Thermodynamics Practice

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1.1Identify a Closed System

Exam I | Problem 1.1 |

A gas is sealed inside a piston-cylinder device. The piston can move, and heat can cross the boundary, but no mass can enter or leave.

What type of system is this?

Solution

This is a closed system or control mass.

Mass does not cross the boundary, but energy can cross as heat or work.

1.2Classify Properties by Size Dependence

Exam I | Problem 1.2 |

Which of the following are extensive properties?

m,\quad P,\quad H,\quad T

Solution

An extensive property depends on system size.

So the extensive properties are

m \text{ and } H

Pressure $P$ and temperature $T$ are intensive properties.

1.3Convert Gauge Pressure to Absolute Pressure

Exam I | Problem 1.3 |

A tire gauge reads $220\ \text{kPa}$ and the atmospheric pressure is $101\ \text{kPa}$.

What is the absolute pressure of the tire?

Solution

Use

P_{abs} = P_{gage} + P_{atm}

P_{abs} = 220 + 101 = 321\ \text{kPa}

1.4Convert Celsius to Kelvin

Exam I | Problem 1.4 |

Convert $27^\circ\text{C}$ to Kelvin.

Solution

Use the absolute temperature relation:

T(K) = T(^\circ C) + 273.15

T = 27 + 273.15 = 300.15\ \text{K}

1.5Choose the Phase Region

Exam I | Problem 1.5 |

A pure substance is at a pressure of $500\ \text{kPa}$ and a temperature below $T_{sat}$ at that pressure.

What phase region is it in?

Solution

If the temperature is below the saturation temperature at the given pressure, the substance is in the compressed liquid or subcooled liquid region.

1.6Compute Quality from Masses

Exam I | Problem 1.6 |

A saturated liquid-vapor mixture contains $3\ \text{kg}$ of saturated liquid and $9\ \text{kg}$ of saturated vapor.

What is the quality $x$?

Solution

Use

x = \frac{m_g}{m_f + m_g}

x = \frac{9}{3+9} = \frac{9}{12} = 0.75

1.7Use the Ideal Gas Law

Exam I | Problem 1.7 |

An ideal gas has

P = 300\ \text{kPa},\qquad T = 300\ \text{K},\qquad R = 0.287\ \text{kJ/(kg\cdot K)}.

What is the specific volume $v$?

Solution

Use the specific ideal gas law:

$$ Pv = RT $$

v = \frac{RT}{P} = \frac{0.287(300)}{300} = 0.287\ \text{m}^3/\text{kg}

1.8Find Constant-Pressure Boundary Work

Exam I | Problem 1.8 |

A gas expands at constant pressure from $0.50\ \text{m}^3$ to $0.80\ \text{m}^3$ while the pressure remains $150\ \text{kPa}$.

What is the boundary work?

Solution

For constant pressure,

$$ W_b = P(V_2 - V_1) $$

W_b = 150(0.80 - 0.50) = 150(0.30) = 45\ \text{kJ}

The work is positive because the system does work on the surroundings.

1.9Apply the Closed-System First Law

Exam I | Problem 1.9 |

A closed system receives $120\ \text{kJ}$ of heat and does $35\ \text{kJ}$ of work. Neglect changes in kinetic and potential energy.

What is the change in internal energy?

Solution

Use the closed-system First Law:

\Delta U = Q - W

\Delta U = 120 - 35 = 85\ \text{kJ}

1.10Compute Mass Flow Rate

Exam I | Problem 1.10 |

Air with density $1.2\ \text{kg/m}^3$ flows through a duct with area $0.050\ \text{m}^2$ at a speed of $20\ \text{m/s}$.

What is the mass flow rate?

Solution

Use

\dot{m} = \rho A V

\dot{m} = 1.2(0.050)(20) = 1.2\ \text{kg/s}

Difficulty

Integrated

2.1Find a Saturated Mixture Property

Exam II | Problem 2.1 |

A saturated mixture has

u_f = 500\ \text{kJ/kg},\qquad u_{fg} = 1500\ \text{kJ/kg},\qquad x = 0.20.

