6 Thermodynamic Cycles Free Online FlashCards

What does a clockwise loop on a P–V diagram generally indicate?

A clockwise loop generally indicates positive net work done by the working fluid; a counterclockwise loop indicates net work input.

Why must a power cycle reject some heat?

A power cycle operating between hot and cold surroundings must reject some heat; it cannot convert all supplied heat into work.

What happens in the four main processes of a Rankine cycle?

In a Rankine cycle, the pump raises liquid pressure, the boiler adds heat, the turbine produces work as steam expands, and the condenser rejects heat and returns the fluid to liquid.

How do refrigerators and heat pumps move heat?

Both devices use work input to move heat from a colder region to a warmer one.

Which components make up a vapor-compression cycle, and where is heat absorbed?

A vapor-compression cycle uses a compressor, condenser, expansion device, and evaporator. The evaporator absorbs heat, while the condenser releases it.

Why is a cycle’s net change in internal energy zero?

Its net change in internal energy is zero because the working substance returns to its initial state.

How are net heat transfer and net work related over a complete cycle?

For a complete cycle, net heat transfer equals net work: $Q_{\text{net}}=W_{\text{net}}$, with heat entering and work done by the system positive.

How is a power cycle’s thermal efficiency defined?

Thermal efficiency is net work output divided by heat input: $\eta=\frac{W_{\text{net,out}}}{Q_{\text{in}}}$.

A cycle receives $1000\ \text{kJ}$ and rejects $600\ \text{kJ}$. What is its net work output?

The net work output is $400\ \text{kJ}$, since $W_{\text{net,out}}=Q_{\text{in}}-Q_{\text{out}}$.

Which energy changes are often neglected in introductory steady-flow device analyses?

Changes in kinetic energy and potential energy are often neglected in introductory analyses.

What typically happens to a fluid’s enthalpy as it produces work in a turbine?

A turbine typically produces work as the fluid’s enthalpy decreases.

How do the coefficients of performance for a refrigerator and heat pump differ?

A refrigerator’s COP is heat removed divided by work input, $\mathrm{COP}_{\text{refrigerator}}=\frac{Q_{\text{in}}}{W_{\text{in}}}$; a heat pump’s COP is heat supplied divided by work input, $\mathrm{COP}_{\text{heat pump}}=\frac{Q_{\text{out}}}{W_{\text{in}}}$.