Thermodynamic Cycles

Abstract

This chapter provides a comprehensive overview of thermodynamic cycles with an emphasis on the application of thermodynamic cycles in the real world. It starts with a brief introduction in Sect. 17.1. Section 17.2 explains Carnot cycle, reversed Carnot cycle, Carnot engine, and Carnot refrigerator. Comprehensive coverage of Rankine cycle including the use of regenerative feedwater heating and reheat cycle is provided in Sect. 17.3. Section 17.4 covers Brayton cycle along with the use of a regenerator. Section 17.5 illustrates the analysis and applications of combined gas and steam cycles. Cogeneration power plants are covered in Sect. 17.6 with numerous citations for further study. Otto Cycle, which is the basis of gasoline engines, is explained in Sect. 17.7. The use of vapor compression cycles in refrigeration, air-conditioning, and heating systems is explained in Sect. 17.8.

Appendices

Practice Problems

Practice Problem 17.1

A Carnot cycle has an efficiency of 62% and produces 15 hp power. Determine

• A. The rate of heat input (Btu/hr) required

• B. The temperature (in °F) of the heat source if the heat sink is at 50 °F

• C. The rate at which heat should be removed from the heat sink

Practice Problem 17.2

A Carnot air-conditioner has a COP_ref./AC of 3.72 and achieves 3 tons of cooling (1 ton of cooling ≡ 12,000 Btu/hr). Determine:

• A. The power input (hp) required

• B. The temperature of the heat sink (°F) if the temperature is to be maintained at 70 °F in the air-conditioned space

• C. The rate at which heat is rejected to the heat sink in Btu/hr

Practice Problem 17.3

The turbine and pump in Example 17.4 have efficiencies of 90% and 80%, respectively. The power produced will remain the same as in Example 17.4. Using the fuel and boiler efficiency data given in Example 17.5, calculate the percentage increase in coal required per hour.

Hint: Refer to isentropic efficiencies for turbines and compressors in Chap. 14. The consequence of isentropic efficiency of pump is similar to the impact of isentropic efficiency in a compressor.

Practice Problem 17.4

An Otto cycle has an efficiency of 55% and takes in air at 14 psia, 70 °F. The volume at the start of the compression process is 3 liters. Determine the volume, pressure, and temperature after compression process.

Practice Problem 17.5

A vapor compression cycle is used in heating 1700 cfm of atmospheric air from 40 to 70 °F. The refrigerant, R-134a, is compressed from 30 to 250 psia. Determine:

• A. The enthalpy of the refrigerant at each state point

• B. The isentropic horsepower required for the compressor

• C. The COP_{heat pump}

Solutions to Practice Problems

Practice Problem 17.1

• (Solution)

• A. Calculate the rate of heat input required using the first part of Eq. 17.2.

$$ {\dot{Q}}_{\mathrm{H}}=\frac{\dot{W}}{\eta_{\mathrm{th},\mathrm{Carnot}}}=\left(\frac{15\ \mathrm{hp}}{0.62}\times \frac{2544\ \mathrm{Btu}/\mathrm{hr}}{1\ \mathrm{hp}}\right)=61,548\ \mathrm{Btu}/\mathrm{hr} $$

• B. Calculate the absolute temperature of the heat sink using Eq. 13.2.

$$ {T}_{\mathrm{L}}=50{}^{\circ}\mathrm{F}+460{}^{\circ}=510{}^{\circ}\mathrm{R} $$

Substitute the known quantities into the second part of Eq. 17.2, and solve for T_H.

$$ {\eta}_{\mathrm{th},\mathrm{Carnot}}=0.62=\frac{T_{\mathrm{H}}-{T}_{\mathrm{L}}}{T_{\mathrm{H}}}=\frac{T_{\mathrm{H}}-510{}^{\circ}\mathrm{R}}{T_{\mathrm{H}}}\Rightarrow {T}_{\mathrm{H}}=1342{}^{\circ}\mathrm{R} $$

Convert the absolute temperature to degree Fahrenheit using Eq. 13.2.

$$ {T}_{\mathrm{H}}=1342{}^{\circ}\mathrm{R}-460{}^{\circ}=882{}^{\circ}\mathrm{F} $$

