How to Calculate and Solve for Volume of Superheated Steam | Enthalpy

The volume of superheated steam is illustrated by the image above.

To compute for volume of superheated steam, three essential parameters are needed and these parameters are Specific Volume of Dry Steam (ug), Superheated Temprature (tsup) and Saturated Temprature (ts).

The formula for calculating the volume of superheated steam:

vsup = ugTsup / Ts

Where:

vsup = Volume of Superheated Steam
ug = Specific Volume of Dry Steam
Tsup = Superheated Temprature
Ts = Saturated Temprature

Let’s solve an example;
Find the volume of superheated steam when the specific volume of dry steam is 32, the superheated temperature is 24 and the saturated temperature is 14.

This implies that;

ug = Specific Volume of Dry Steam = 32
Tsup = Superheated Temprature = 24
Ts = Saturated Temprature = 14

vsup = ugTsup / Ts
vsup = 32 x 24 / 14
vsup = 768 / 14
vsup = 54.85

Therefore, the volume of superheated steam is 54.85 m3.

Calculating the Specific Volume of Dry Steam when the Volume of Superheated Steam, the Superheated Temperature and the Saturated Temperature is Given.

ug = vsup x Ts / Tsup

Where:

ug = Specific Volume of Dry Steam
vsup = Volume of Superheated Steam
Tsup = Superheated Temprature
Ts = Saturated Temprature

Let’s solve an example;
Find the specific volume of dry steam when the volume of superheated steam is 10, the superheated temperature is 4 and the saturated temperature is 8.

This implies that;

vsup = Volume of Superheated Steam = 10
Tsup = Superheated Temprature = 4
Ts = Saturated Temprature = 8

ug = vsup x Ts / Tsup
ug = 10 x 8 / 4
ug = 80 / 4
ug = 20

Therefore, the specific volume of dry steam is 20 m3.

Continue reading How to Calculate and Solve for Volume of Superheated Steam | Enthalpy

How to Calculate and Solve for Volume of Wet and Dry Steam | Enthalpy

The volume of wet and dry steam is illustrated by the image above.

To compute for the volume of wet and dry steam, three essential parameters are needed and these parameters are Dryness Fraction (x), Specific Volume of Dry Steam (ug) and Specific Volume of Water (uf).

The formula for calculating volume of wet and dry steam:

v = xug + (1 – x)uf

Where:

v = Volume of Wet and Dry Steam
x = Dryness Fraction
ug = Specific Volume of Dry Steam
uf = Specific Volume of Water

Let’s solve an example;
Given that the dryness fraction is 22, the specific volume of dry steam is 12 and the specific volume of water is 8. Find the volume of wet and dry steam?

This implies that;

x = Dryness Fraction = 22
ug = Specific Volume of Dry Steam = 12
uf = Specific Volume of Water = 8

v = xug + (1 – x)uf
v = 22 x 12 + (1 – 22) x 8
v = 22 x 12 + (-21) x 8
v = 264 + -168
v = 96

Therefore, the volume of wet and dry steam is 96 m3.

Continue reading How to Calculate and Solve for Volume of Wet and Dry Steam | Enthalpy

How to Calculate and Solve for Enthalpy of Superheated Steam | Enthalpy

The image above represents enthalpy of superheated steam.

To compute for enthalpy of superheated steam, five essential parameters are needed and these parameters are Constant Pressure (Cps), Enthalpy of Water (hf), Enthalpy of Vapourization (hfg), Superheated Temprature (Tsup) and Saturated Temprature (Ts).

The formula for calculating enthalpy of superheated steam:

hsup = hf + hfg + Cps(Tsup – Ts)

Where:

hsup = Enthalpy of Superheated Steam
Cps = Constant Pressure
hf = Enthalpy of Water
hfg = Enthalpy of Vapourization
Tsup = Superheated Temprature
Ts = Saturated Temprature

Let’s solve an example;
Given that the constant pressure is 2, the enthalpy of water is 4, the enthalpy of vapourization is 6, the superheated temperature is 8 and the saturated temperatue is 10. Find the enthalpy of superheated steam?

This implies that;

Cps = Constant Pressure = 2
hf = Enthalpy of Water = 4
hfg = Enthalpy of Vapourization = 6
Tsup = Superheated Temprature = 8
Ts = Saturated Temprature = 10

hsup = hf + hfg + Cps(Tsup – Ts)
hsup = 4 + 6 + 2 x (8 – 10)
hsup = 10 + 2 x (-2)
hsup = 10 + (-4)
hsup = 6

Therefore, the enthalpy of superheated steam is 6 J/Kg.

Calculating the Constant Pressure when the Enthalpy of Superheatred Steam, the Enthalpy of Water, the Enthalpy of Vapourization, the Superheated Temperature and the Saturated Temperature is Given.