What is the specific internal energy $u$?

Solution

Use the saturated mixture relation:

u = u_f + x u_{fg}

u = 500 + 0.20(1500) = 500 + 300 = 800\ \text{kJ/kg}

2.2Approximate a Compressed-Liquid Enthalpy

Exam II | Problem 2.2 |

At a given temperature, a liquid has

h_f(T) = 340.0\ \text{kJ/kg},\qquad v_f(T) = 0.0010\ \text{m}^3/\text{kg}

and

P - P_{sat}(T) = 500\ \text{kPa}.

Approximate the compressed-liquid enthalpy $h$.

Solution

Use the compressed-liquid approximation:

h \approx h_f(T) + v_f(T)\left[P - P_{sat}(T)\right]

h \approx 340.0 + 0.0010(500)

Since $1\ \text{kPa}\cdot\text{m}^3/\text{kg} = 1\ \text{kJ/kg}$,

h \approx 340.5\ \text{kJ/kg}

2.3Use an Isentropic Ideal-Gas Relation

Exam II | Problem 2.3 |

Air with $k = 1.4$ is compressed isentropically from $P_1 = 100\ \text{kPa}$ and $T_1 = 300\ \text{K}$ to $P_2 = 400\ \text{kPa}$.

What is the final temperature $T_2$?

Solution

For an isentropic ideal gas,

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k}

Substitute the values:

T_2 = 300\left(\frac{400}{100}\right)^{0.4/1.4}

T_2 \approx 300(4^{0.2857}) \approx 446\ \text{K}

2.4Evaluate Polytropic Boundary Work

Exam II | Problem 2.4 |

An ideal gas undergoes an isothermal expansion. At the initial state,

P_1 = 200\ \text{kPa},\qquad V_1 = 0.30\ \text{m}^3,

and the final volume is $V_2 = 0.60\ \text{m}^3$.

What is the boundary work?

Solution

For an ideal-gas isothermal process, $n=1$ and

W_b = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)

Compute:

W_b = 200(0.30)\ln(2)

W_b = 60(0.693) \approx 41.6\ \text{kJ}

2.5Solve a Spring-Loaded Energy Balance

Exam II | Problem 2.5 |

A piston-cylinder device compresses a linear spring from $x_1 = 0$ to $x_2 = 0.20\ \text{m}$. The spring constant is $k = 8.0\ \text{kN/m}$.

During the process, the closed system receives $Q = 1.0\ \text{kJ}$ of heat and no other work occurs.

Find the spring work and the change in internal energy.

Solution

For a linear spring,

W = \frac{1}{2}k(x_2^2 - x_1^2)

W = \frac{1}{2}(8000)(0.20^2) = 160\ \text{J} = 0.16\ \text{kJ}

Now apply the First Law:

\Delta U = Q - W

\Delta U = 1.0 - 0.16 = 0.84\ \text{kJ}

2.6Analyze a Turbine

Exam II | Problem 2.6 |

A steady adiabatic turbine has negligible kinetic and potential energy changes. The mass flow rate is $3\ \text{kg/s}$, the inlet enthalpy is $h_1 = 3200\ \text{kJ/kg}$, and the outlet enthalpy is $h_2 = 2500\ \text{kJ/kg}$.

What is the power output of the turbine?

Solution

For an adiabatic turbine with negligible kinetic and potential energy changes,

\dot{W}_{out} = \dot{m}(h_1 - h_2)

\dot{W}_{out} = 3(3200 - 2500) = 3(700) = 2100\ \text{kW}

2.7Analyze a Throttling Valve

Exam II | Problem 2.7 |

A throttling valve drops the pressure of a fluid from $900\ \text{kPa}$ to $200\ \text{kPa}$.

If the inlet enthalpy is $h_1 = 245\ \text{kJ/kg}$, what is the outlet enthalpy?