• C. Calculate the rate at which heat should be removed from the heat sink using the middle part of Eq. 13.2, and substitute the known values.

$$ {\displaystyle \begin{array}{l}{\eta}_{\mathrm{th},\mathrm{Carnot}}=0.62=\frac{{\dot{Q}}_H-{\dot{Q}}_L}{{\dot{Q}}_H}=\frac{61548\frac{\mathrm{Btu}}{\mathrm{hr}}-{\dot{Q}}_L}{61548\frac{\mathrm{Btu}}{\mathrm{hr}}}\\ {}\Rightarrow {\dot{Q}}_L=23,388\ \mathrm{Btu}/\mathrm{hr}\end{array}} $$

Practice Problem 17.2

• (Solution)

• A. Calculate the power input required using the first part of Eq. 17.4.

$$ {\displaystyle \begin{array}{l}\dot{W}=\frac{{\dot{Q}}_{\mathrm{L}}}{{\mathrm{COP}}_{\mathrm{refg}./\mathrm{AC}}}=\frac{\left(3\ \mathrm{tons}\right)\left(\frac{12000\ \mathrm{Btu}/\mathrm{hr}}{\mathrm{ton}}\right)\left(\frac{0.000393\ \mathrm{hp}}{\mathrm{Btu}/\mathrm{hr}}\right)}{3.72}\\ {}\kern6em =3.80\ \mathrm{hp}\end{array}} $$

• B. Convert the temperature of the air-conditioned space to absolute value using Eq. 13.2.

$$ {T}_{\mathrm{L}}=70{}^{\circ}\mathrm{F}+460{}^{\circ}=530{}^{\circ}\mathrm{R} $$

Use the last part of Eq. 17.4, and substitute the known values to obtain the absolute temperature of the heat sink.

$$ {\displaystyle \begin{array}{c}{\mathrm{COP}}_{\mathrm{refg}./\mathrm{AC}}=\frac{T_{\mathrm{L}}}{T_{\mathrm{H}}-{T}_{\mathrm{L}}}\Rightarrow \\ {}\kern0ex {T}_{\mathrm{H}}=\frac{T_{\mathrm{L}}}{{\mathrm{COP}}_{\mathrm{refg}./\mathrm{AC}}}+{T}_{\mathrm{L}}=\frac{530^{\circ}\mathrm{R}}{3.72}+{530}^{\circ}\mathrm{R}={672.5}^{\circ}\mathrm{R}\end{array}} $$

Convert the absolute temperature of the heat sink to °F using Eq. 13.2.

$$ {T}_{\mathrm{H}}=672.5{}^{\circ}\mathrm{R}-460{}^{\circ}=212.5{}^{\circ}\mathrm{F} $$

• C. Calculate the rate of heat rejection to the heat sink using the energy balance equation for the reversed Carnot engine. The rate of work input in Btu/hr from the solution to part(A) is:

$$ \dot{W}=\frac{{\dot{Q}}_{\mathrm{L}}}{{\mathrm{COP}}_{\mathrm{refg}./\mathrm{AC}}}=\frac{\left(3\ \mathrm{tons}\right)\left(\frac{12000\ \mathrm{Btu}/\mathrm{hr}}{\mathrm{ton}}\right)}{3.72}=9677\ \mathrm{Btu}/\mathrm{hr} $$

$$ {\dot{Q}}_{\mathrm{H}}={\dot{Q}}_{\mathrm{L}}+\dot{W}=36000\frac{\mathrm{Btu}}{\mathrm{hr}}+9677\frac{\mathrm{Btu}}{\mathrm{hr}}=45,677\ \mathrm{Btu}/\mathrm{hr} $$

Practice Problem 17.3

• (Solution)

From the solution to Example 17.4, the following information is available:

$$ {\displaystyle \begin{array}{l}{w}_{\mathrm{p}}=1.50\ \mathrm{Btu}/\mathrm{lbm},\\ {}{w}_{\mathrm{t}}=275\ \mathrm{Btu}/\mathrm{lbm},\mathrm{and}\ \\ {}{\dot{W}}_{\mathrm{net}}=5.21\ \mathrm{MW}\end{array}} $$

Calculate the actual turbine work produced using Eq. 14.13.

