Thermodynamics: PV Work: compression-expansion work

derivation of PV work formulas


1. isobaric conditions: p = c


W = F * d

F = A * p


substituting,

W = A * p * d


but:

V = area * height

dV = A * d



W = p * dV  ---> PV work for constant pressure process



2. isothermal conditions: T = c


W = S(V1 V2) p * dV


but:

pV = RT


and

p = RT/V


substituting,


W = S(V1 V2) p * dV

W = S(V1 V2) RT/V * dV


because temperature T = constant,

W = RT * S(V1 V2) 1/V * dV


but the

Integral of 1/v * dv = ln V


putting the limits (V1 V2)

W = RT * (ln V2 - ln V1)


but from logarithmic properties

ln (u/v) = ln u - ln v

ln (uv) = ln u + ln v

ln u^n = n ln u



finally getting


W = RT * ln (V2/V1)  ---> PV work for constant temperature process



where:

W = Work done by compression or expansion(- if done on the surroundings)

F = Force

d = distance

p = absolute pressure

A = area

V = volume

dV = change in volume

T = absolute temperature

V2 = volume at point2

V1 = volume at point1

R = gas constant

S(V1 V2) = integral from V1 to V2

CHEMICAL ENGINEERING: Chemistry - Molecular weight, molar mass, molecular mass, moles, Avogadro's number, molecules

Molecular Weight (also called molecular mass or molar mass)

- is the sum of the atomic weights of all the atoms in a molecule.

- it has a unit of amu. One atomic mass unit (1 amu) is 1/12 the mass of the carbon-12 isotope, which is assigned the value 12.



water: H2O

1 molecule of water H2O = 2 atoms of hydrogen + 1 atom of oxygen

components:

2 H * 1 amu = 2 amu
1 O * 16 amu = 16 amu

MW of water = sum of components

MW of water = 2 amu + 16 amu

MW of water = 18 amu



dry air:

MW of air = 29 amu



Moles

A mole (mol) of any substance is the amount of that substance that contains Avogadro's number of atoms or molecules.  Avogadro's number is defined as the number of carbon atoms in 12 g of 12C.  It has a value of 6.022 x10^23 molecules/mol.

A mole of a substance is Avogadro's number of that substance.


n = N/Na

n = m/MW

m = n * MW


where:

n = number of moles

m = mass of substance

MW = molecular weight of substance

N = number of molecules

Na = Avogadro's Number, 6.022 x10^23 molecules/mol




Avogadro's Number

- the number of atoms needed such that the number of grams of a substance equals the atomic mass of the substance, 6.022 x10^23 /mol

- an Avogadro's number of substance is called a mole.

- for example, a mole of carbon-12 atoms is 12 grams, a mole of hydrogen atoms is 1 gram, a mole of hydrogen molecules is 2 grams




Units of molar mass --> g/mol

The most common unit of molar mass is g/mol because in that unit the numerical value equals the average molecular mass in units of u.



1 mole of Water H2O


average atomic mass of Hydrogen =  1 u

average atomic mass of Oxygen = 16 u

molecular mass of water = (2 * 1 u) + 16 u = 18 u

Thus,

1 mole of water has a mass of 18 grams.




Problem 1:

Ten kg of Carbon dioxide has a volume of 998 L at a pressure of 200 kPa. Determine the number of moles of CO2 present. 

find:

n = number of moles of CO2


given:

m = 10 kg = 10,000 g


solution:

Mw of CO2 = 12 + 2*16

MW of CO2 = 12 + 32

MW of CO2 = 44 g/mol


n = m/MW

n = 10,000 g/44 g/mol

n = 227.3 mol




Problem 2:

How many molecules are present in 5 moles of H2O?


find:

N = number of molecules


given:

n = 5 moles


solution:

n = N/Na

5 = N/6.022 x10^23

N = 5 * 6.022 x10^23

N = 30 x10^23

N = 3.0 x10^24 molecules

MECHANICAL ENGINEERING: Thermodynamics - Closed system, piston-cylinder, Boyle's law, Constant Temperature, Parabolic curve

CLOSED SYSTEM: piston-cylinder

conditions:

Temperature is constant, T = c


Boyle's law:  Absolute pressure is inversely proportional to the volume of a gas

pV = constant

pV = k

p1V1 = p2V2


curve: Parabolic


where:

p = pressure of the gas. p1, p2 are the pressures of the gas at points 1, 2

V = Volume of the gas. V1, V2 are the volumes of the gas at points 1, 2

k = constant


1. A certain gas in a piston is compressed to a volume of 12 L. Its original volume was 21 L at 10 bar. Find the new pressure if the temperature remains constant.

find:

p2 = pressure corresponding to the compressed volume


given:

V2 = 12 L

V1 = 21 L

p1 = 10 bar


solution:

p1V1 = p2V2

10 * 21 = p2 * 12

p2 = 210/12

p2 = 17.5 bar

MECHANICAL ENGINEERING: Thermodynamics - Atmospheric, Barometric, Gage, Absolute, and Vacuum pressure

