Thermodynamics

Question 1

What type of properties are INDEPENDENT of mass?

Answer 1

Intensive Properties

Question 2

What type of properties are PROPORTIONAL to mass?

Answer 2

Extensive Properties

Question 3

What type of properties are expressed on a per-mass basis; shown in lowercase?

Answer 3

Specific Properties

Question 4

For a single-phase pure component, how many intensive, independent properties are required to determine all the rest?

Answer 4

2

Question 5

When heat is added to a gas at constant volume, we have

Answer 5

Qv = Cv*∆T = ∆U + W = ∆U

because no work is done. Therefore,
dU = Cv*dT and Cv = dU/dT

Question 6

When heat is added to a gas at constant pressure, we have

Answer 6

When heat is added at constant pressure, we have
Qp = Cp∆T = ∆U + W = ∆U + P∆V

CpdT = dU + PdV = CVdT + PdV

Question 7

Quality x, for liquid-vapor systems at saturation, is defined as the mass fraction of the vapor phase:

Answer 7

x = m_g / (m_g + m_f)

where
m_g = mass of vapor
m_f = mass of liquid

Question 8

Two-Phase (Vapor-Liquid) system expressions:

Answer 8

v = xv_g + (1 – x) v_f    or   v = v_f + xv_fg
u = xu_g + (1 – x) u_f   or   u = u_f + xu_fg
h = xh_g + (1 – x) h_f   or   h = h_f + xh_fg
s = xs_g + (1 – x) s_f    or   s = s_f + xs_fg

where
_f = specific property of saturated liquid
_g = specific property of saturated vapor
_fg = specific property change upon vaporization = vg – vf

Question 9

Ideal Gas Law

Answer 9

Pv = RT or PV = mRT and P1v1/T1 = P2v2/T2

where
P = pressure
v = specific volume
m = mass of gas
R = gas constant
T = absolute temperature
V = volume

Question 10

Gas constant, R

Answer 10

R is specific to each gas but can be found from

R = R_ / (mol wt)

where R_ = universal gas constant

For ideal gases, Cp – Cv = R

Question 11

For cold air standard, heat capacities are assumed to be constant at their room temperature values. In that case, the following are true:

Answer 11

∆u = Cv∆T; ∆h = Cp ∆T
∆s = Cp ln (T2 /T1) – R ln (P2 /P1)
∆s = Cv ln (T2 /T1) + R ln (v2 /v1)

Also, for constant entropy processes:

(P2/P1) = (v1/v2)^k
(T2/T1) = (P2/P1)^((k-1)/K)
(T2/T1) = (v1/v2)^(k-1_

where:

k = Cp/Cv

Question 12

Mole Fraction of an Ideal Gas Mixture (xi)

where:
i = 1, 2, …, n constituents. Each constituent is an ideal gas.
Ni = number of moles of component i
N = total moles in the mixture

Answer 12

xi = Ni/N

N = ΣNi

Σxi = 1

Question 13

Mass Fraction of an Ideal Gas Mixture (yi)

where:
i = 1, 2, …, n constituents. Each constituent is an ideal gas.

Answer 13

yi = mi/m

m = Σmi

Σyi = 1

Question 14

Molecular Weight of an Ideal Gas (M)

Answer 14

M = m/N = ΣxiMi

Question 15

To convert mole fractions (xi) to mass fractions (yi):

Answer 15

yi = xiMi / ΣxiMi

Question 16

To convert mass fractions to mole fractions:

Answer 16

xi = yi/Mi / Σyi/Mi

Question 17

Partial Pressures & Volumes of an Ideal Gas:

where:

P, V, T = pressure, volume, and temperature of the mixture
Ri = R/Mi

Answer 17

Pi = miRiT / V and P = ΣPi

Vi = miRiT / P and V = ΣVi

Question 18

Expressions of Internal energy (u), enthalpy (h), and entropy (s) for mole fractions can be evaluated at:

Answer 18

ui and hi are evaluated at T

si is evaluated at T and Pi

Question 19

Compressibility Factor

The generalized compressibility chart provides reasonable estimates for the compressibility factor Z based on dimensionless reduced pressure P_R and reduced temperature T_R.

Where, P_C and T_C are the critical pressure and temperature respectively expressed in absolute units.

Answer 19

P_R = P / P_C
T_R = T / T_C

Question 20

Equations of state (EOS) are used to quantify PvT behavior.

