Thermodynamics Practice Exam — Free

This is a closed system, so the first law is $Q - W = \Delta U$ with $W$ the boundary work done by the air.

Mass of air from the ideal-gas law (with $P$ in kPa and $V$ in m³ the result is in kg):

$$m = \frac{P_1V_1}{RT_1} = \frac{100(0.50)}{0.287(300)} = 0.5807\,\text{kg}$$

Final state from the polytropic relations with $n = 1.3$:

$$V_2 = V_1\left(\frac{P_1}{P_2}\right)^{1/n} = 0.50(0.20)^{0.7692} = 0.14498\,\text{m}^3,\qquad T_2 = T_1\left(\frac{P_2}{P_1}\right)^{(n-1)/n} = 300(5)^{0.2308} = 434.9\,\text{K}$$

Boundary work for a polytropic process ($n \ne 1$):

$$W = \frac{P_2V_2 - P_1V_1}{1 - n} = \frac{500(0.14498) - 100(0.50)}{1 - 1.3} = \frac{72.49 - 50.0}{-0.3} = -74.96\,\text{kJ}$$

The negative sign means work is done on the air, as expected in a compression. Internal energy change of an ideal gas with constant $c_v$:

$$\Delta U = mc_v(T_2 - T_1) = 0.5807(0.718)(434.9 - 300) = 56.25\,\text{kJ}$$

First law:

$$Q = \Delta U + W = 56.25 + (-74.96) = -18.7\,\text{kJ}$$

About 18.7 kJ of heat leaves the air: the compression work (75.0 kJ) exceeds the rise in internal energy, so the difference must be rejected. This is consistent with $n = 1.3$ lying between isothermal ($n = 1$, all work rejected as heat) and isentropic ($n = k = 1.4$, no heat transfer).

The tempting answer +131 kJ treats the work as positive and adds it to $\Delta U$; −75.0 kJ is the work, not the heat; +56.3 kJ is $\Delta U$ alone, which would be the heat only if no work were done.

$Q = -18.7\,\text{kJ}$ (18.7 kJ of heat rejected by the air)

First find the ideal (isentropic) exit state. For an ideal gas with constant specific heats, an isentropic process between the same two pressures obeys

$$T_{2s} = T_1\left(\frac{P_2}{P_1}\right)^{(k-1)/k} = 300\,(8)^{0.4/1.4} = 300(1.811) = 543.4\,\text{K}$$

Isentropic work input (steady flow, adiabatic, so $w = \Delta h = c_p\,\Delta T$):

$$w_s = c_p(T_{2s} - T_1) = 1.005(543.4 - 300) = 244.7\,\text{kJ/kg}$$

For a compressor the isentropic efficiency is the ideal work divided by the actual work, $\eta_c = w_s/w_a$, because a real compressor needs more work than the ideal one:

$$w_a = \frac{w_s}{\eta_c} = \frac{244.7}{0.82} = 298.4\,\text{kJ/kg}$$

The actual exit temperature follows from the same energy balance, $w_a = c_p(T_{2a} - T_1)$:

$$T_{2a} = T_1 + \frac{w_a}{c_p} = 300 + \frac{298.4}{1.005} = 596.9\,\text{K}$$

The actual exit temperature is higher than the isentropic one (597 K versus 543 K) because the irreversibilities deposit the extra work as internal energy in the air. Dividing by the efficiency rather than multiplying is the step to get right: multiplying would give an exit temperature below the isentropic value, which is impossible for an adiabatic compressor.

$T_{2a} \approx 597\,\text{K}$, $w_a \approx 298\,\text{kJ/kg}$

Number the states: 1 = saturated liquid leaving the condenser (10 kPa), 2 = compressed liquid leaving the pump (3 MPa), 3 = turbine inlet, 4 = turbine exit (10 kPa).

(a) The ideal turbine is isentropic, so $s_4 = s_3 = 6.9235$ kJ/(kg·K). This lies between $s_f$ and $s_g$ at 10 kPa, so state 4 is a wet mixture:

$$x_4 = \frac{s_4 - s_f}{s_{fg}} = \frac{6.9235 - 0.6492}{7.4996} = 0.8366$$ $$h_4 = h_f + x_4 h_{fg} = 191.81 + 0.8366(2392.1) = 2193.1\,\text{kJ/kg}$$

(b) The pump handles an incompressible liquid, so $w_{pump} = v_f(P_2 - P_1)$; with pressures in kPa the result is in kJ/kg:

$$w_{pump} = 0.001010(3000 - 10) = 3.02\,\text{kJ/kg},\qquad h_2 = h_1 + w_{pump} = 191.81 + 3.02 = 194.83\,\text{kJ/kg}$$

Turbine work and net work:

$$w_{turb} = h_3 - h_4 = 3231.7 - 2193.1 = 1038.6\,\text{kJ/kg},\qquad w_{net} = 1038.6 - 3.02 = 1035.6\,\text{kJ/kg}$$

(c) Heat added in the boiler is $q_{in} = h_3 - h_2 = 3231.7 - 194.83 = 3036.9$ kJ/kg, so

$$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{1035.6}{3036.9} = 0.341$$

The same value follows from $\eta = 1 - q_{out}/q_{in}$ with $q_{out} = h_4 - h_1 = 2001.3$ kJ/kg. The pump work is under 0.3% of the turbine work, which is why it is often neglected in quick estimates, and an exit quality of 0.84 is low enough to erode turbine blades, which is why real plants superheat further or add reheat.

(a) $x_4 = 0.837$; (b) $w_{pump} = 3.02$ kJ/kg, $w_{net} = 1036$ kJ/kg; (c) $\eta_{th} = 34.1\%$