$$ {w}_{\mathrm{t},\mathrm{actual}}={\eta}_{\mathrm{t}}\times {w}_{\mathrm{t},\mathrm{ideal}}=(0.90)\left(275\frac{\mathrm{Btu}}{\mathrm{lbm}}\right)=247.5\ \mathrm{Btu}/\mathrm{lbm} $$

Calculate the actual work consumed by the pump using Eq. 14.14 (same equation can be applied to both pumps and compressors).

$$ {w}_{\mathrm{p},\mathrm{actual}}=\frac{w_{\mathrm{p},\mathrm{ideal}}}{\eta_{\mathrm{p}}}=\frac{1.50\frac{\mathrm{Btu}}{\mathrm{lbm}}}{0.80}=1.88\ \mathrm{Btu}/\mathrm{lbm} $$

Calculate the actual net (energy) work produced.

$$ {w}_{\mathrm{net},\mathrm{actual}}={w}_{\mathrm{t},\mathrm{actual}}-{w}_{\mathrm{p},\mathrm{actual}}=247.5\frac{\mathrm{Btu}}{\mathrm{lbm}}-1.88\frac{\mathrm{Btu}}{\mathrm{lbm}}=245.62\ \mathrm{Btu}/\mathrm{lbm} $$

Using Eq. 17.8, calculate the new mass flow rate of steam required due to turbine and pump efficiencies. The power output remains the same as before.

$$ {\dot{m}}_{\mathrm{s},\mathrm{actual}}=\frac{{\dot{W}}_{\mathrm{net}}}{w_{\mathrm{net},\mathrm{actual}}}=\frac{5.21\ \mathrm{MW}\times \frac{3412142\frac{\mathrm{Btu}}{\mathrm{hr}}}{1\ \mathrm{MW}}}{245.62\frac{\mathrm{Btu}}{\mathrm{lbm}}}=72,377\ \mathrm{lbm}/\mathrm{hr} $$

Using Eq. 14.14 (same equation can be applied to pumps and compressors), calculate the enthalpy of water leaving the pump since it would have changed due to the efficiency of the pump. From the solution to Example 17.4, the enthalpy of water entering the pump is h₁ = 180.18 Btu/lbm. The actual enthalpy of water leaving the pump (same as the enthalpy of water entering the boiler) can be obtained by adding actual pump work to the enthalpy of water entering the pump.

$$ {h}_{2^{\prime }}=180.18\frac{\mathrm{Btu}}{\mathrm{lbm}}+1.88\frac{\mathrm{Btu}}{\mathrm{lbm}}=182.06\ \mathrm{Btu}/\mathrm{lbm} $$

From the solution to Example 17.4, the enthalpy of steam leaving the boiler is h₃ = 1290 Btu/lbm. Calculate the actual heat input to the boiler per unit mass of steam using Eq. 17.11.

$$ {q}_b={h}_3-{h}_{2\hbox{'}}=1290\frac{\mathrm{Btu}}{\mathrm{lbm}}-182.06\frac{\mathrm{Btu}}{\mathrm{lbm}}=1107.94\ \mathrm{Btu}/\mathrm{lbm} $$

The preceding result is almost the same as that was obtained assuming ideal (or isentropic) pump. Hence, we can safely neglect the effect of pump efficiency on the heat input to the boiler. Calculate the actual rate of heat input to the boiler using Eq. 17.9.

$$ {\displaystyle \begin{array}{l}{\dot{Q}}_{\mathrm{b},\mathrm{actual}}={\dot{m}}_{\mathrm{s},\mathrm{actual}}{q}_{\mathrm{b},\mathrm{actual}}=\left(72377\frac{\mathrm{lbm}}{\mathrm{hr}}\right)\left(1107.64\frac{\mathrm{Btu}}{\mathrm{lbm}}\right)\\ {}\kern8.5em =8.036\times {10}^7\ \mathrm{Btu}/\mathrm{hr}\end{array}} $$

Calculate the actual mass rate of fuel required by using the net heating value of the fuel and boiler efficiency, same as in Example 17.4.