Cps = hsup – hf – hfg / Tsup + Ts

Where:

Cps = Constant Pressure
hsup = Enthalpy of Superheated Steam
hf = Enthalpy of Water
hfg = Enthalpy of Vapourization
Tsup = Superheated Temprature
Ts = Saturated Temprature

Let’s solve an example;
Find the Constant Pressure when the enthalpy of superheated steam is 30, the enthalpy of water is 10, the enthalpy of vapourization is 4, the superheated temperature is 8 and the saturated temperature is 2.

This implies that;

hsup = Enthalpy of Superheated Steam = 30
hf = Enthalpy of Water = 10
hfg = Enthalpy of Vapourization = 4
Tsup = Superheated Temprature = 8
Ts = Saturated Temprature = 2

Cps = hsup – hf – hfg / Tsup + Ts
Cps = 30 – 10 – 4 / 8 + 2
Cps = 16 / 10
Cps = 1.6

Therefore, the constant pressure is 1.6.

Continue reading How to Calculate and Solve for Enthalpy of Superheated Steam | Enthalpy

How to Calculate and Solve for Enthalpy of Wet Stream | Enthalpy

The image above represents enthalpy for wet steam.

To compute for enthalpy for wet steam, three essential parameters are needed and these parameters are Dryness Fraction (x), Enthalpy of Water (hf) and Enthalpy of Vapourization (hfg).

The formula for calculating enthalpy for wet steam:

h = hf + xhfg

Where:

h = Enthalpy of Wet Steam
x = Dryness Fraction
hf = Enthalpy of Water
hfg = Enthalpy of Vapourization

Let’s solve an example;
Find the enthalpy of wet steam when the dryness fraction is 4, the enthalpy of water is 6 and the enthalpy of vapourization is 8.

This implies that;

x = Dryness Fraction = 4
hf = Enthalpy of Water = 6
hfg = Enthalpy of Vapourization = 8

h = hf + xhfg
h = 6 + 4 x 8
h = 6 + 32
h = 38

Therefore, the Enthalpy of Wet Steam is 38 J/Kg.

Calculating the Dryness Fraction when the Enthalpy of Wet Steam, the Enthalpy of Water and the Enthalpy of Vapourization is Given.

x = h – hf / hfg

Where;

x = Dryness Fraction
h = Enthalpy of Wet Steam
hf = Enthalpy of Water
hfg = Enthalpy of Vapourization

Let’s solve an example;
Find the dryness fraction when the enthalpy of wet steam is 14, the enthalpy of water is 8 and the enthalpy of vapourization is 2.

This implies that;

h = Enthalpy of Wet Steam = 14
hf = Enthalpy of Water = 8
hfg = Enthalpy of Vapourization = 2

x = h – hf / hfg
x = 14 – 8 / 2
x = 6 / 2
x = 3

Therefore, the dryness fraction is 3.

Continue reading How to Calculate and Solve for Enthalpy of Wet Stream | Enthalpy

How to Calculate and Solve for Dryness Fraction | Enthalpy

The image above represents the dryness fraction.

To compute for dryness fraction, two essential parameters are needed and these parameters are Mass of Dry Steam Contained in Steam Considered (ms) and Weight of Water Particles in Suspension in Steam Considered (mw).

The formula for calculating dryness fraction:

x = ms ms + mw

Where:

x = Dryness Fraction
ms = Mass of Dry Steam Contained in Steam Considered
mw = Weight of Water Particles in Suspension in Steam Considered

Let’s solve an example;
Find the dryness fraction when the mass of dry steam contained in steam considered is 21 and the weigth of water particles in suspension in steam considered is 10.

This implies that;

ms = Mass of Dry Steam Contained in Steam Considered = 21
mw = Weight of Water Particles in Suspension in Steam Considered = 10

x = msms + mw
x = 21 / 21 + 10
x = 21 / 31
x = 0.677

Therefore, the dryness fraction is 0.677.

Continue reading How to Calculate and Solve for Dryness Fraction | Enthalpy

How to Calculate and Solve for Crack Opening Displacement | Polymer Deformation

The image above represents crack opening displacement.

To compute for crack opening displacement, two essential parameters are needed and these parameters are Shear Modulus (G) and Yield Strength (σys).

The formula for calculating crack opening displacement:

d = G / 2σys

Where:

d = Crack Opening Displacement (G)
G = Shear Modulus
σys = Yield Strength

Let’s solve an example;
Find the crack opening displacement when the shear modulus is 12 and the yield strength is 4.

This implies that;

G = Shear Modulus = 12
σys = Yield Strength = 4

d = G / 2σys
d = 12 / 2(4)
d = 12 / 8
d = 1.5

Therefore, the crack opening displacement is 1.5 m.

Calculating the Shear Modulus when the Crack Opening Displacement and the Yield Strength is Given.

G = d x σys

Where:

G = Shear Modulus
d = Crack Opening Displacement (G)
σys = Yield Strength

Let’s solve an example;
Find the shear modulus when the crack opening displacement is 10 and the yield strength is 4.