Solution

For throttling,

$$ h_1 = h_2 $$

h_2 = 245\ \text{kJ/kg}

2.8Compute Ideal-Gas Entropy Change

Exam II | Problem 2.8 |

Air has

c_p = 1.005\ \text{kJ/(kg\cdot K)},\qquad R = 0.287\ \text{kJ/(kg\cdot K)}.

It changes from $T_1 = 300\ \text{K}$ and $P_1 = 100\ \text{kPa}$ to $T_2 = 450\ \text{K}$ and $P_2 = 200\ \text{kPa}$.

Find $s_2 - s_1$.

Solution

Use the constant-specific-heat ideal-gas entropy relation:

s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)

Substitute:

s_2 - s_1 = 1.005\ln\left(\frac{450}{300}\right) - 0.287\ln(2)

s_2 - s_1 \approx 1.005(0.4055) - 0.287(0.6931)

s_2 - s_1 \approx 0.4075 - 0.1988 = 0.2087\ \text{kJ/(kg\cdot K)}

Difficulty

Applied

3.1Mix Two Streams

Final | Problem 3.1 |

An adiabatic mixing chamber has two inlet streams of the same substance:

\dot{m}_1 = 1\ \text{kg/s},\quad h_1 = 300\ \text{kJ/kg}

and

\dot{m}_2 = 2\ \text{kg/s},\quad h_2 = 500\ \text{kJ/kg}.

Find the outlet mass flow rate and outlet enthalpy.

Solution

Apply mass conservation:

\dot{m}_{out} = \dot{m}_1 + \dot{m}_2 = 1 + 2 = 3\ \text{kg/s}

For an adiabatic mixing chamber with no work and negligible kinetic and potential energy changes:

\sum \dot{m}_{in}h_{in} = \dot{m}_{out}h_{out}

1(300) + 2(500) = 3h_{out}

1300 = 3h_{out}

h_{out} = 433.3\ \text{kJ/kg}

3.2Analyze a Heat Exchanger

Final | Problem 3.2 |

An adiabatic heat exchanger has a hot stream and a cold stream. The hot stream has

\dot{m}_h = 2\ \text{kg/s},\quad h_{h,in} = 350\ \text{kJ/kg},\quad h_{h,out} = 250\ \text{kJ/kg}.

The cold stream has

\dot{m}_c = 3\ \text{kg/s},\quad h_{c,in} = 100\ \text{kJ/kg}.

Find the cold-stream outlet enthalpy.

Solution

With negligible heat loss to the surroundings, the enthalpy lost by the hot stream equals the enthalpy gained by the cold stream:

\dot{m}_h(h_{h,in}-h_{h,out}) = \dot{m}_c(h_{c,out}-h_{c,in})

Substitute:

2(350-250) = 3(h_{c,out}-100)

200 = 3(h_{c,out}-100)

h_{c,out}-100 = 66.7

h_{c,out} = 166.7\ \text{kJ/kg}

3.3Evaluate a Heat Engine

Final | Problem 3.3 |

A heat engine absorbs $900\ \text{kJ}$ from a hot reservoir and rejects $540\ \text{kJ}$ to a cold reservoir during one cycle.

Find the net work output and the thermal efficiency.

Solution

For a cycle,

W_{net,out} = Q_H - Q_L

W_{net,out} = 900 - 540 = 360\ \text{kJ}

Thermal efficiency is

\eta_{th} = \frac{W_{net,out}}{Q_H}

Thus

\eta_{th} = \frac{360}{900} = 0.40

So the efficiency is $40\%$.

3.4Evaluate a Refrigerator

Final | Problem 3.4 |

A refrigerator removes $420\ \text{kJ}$ of heat from the cold space and requires $140\ \text{kJ}$ of work input.

Find $Q_H$, $COP_R$, and $COP_{HP}$.