$$ {\dot{m}}_{\mathrm{fuel},\mathrm{actual}}=\frac{{\dot{Q}}_{\mathrm{b},\mathrm{actual}}}{{\mathrm{NHV}}_{\mathrm{fuel}}\times {\eta}_{\mathrm{b}}}=\frac{8.036\times {10}^7\frac{\mathrm{Btu}}{\mathrm{hr}}}{12200\frac{\mathrm{Btu}}{\mathrm{lbm}}\times 0.85}=7749\ \mathrm{lbm}/\mathrm{hr} $$

Calculate the percentage increase in the fuel requirement per hour.

$$ \%\mathrm{increase}=\left(\frac{7749\frac{\mathrm{lbm}}{\mathrm{hr}}-6947\frac{\mathrm{lbm}}{\mathrm{hr}}}{6947\frac{\mathrm{lbm}}{\mathrm{hr}}}\right)100=11.54\% $$

Practice Problem 17.4

• (Solution)

Calculate the compression ratio using Eq. 17.35.

$$ {\eta}_{\mathrm{Otto}}=1-{r}^{1-k}\Rightarrow r={\left(1-{\eta}_{\mathrm{Otto}}\right)}^{\frac{1}{1-k}}={\left(1-0.55\right)}^{\frac{1}{1-1.4}}=7.36 $$

Calculate the volume after the compression process using the definition of compression ratio (Eq. 17.34).

$$ r=\frac{V_1}{V_2}\Rightarrow {V}_2=\frac{V_1}{r}=\frac{3\ \mathrm{L}}{7.36}=0.41\ \mathrm{L} $$

Calculate the pressure after the isentropic compression process using Eq. 17.36.

$$ {P}_2={P}_1{\left(\frac{V_1}{V_2}\right)}^k=\left(14\ \mathrm{psia}\right){(7.36)}^{1.4}=229\ \mathrm{psia} $$

Calculate the absolute temperature of intake air using Eq. 13.2.

$$ {T}_1=460{}^{\circ}+70{}^{\circ}\mathrm{F}=530{}^{\circ}\mathrm{R} $$

Calculate the absolute temperature after the isentropic compression process using Eq. 13.37.

$$ {T}_2={T}_1{\left(\frac{V_1}{V_2}\right)}^{k-1}=\left({T}_1\right){r_p}^{k-1}=\left({530}^{\circ}\mathrm{R}\right){7.36}^{1.4-1}={1178}^{\circ}\mathrm{R}\kern0.5em \left({718}^{\circ}\mathrm{F}\right) $$

Practice Problem 17.5

• (Solution)

• A. Draw the P – h diagram as a reference for state points and for heat/work flow.

Also, determine the enthalpies at state points using the P – h diagram for R-134a as shown.

From the preceding illustration,

$$ {h}_1=104\ \mathrm{Btu}/\mathrm{lbm} $$

$$ {h}_2=120\ \mathrm{Btu}/\mathrm{lbm} $$

$$ {h}_4={h}_3=55\ \mathrm{Btu}/\mathrm{lbm} $$

• B. Obtain the heating load of the heat pump by calculating the rate of heating required to heat the air.

Calculate the individual gas constant for air using Eq. 13.16.

$$ {R}_{\mathrm{air}}=\frac{\overline{R}}{M_{\mathrm{air}}}=\frac{10.73\frac{\mathrm{psia}\hbox{-} {\mathrm{ft}}^3}{\mathrm{lbm}\mathrm{ol}\hbox{-} {}^{\circ}\mathrm{R}}\ }{29\frac{\mathrm{lbm}}{\mathrm{lbm}\mathrm{ol}}}=0.37\ \mathrm{psia}\hbox{-} {\mathrm{ft}}^3/\mathrm{lbm}\hbox{-} {}^{\circ}\mathrm{R} $$

Calculate the average temperature of air and its absolute value using Eq. 13.2. T_avg = (40 ° F + 70 ° F)/2 = 55 ° F ⇒ T_air = 460 °  + 55 ° F = 515 ° R

Calculate the density of air at its average temperature using the ideal gas law (Eq. 13.12).

