Exam 1 Vocabulary

Question 1

Closed system

Answer 1

A fixed set of identifiable particles of constant mass; the same set of particles is followed through the analysis. (abbreviation: sys)

Question 2

Control volume (open system)

Answer 2

A region of space which may or may not be moving (often taken to be fixed in space) and through which fluid may flow. Therefore, the identity of the particles in the control volume may vary from instant to instant. (abbreviation: CV)

Question 3

Control surface

Answer 3

The geometric bonding surface of the control volume. (abbreviation: CS)

Question 4

Extensive property

Answer 4

A property that depends on the amount of matter in a sample. (ex. mass, volume, energy, entropy)

Question 5

Specific property

Answer 5

Corresponding extensive property per unit mass. Intensive property. Independent of amount or size. (ex. temperature, density, pressure, specific heat capacity)

Question 6

Conservation of Mass

Answer 6

Mass can neither be created nor destroyed. Mass of a system is constant.

Question 7

Newton’s Second Law

Answer 7

Force = time rate of change of linear momentum. F=ma or F=mv(dot)

Question 8

Absolute temperature scales

Answer 8

Kelvin and Rankine. Absolute means it starts from absolute zero and measures temperatures in increments directly proportional to the average kinetic energy of gas molecules.

Question 9

First Law of Thermodynamics

Answer 9

Energy can be neither created not destroyed, it can only change in form. Energy eqn –> Q(diff) - W(diff) = E(diff wrt time) Q = heat transfer, W = work, E = energy. Use momentum equation. U is internal energy. In control volume, use internal energy (u), kinetic energy (V^2/2), and potential energy (gz).

Question 10

Second Law of Thermodynamics

Answer 10

Highlights irreversibility of natural processes when considered in terms of heat transfer. Heat generally goes to where it’s cooler, but not the other way around unless there are external interventions.
Entropy eqn –> S(diff wrt time) >/ Q(diff)/T. S = total entropy, Q = heat transfer, T = temperature. Use momentum equation.

Question 11

The Isentropic Case of the Second Law of Thermodynamics

Answer 11

s2-s1 = 0 assuming a thermally and calorically perfect gas. Entropy of system remains unchanged. Entropy may change WITHIN the closed system as long as net change is zero.

Question 12

Reynold’s Transport Theorem

Answer 12

Analyzes the change of extensive properties within a control volume as it moves through space. How to use: 1. identify extensive property to analyze
2. define a control volume
3. apply RTT to write general eqn governing the change of the extensive property within the CV
4. apply additional assumptions or simplifications
5. solve resulting eqn to obtain information about the behavior of system over time

Question 13

Q(diff)

Answer 13

Derivative of heat transfer. Temperature x Infinitesimal change in entropy. Heat transfer is a path dependent quantity. Heat transfer rate to CV.

Question 14

Ws(diff)

Answer 14

Ws = shaft work done in a system. Represents an infinitesimal amount of work done in a system. Pressure x Infinitesimal change in volume. Shear and shaft work rate by CV. Total work rate due both to shaft and shear work.

Question 15

Specific enthalpy

Answer 15

Denoted as h. Used to describe the energy content of a substance per unit mass. Sum of internal energy and the product of pressure times specific volume. Change in enthalpy = change inn work + change in heat transfer

Question 16

Entropy

Answer 16

Quantifies the order of randomness or chaos in a system. Denoted by S. Represents the amount of energy in a system that is not able to do work.

Question 17

Thermally/calorically perfect assumption

Answer 17

Specific heats remain constant wrt temperature. Cp and Cv are zero. For isentropic processes S(diff) = 0; P/rho^gamma = constant.

Question 18

Ratio of specific heats

Answer 18

Which variable? GAMMA = Cp/Cv

Question 19

Isentropic process

Answer 19

What can you assume? Assume no heat transfer. Constant entropy, process is reversible and adiabatic. Constant ratio of specific heats (gamma).

Question 20

Adiabatic process

Answer 20

What can you assume? Assume no heat transfer. Internal energy can change. Constant ratio of specific heats (gamma).

Question 21

Mach number

Answer 21

Eqn? M = V/a. a = sqrt(gammaRT)

Question 22

Subsonic

Answer 22

Flow velocity is less than the speed of sound

Question 23

Sonic

Answer 23

Flow velocity is equal to the speed of sound

Question 24

Supersonic

Answer 24

Flow velocity exceeds the speed of sound

Question 25

Transonic

Answer 25

Flow velocity does not necessarily exceed the speed of sound, but is very close to it, resulting in a mixture of supersonic and subsonic properties

Question 26

Hypersonic

Answer 26

Flow velocity is much higher than the speed of sound

Question 27

Static property

Answer 27

Properties measured when moving along with the flow at the local fluid velocity.

Question 28

Stagnation property

Answer 28

No movement. Stagnation point is a point in a flowing fluid where the velocity is reduced to zero.

Question 29

Mass flow rate m(diff)

Answer 29

Density x cross-sectional area x velocity. Density x Q. Described rate at which mass flows through a system per unit time.

Question 30

Volumetric flow rate