This implies that;

d = Crack Opening Displacement (G) = 10
σys = Yield Strength = 4

G = d x σys
G = 10 x 4
G = 40

Therefore, the shear modulus is 40.

Continue reading How to Calculate and Solve for Crack Opening Displacement | Polymer Deformation

How to Calculate and Solve for Fracture Toughness on Knoop Indentor | Polymer Deformation

The image above represents fracture toughness on knoop indentor.

To compute for fracture toughness on knoop indentor, five essential parameters are needed and these parameters are Indentor Constant (d), Elastic Modulus, (E), Indentation Hardness Value (H), Load (P) and a.

The formula for calculating fracture toughness on knoop indentor:

Kc = d(E)(P) / H(a)3/2

Where:

Kc = Fracture Toughness on Knoop Indentor
d = Indentor Constant
E = Elastic Modulus
H = Indentation Hardness Value
P = Load
a

Let’s solve an example;
Find the fracture toughness on knoop indentor when the indentor constant is 2, the elastic modulus is 4, the indentation hardness value is 6, the load is 8 and a is 3.

This implies that;

d = Indentor Constant = 2
E = Elastic Modulus = 4
H = Indentation Hardness Value = 6
P = Load = 8
a = 3

Kc = d(E)(P) / H(a)3/2
Kc = (2)(4)(8) / (6)(3)3/2
Kc = 64 / (6)(5.196)
Kc = 64 / 31.176
Kc = 2.05

Therefore, the fracture toughness on knoop indentor is 2.05 J/m³.

Continue reading How to Calculate and Solve for Fracture Toughness on Knoop Indentor | Polymer Deformation

How to Calculate and Solve for Crack Opening Displacement | Polymer Deformation

The image above represents crack opening displacement.

To compute for crack opening displacement, three essential parameters are needed and these parameters are Fracture Toughness of Material (K), Elastic Modulus (E) and Yield Strength (σys).

The formula for calculating crack opening displacement:

d = / 2Eσys

Where:

d = Crack Opening Displacement (E)
K = Fracture Toughness of Material
E = Elastic Modulus
σys = Yield Strength

Let’s solve an example;
Find the crack opening displacement when the fracture toughness of material is 21, the elastic modulus is 11 and the yield strength is 7.

This implies that;

K = Fracture Toughness of Material = 21
E = Elastic Modulus = 11
σys = Yield Strength = 7

d = / 2Eσys
d = (21)² / 2(11)(7)
d = (441) / (154)
d = 2.86

Therefore, the crack opening displacement is 2.86 m.

Continue reading How to Calculate and Solve for Crack Opening Displacement | Polymer Deformation

How to Calculate and Solve for Rod Toughness | Polymer Deformation

The image above represents rod toughness.

To compute for rod toughness, three essential parameters are needed and these parameters are Short Rod Calibration Constant (A), Maximum Load (Pc) and Short Rod Diameter (B).

The formula for calculating rod toughness:

KIC = APcB-3/2

Where:

KIC = Rod Toughness
A = Short Rod Calibration Constant
Pc = Maximum Load
B = Short Rod Diameter

Let’s solve an example;
Find the rod toughness when the short rod calibration constant is 12, the maximum load is 14 and the short rod diameter is 18.

This implies that;

A = Short Rod Calibration Constant = 12
Pc = Maximum Load = 14
B = Short Rod Diameter = 18

KIC = APcB-3/2
KIC = (12)(14)(18)-3/2
KIC = (12)(14)(0.013)
KIC = 2.199

Therefore, the rod toughness is 2.199 J/m³.

Continue reading How to Calculate and Solve for Rod Toughness | Polymer Deformation

How to Calculate and Solve for Apparent Length of Profile | Polymer Deformation

The image above represents apparent length of profile.

To compute for apparent length of profile, three essential parameters are needed and these parameters are Constant for Dimension of Length (Lo), Measurement Unit (η) and Fractal Dimension (D).

The formula for calculating apparent length of profile:

L(η) = Loη-(D – 1)

Where:

L(η) = Apparent Length of Profile
Lo = Constant for Dimension of Length
η = Measurement Unit
D = Fractal Dimension

Let’s solve an example;
Find the apparent length of profile when the constant for dimension of length is 10, the measurement unit is 2 and the fractal dimension is 4.

This implies that;

Lo = Constant for Dimension of Length =  10
η = Measurement Unit = 2
D = Fractal Dimension = 4

L(η) = Loη-(D – 1)
L(η) = 10(2)-(4 – 1)
L(η) = 10(2)-(3)
L(η) = 10(2)-3
L(η) = 10(0.125)
L(η) = 1.25

Therefore, the apparent length of profile is 1.25 m.

Continue reading How to Calculate and Solve for Apparent Length of Profile | Polymer Deformation