \(V_1\) and \(V_2\) are the initial and final velocities, and \(z_1\) and \(z_2\) are the initial and final elevations.

Substitue given values and simplify

Given that the elevation change is negligible (\(z_2 = z_1\)), mass flow rate is \(5 \mathrm{kg/s}\), \(h_1 = h_{g@10°C}\) (saturated vapor), \(h_2 = 281.39 \mathrm{kJ/kg}\), \(V_1\) is negligible (approximately zero), \(V_2 = 50 \mathrm{m/s}\), and \(\dot{W} = 132.4 \mathrm{kW}\), substitute these values into the first-law expression: \(\dot{Q} - 132.4 = 5(281.39 - h_{g@10°C} + \frac{50^2 - 0^2}{2 \times 1000})\)

Find the initial enthalpy of saturated vapor at \(10^{\circ}\mathrm{C}\)

Using the refrigerant 134a property tables, we find the enthalpy of saturated vapor at \(10^{\circ}\mathrm{C}\): \(h_{g@10°C} = 249.9 \mathrm{kJ/kg}\)

Compute the rate of heat transfer

Substituting the initial enthalpy into our equation and solving for \(\dot{Q}\): \(\dot{Q} - 132.4 = 5(281.39 - 249.9 + \frac{50^2}{2 \times 1000})\), so \(\dot{Q} = 132.4 + 5(31.49 + 1.25) = 132.4 + 5 \times 32.74 = \boxed{297 \mathrm{kW}}\) The rate of heat transfer associated with this process is \(297 \mathrm{kW}\).