Answer 20

For ideal gas EOS (applicable only to ideal gases):

P = (RT / v)

For generalized compressibility EOS (applicable to all systems as gases, liquids, and/or solids):

P = (RT / v) * Z

Question 21

What are the (3) types of thermodynamic systems?

Answer 21

Open
Closed
Isolated

An OPEN system can exchange both energy and matter with its surroundings. The stovetop example would be an open system, because heat and water vapor can be lost to the air.

A CLOSED system, on the other hand, can exchange only energy with its surroundings, not matter. If we put a very tightly fitting lid on the pot from the previous example, it would approximate a closed system.

An ISOLATED system is one that cannot exchange either matter or energy with its surroundings. A perfect isolated system is hard to come by, but an insulated drink cooler with a lid is conceptually similar to a true isolated system. The items inside can exchange energy with each other, which is why the drinks get cold and the ice melts a little, but they exchange very little energy (heat) with the outside environment.

Question 22

What are the components of a thermodynamic universe?

Answer 22

A SYSTEM and its SURROUNDINGS, separated by a BOUNDARY.

Question 23

What is the 1st Law of Thermodynamics?

Answer 23

aka the Law of Conservation of Energy

The net energy crossing the system boundary is equal to the change in energy inside the system.

“energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another”

HEAT, Q (q = Q/m), is energy transferred due to temperature difference and is considered positive if it is inward or added to the system.

WORK, W (w = W/m), is considered positive if it is outward or work done by the system.

Question 24

Closed Thermodynamic System

Answer 24

No mass crosses the system boundary:
Q – W = ∆U + ∆KE + ∆PE
where
∆U = change in internal energy
∆KE = change in kinetic energy
∆PE = change in potential energy

Question 25

Special Cases of Closed Systems (With No Change in Kinetic or Potential Energy)

Constant system pressure process (Charles’s Law):

Answer 25

w_b = P∆v

T/v = constant for ideal gas

Question 26

Special Cases of Closed Systems (With No Change in Kinetic or Potential Energy)

Constant volume process:

Answer 26

w_b = 0

T/P = constant for ideal gas

Question 27

Special Cases of Closed Systems (With No Change in Kinetic or Potential Energy)

Isentropic process:

where:
Pv^k = constant for ideal gas

Answer 27

w = (P2v2 - P1v1) / (1-k) = R(T2-T1) / (1-k)

Question 28

Special Cases of Closed Systems (With No Change in Kinetic or Potential Energy)

Constant temperature process (Boyle’s Law):

where:
P_v = constant for ideal gas

Answer 28

w_b = RTln(v2/v1) = RTln(P1/P2)

Question 29

Special Cases of Closed Systems (With No Change in Kinetic or Potential Energy)

Polytropic process:

where:
Pv^n = constant for ideal gas

where n is the polytropic exponent or polytropic index

Answer 29

w = (P2v2 - P1v1) / (1-n)

n does not equal 1

Question 30

Open Thermodynamic Systems

Answer 30

Mass does cross the system boundary. 
Flow work (Pv) is done by mass entering the system.

The reversible flow work can be expressed:
w_rev = – ⌠vdP + ∆KE + ∆PE

The First Law applies whether or not processes are reversible.

Question 31

Open System First Law (energy balance):

Answer 31

d(m_s,u_s)/dt = Σmi(hi + ((Vi^2) / 2) + gZi) - Σme(he + ((Ve^2) / 2) + gZe) + Q_in - W_net

where
m = mass flow rate (subscripts i and e refer to inlet and exit states of system)
g = acceleration of gravity
Z = elevation
V = velocity
ms = mass of fluid within the system
us = specific internal energy of system
Qin = rate of heat transfer (ignoring kinetic and potential energy of the system)
Wnet = rate of net or shaft work

Question 32

Special Cases of Open Systems (With No Change in Kinetic or Potential Energy)

Constant volume process:

Answer 32

w_rev = – v (P2 – P1)

Question 33

Special Cases of Open Systems (With No Change in Kinetic or Potential Energy)

Constant system pressure process:

Answer 33

w_rev = 0

Question 34

Special Cases of Open Systems (With No Change in Kinetic or Potential Energy)

Constant temperature process:
Pv = constant for ideal gas

Answer 34

w_rev = RTln(v2/v1) = RTln(P1/P2)