Atmospheric pressure --> Patm

= 14.7 psi

= 101.325 kPa

= 760 mmHg

= 760 torr

= 29.92 inHg

= 1013.25 millibars

= 1.01325 bar


When gage pressure is above atmospheric:


Pabs = Patm + Pgage


When gage pressure is below atmospheric (vacuum):


Pabs = Patm - Pgage


where:

Pabs = absolute pressure

Pgage =  gage pressure

Patm = atmospheric pressure


1. An oxygen cylinder has a pressure of 2400 psig at 70 F. What is the absolute pressure of the oxygen inside?

find:

Pabs = absolute pressure of oxygen inside the cylinder


given:

Pgage = 2400 psig


solution:

gage pressure is above atmospheric:

Pabs = Patm + Pgage

Pabs = 2400 + 14.7

Pabs = 2414.7 psia



2. The gauge at the vacuum pump reads 7 in. of Hg. What is the absolute pressure?

find:

Pabs = absolute pressure at the vacuum pump


given:

Pgage = 7 inHg


solution:

gage pressure is below atmospheric (vacuum):

Pabs = Patm - Pgage

Pabs = 29.92 - 7

Pabs = 22.92 inHg

MECHANICAL ENGINEERING: Pump work - work done by a pump in lifting a fluid

Wp = F * h


where:

Wp = pump work, the work done by the pump

F = weight of fluid

h = lifting height, measured from the centroid


1. Find the work needed to pump all the water in an 8-ft radius hemispherical tank to a height of 10 ft above the top of the tank.


find:

Wp = pump work


given:

Hemispherical tank (Centroid: y = 3/8 r)

D = Water Density = 62.4 lb/cu.ft

r = 8 ft

h1 = 10 ft


solution:

V = 2/3 pi * r^3

V = 2/3 * 3.14 * 8^3

V = 1072 cu.ft


F = D * V

F = 62.4 * 1072

F = 66,893 lb


h = h1 + yCentroid

h = 10 + 3/8 r

h = 10 + 3/8 (8)

h = 10 + 3

h = 13 ft


Wp = F * h

Wp = 66,893 * 13

Wp = 869,609 lb.ft

Thermodynamics - Specific heats at constant pressure, volume, Specific heat ratio, Gas constants

Specific heat

The ratio of the amount of heat required to raise the temperature of a unit mass of a substance by one unit of temperature to the amount of heat required to raise the temperature of a similar mass of a reference material, usually water, by the same amount.


Specific heat ratio

The specific heat ratio of a gas is the ratio of the specific heat at constant pressure, Cp, to the specific heat at constant volume, Cv. It is sometimes referred to as the adiabatic index or the heat capacity ratio or the isentropic expansion factor or the adiabatic exponent or the isentropic exponent.

For an ideal gas, the heat capacity is constant with temperature.


k = Enthalpy/Internal energy

k = H/U


Enthalpy

H = Cp * T


Internal energy

U = Cv * T


Specific heat ratio (k)

k = H/U

k = Cp * T/CV * T

k = Cp/Cv


Specific heat and Gas constant R

Cp = Cv + R

R = Cp - Cv



---Derivation---

Heating a gas at constant pressure increases the internal energy of the gas and Work is done, whereas supplying the same amount of heat at constant volume only increases the internal energy, no work is done.



constant pressure process:

du = dq - w

dq = du + w

w = pdV

dq = du + pdV


from ideal gas relations

pV = mRT


but at constant pressure

pdV = mRdT


thus

dq = du + mRdT ---> equation1


and

dq = m * cp * dT


equation1 becomes

m * cp * dT = du + mRdT ---> equation2





constant volume process:

du = dq - w


at constant volume, w = 0

w = pdV

w = p(v2 - v1)

but v2 = v1

w = p(0)

w = 0


du = dq + 0

du = dq

dq = m * Cv * dT

du = m * Cv * dT ---> equation3


equation3 in equation2

m * cp * dT = du + mRdT

m * cp * dT = (m * Cv * dT) + mRdT


factoring

(m * dT) (Cp) = (m * dT) (Cv + R)


(m * dT) cancels and thus leaving

Cp = Cv + R

MECHANICAL ENGINEERING: Springs - connected in series, parallel

Hooke's law on springs:
Stretch is directly proportional to the Force

F = k S

where:

F = force applied

k = spring constant

S = stretch of spring




Springs connected in series


|
|               
|  
|-www-s1----www-s2----www-s3------> F
|               
|                
|                


S = s1 + s2 + s3



where:

S = total stretch of spring, equal to the sum of individual stretches of each spring s1, s2, s3

F = force applied 




Springs connected in parallel


|
|               
|   
|---wwww-k1----f1-|
|---wwww-k2----f2-|---> F
|---wwww-k3----f3-|
|               
|                
|                


if the springs are connected by a rigid bar, then

S = s1 = s2 = s3



sum of forces along x = 0; then 

F = f1 + f2 + f3


f1 = k1 * s1


where:

k1, k2, k3 = spring constants

S = total stretch of spring, equal to the sum of invidual stretches of each spring s1, s2, s3

F = force applied, equal to the individual forces on each springs