$$ \rho =\frac{m}{V}=\frac{P}{R_{\mathrm{air}}{T}_{\mathrm{air}}}=\frac{14.7\ \mathrm{psia}}{\left(0.37\frac{\mathrm{psia}\hbox{-} {\mathrm{ft}}^3}{\mathrm{lbm}\hbox{-} {}^{\circ}\mathrm{R}}\right)\left(515{}^{\circ}\mathrm{R}\right)}=0.0771\ \mathrm{lbm}/{\mathrm{ft}}^3 $$

Calculate the mass flow rate of air using the continuity equation (Eq. 3.5).

$$ {\dot{m}}_{\mathrm{air}}={\rho}_{\mathrm{air}}{\dot{V}}_{\mathrm{air}}=\left(0.0771\frac{\mathrm{lbm}}{{\mathrm{ft}}^3}\right)\left(1700\frac{{\mathrm{ft}}^3}{\min}\right)=131.1\ \mathrm{lbm}/\min $$

From Table 13.2, the specific heat of air at constant pressure is 0.24 Btu/lbm-°R

Energy balance at the high-temperature heat sink results in the following set of equations:

$$ \mathrm{Rate}\ \mathrm{of}\ \mathrm{heat}\ \mathrm{rejected}\ \mathrm{at}\ \mathrm{the}\ \mathrm{heat}\ \mathrm{sink}=\mathrm{Rate}\ \mathrm{of}\ \mathrm{heat}\ \mathrm{absorption}\ \mathrm{by}\ \mathrm{air} $$

$$ {\displaystyle \begin{array}{l}{\dot{Q}}_{\mathrm{out}}={\dot{m}}_{\mathrm{air}}{c}_{\mathrm{p},\mathrm{air}}\Delta {T}_{\mathrm{air}}=\left(131.1\frac{\mathrm{lbm}}{\min}\right)\left(0.24\frac{\mathrm{Btu}}{\mathrm{lbm}\hbox{-} {}^{\circ}\mathrm{R}}\right)\left(70{}^{\circ}\mathrm{F}-40{}^{\circ}\mathrm{F}\right)\\ {}\kern7.75em =943.9\ \mathrm{Btu}/\min \end{array}} $$

Calculate the heat rejection per unit mass of refrigerant using Eq. 17.44.

$$ {q}_{\mathrm{out}}={h}_2-{h}_3=120\frac{\mathrm{Btu}}{\mathrm{lbm}}-55\frac{\mathrm{Btu}}{\mathrm{lbm}}=65\ \mathrm{Btu}/\mathrm{lbm} $$

Using the information of heating load required and the equation for the rate of heat rejection from the condenser (Eq. 17.45), calculate the mass flow rate of refrigerant.

$$ {\dot{m}}_{\mathrm{R}}=\frac{{\dot{Q}}_{\mathrm{out}}}{q_{\mathrm{out}}}=\frac{943.9\frac{\mathrm{Btu}}{\min }}{65\frac{\mathrm{Btu}}{\mathrm{lbm}}}=14.52\ \mathrm{lbm}/\min $$

Calculate the isentropic horsepower input required for the compressor by using Eq. 17.43.

$$ {\displaystyle \begin{array}{l}{\dot{W}}_{\mathrm{c},\mathrm{ideal}}={\dot{m}}_{\mathrm{R}}\left({h}_2-{h}_1\right)=\left(14.52\frac{\mathrm{lbm}}{\min}\right)\left(120\frac{\mathrm{Btu}}{\mathrm{lbm}}-104\frac{\mathrm{Btu}}{\mathrm{lbm}}\right)\times \\ {}\kern17.75em \left(\frac{0.0236\ \mathrm{hp}}{\mathrm{Btu}/\min}\right)\\ {}\kern8em =5.48\ \mathrm{hp}\end{array}} $$

• C. Calculate the coefficient of performance for the heat pump using Eq. 17.47.

$$ {\mathrm{COP}}_{\mathrm{heat}\ \mathrm{pump}}=\frac{{\dot{Q}}_{\mathrm{out}}}{{\dot{W}}_{\mathrm{c}}}=\frac{943.9\frac{\mathrm{Btu}}{\min }}{\left(14.52\frac{\mathrm{lbm}}{\min}\right)\left(120\frac{\mathrm{Btu}}{\mathrm{lbm}}-104\frac{\mathrm{Btu}}{\mathrm{lbm}}\right)}=4.06 $$