Pomodoro
Showing all 27 problems
Progress
0 of 27 completed
Difficulty
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.
Which of the following are extensive properties?
Solution
An extensive property depends on system size.
So the extensive properties are
Pressure $P$ and temperature $T$ are intensive properties.
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
So
Convert $27^\circ\text{C}$ to Kelvin.
Solution
Use the absolute temperature relation:
So
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.
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
So
An ideal gas has
What is the specific volume $v$?
Solution
Use the specific ideal gas law:
So
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,
So
The work is positive because the system does work on the surroundings.
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:
So
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
So
Difficulty
A saturated mixture has
What is the specific internal energy $u$?
Solution
Use the saturated mixture relation:
So
At a given temperature, a liquid has
and
Approximate the compressed-liquid enthalpy $h$.
Solution
Use the compressed-liquid approximation:
So
Since $1\ \text{kPa}\cdot\text{m}^3/\text{kg} = 1\ \text{kJ/kg}$,
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,
Substitute the values:
An ideal gas undergoes an isothermal expansion. At the initial state,
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
Compute:
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,
So
Now apply the First Law:
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,
So
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,
So
Air has
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:
Substitute:
Difficulty
An adiabatic mixing chamber has two inlet streams of the same substance:
and
Find the outlet mass flow rate and outlet enthalpy.
Solution
Apply mass conservation:
For an adiabatic mixing chamber with no work and negligible kinetic and potential energy changes:
So
An adiabatic heat exchanger has a hot stream and a cold stream. The hot stream has
The cold stream has
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:
Substitute:
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,
So
Thermal efficiency is
Thus
So the efficiency is $40\%$.
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:
So
The refrigerator COP is
The heat pump COP is
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:
The cold reservoir gains entropy:
So the universe changes by
Because $\Delta S_{universe} > 0$, the process satisfies the Second Law.
Difficulty
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,
Substitute the values:
Since
we get
So the efficiency is about $56.5\%$.
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,
Substitute the values:
Since
we get
So the efficiency is about $40\%$.
An ideal Rankine cycle has the following enthalpies:
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:
Net work output:
Heat input:
Thermal efficiency:
So the efficiency is about $33.1\%$.
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:
because heat leaves the system, and
because work is done on the system, not by the system.
Now apply the First Law:
So