Solution

Use the refrigerator energy balance:

W_{net,in} = Q_H - Q_L

Q_H = Q_L + W_{net,in} = 420 + 140 = 560\ \text{kJ}

The refrigerator COP is

COP_R = \frac{Q_L}{W_{net,in}} = \frac{420}{140} = 3

The heat pump COP is

COP_{HP} = \frac{Q_H}{W_{net,in}} = \frac{560}{140} = 4

3.5Check Entropy Generation

Final | Problem 3.5 |

Six hundred kilojoules of heat flows directly from a $600\ \text{K}$ reservoir to a $300\ \text{K}$ reservoir.

What is the entropy change of the universe, and does this process satisfy the Second Law?

Solution

The hot reservoir loses entropy:

\Delta S_H = -\frac{600}{600} = -1\ \text{kJ/K}

The cold reservoir gains entropy:

\Delta S_C = \frac{600}{300} = 2\ \text{kJ/K}

So the universe changes by

\Delta S_{universe} = -1 + 2 = 1\ \text{kJ/K}

Because $\Delta S_{universe} > 0$, the process satisfies the Second Law.

Difficulty

Challenge

4.1Evaluate an Otto Cycle

Final | Problem 4.1 |

An air-standard Otto cycle has a compression ratio of $r = 8$ and $k = 1.4$.

What is the thermal efficiency?

Solution

For the ideal Otto cycle,

\eta_{Otto} = 1 - \frac{1}{r^{k-1}}

Substitute the values:

\eta_{Otto} = 1 - \frac{1}{8^{0.4}}

Since

8^{0.4} \approx 2.30

we get

\eta_{Otto} \approx 1 - \frac{1}{2.30} \approx 0.565

So the efficiency is about $56.5\%$.

4.2Evaluate a Brayton Cycle

Final | Problem 4.2 |

An air-standard Brayton cycle has a pressure ratio of $r_p = 6$ and $k = 1.4$.

What is the thermal efficiency?

Solution

For the ideal Brayton cycle,

\eta_{Brayton} = 1 - \frac{1}{r_p^{(k-1)/k}}

Substitute the values:

\eta_{Brayton} = 1 - \frac{1}{6^{0.2857}}

Since

6^{0.2857} \approx 1.67

we get

\eta_{Brayton} \approx 1 - \frac{1}{1.67} \approx 0.40

So the efficiency is about $40\%$.

4.3Compute Rankine Cycle Efficiency

Final | Problem 4.3 |

An ideal Rankine cycle has the following enthalpies:

h_1 = 200,\quad h_2 = 210,\quad h_3 = 3200,\quad h_4 = 2200

all in $\text{kJ/kg}$.

Find the pump work input, turbine work output, net work output, and thermal efficiency.

Solution

Use the Rankine cycle relations:

w_{pump,in} = h_2 - h_1 = 210 - 200 = 10\ \text{kJ/kg}

w_{turb,out} = h_3 - h_4 = 3200 - 2200 = 1000\ \text{kJ/kg}

Net work output:

w_{net,out} = w_{turb,out} - w_{pump,in} = 1000 - 10 = 990\ \text{kJ/kg}

Heat input:

q_{in} = h_3 - h_2 = 3200 - 210 = 2990\ \text{kJ/kg}

Thermal efficiency:

\eta_{Rankine} = \frac{w_{net,out}}{q_{in}} = \frac{990}{2990} \approx 0.331

So the efficiency is about $33.1\%$.

4.4Apply the Workflow to a Sealed Tank

Final | Problem 4.4 |

A rigid, sealed tank contains gas. It loses $5\ \text{kJ}$ of heat to the surroundings and receives $12\ \text{kJ}$ of electrical work input. Neglect kinetic and potential energy changes.

Using the closed-system First Law and the sign convention in the note, find $\Delta U$.

Solution

The tank is a closed system because no mass crosses the boundary.

Using the note's sign convention:

Q = -5\ \text{kJ}

because heat leaves the system, and

W = -12\ \text{kJ}

because work is done on the system, not by the system.

Now apply the First Law:

\Delta U = Q - W

\Delta U = -5 - (-12) = 7\ \text{kJ}