## How to Calculate and Solve for Expected Discharge | Flood Formulae | Irrigation Water Requirement

The image above represents expected discharge.

To compute for expected discharge, three essential parameters are needed and these parameters are Constant (c), Constant (n) and Drainage Basin Area (Ad).

The formula for calculating expected discharge:

Where:

Qp = Expected Discharge | Flood Formulae
c = Constant
n = Constant

Let’s solve an example;
Find the expected discharge when the constant is 8, the constant is 2 and the drainage basin area is 4.

This implies that;

c = Constant = 8
n = Constant = 2
Ad = Drainage Basin Area = 4

Qp = (8) . (4)2
Qp = (8) . (16)
Qp = 128

Therefore, the expected discharge is 128.

Calculating the Constant when the Expected Discharge, the Constant and the Drainage Basin Area is Given.

Where;

c = Constant
Qp = Expected Discharge | Flood Formulae
n = Constant

Let’s solve an example;
Find the constant when the expected discharge is 30, the constant is 3 and the drainage basin area is 2.

This implies that;

Qp = Expected Discharge | Flood Formulae = 30
n = Constant = 3
Ad = Drainage Basin Area = 2

c = 30 / 23
c = 30 / 8
c = 3.75

Therefore, the constant is 3.75.

## How to Calculate and Solve for Flexural Strength with Relation to Volume Fraction | Ceramics

The image above represents Flexural strength with relation to volume.

To compute for flexural strength with relation to volume, three essential parameters are needed and these parameters are Initial Stress (σo), Constant (n) and Volume fraction porosity (P).

The formula for calculating flexural strength with relation to volume:

σfs = σo exp (-nP)

Where:

σfs = Flexural Strength
σo = Initial Stress
n = Constant
P = Volume Fraction Porosity

Let’s solve an example;
Find the flexural strength when the initial stress is 11, the constant is 8 and the volume fraction porosity is 22.

This implies that;

σo = Initial Stress = 11
n = Constant = 8
P = Volume Fraction Porosity = 22

σfs = σo exp (-nP)
σfr = (11)exp(-(8)(22))
σfr = (11)exp(-176)
σfr = (11)(3.665e-77)
σfr = 4.032e-76

Therefore, the flexural strength is 4.032e-76 Pa.

Calculating the Initial Stress when the Flexural Strength, the Constant and the Volume Fraction Porosity is Given.

σo = σfr / exp (-nP)

Where;

σo = Initial Stress
σfs = Flexural Strength
n = Constant
P = Volume Fraction Porosity

Let’s solve an example;
Find the initial stress when the flexural strength is 20, the constant is 10 and the volume fraction porosity is 8.

This implies that;

σfs = Flexural Strength = 20
n = Constant = 10
P = Volume Fraction Porosity = 8

σo = σfr / exp (-nP)
σo = 20 / exp (-10 x 8)
σo = 20 / exp (-80)
σo = 20 / – 5.54e+34
σo = – 3.61e-34

Therefore, the initial stress is 3.61e-34.

## How to Calculate and Solve for Relative Freezing Time | Foundry Technology

The image above represents relative freezing time.

To compute for relative freezing time, four essential parameters are needed and these parameters are Constant (L), Constant (C), Volume of Riser/Volume of Casting (y) and Relative Contraction of Freezing (B).

The formula for calculating relative freezing time:

x = L / y – B + C

Where:

x = Relative Freezing Time
y = Volume of Riser / Volume of Casting
B = Relative Contraction on Freezing
C = Constant
L = Constant

Let’s solve an example;
Find the relative freezing time when the volume of riser/volume of casting is 26, the relative contraction on freezing is 21, the constant is 8 and the constant is 11.

This implies that;

y = Volume of Riser / Volume of Casting = 26
B = Relative Contraction on Freezing = 21
C = Constant = 11
L = Constant = 8

x = L / y – B + C
x = 8 / 26 – 21 + 11
x = 8 / 5 + 11
x = 1.6 + 11
x = 12.6

Therefore, the relative freezing time is 12.6 s.

Calculating the Constant (L) when the Relative Freezing Time, Constant (C), the Volume of riser/Volume of Casting and the Relative Contraction on Freezing is Given.

L = x (y – B) – C

Where;

L = Constant
x = Relative Freezing Time
y = Volume of riser / Volume of Casting
B = Relative Contraction on Freezing
C = Constant

Let’s solve an example;
Find the Constant when the relative freezing time is 24, the volume of riser / volume of casting is 14, the relative contraction on freezing is 8 and the constant is 10.

This implies that;

x = Relative Freezing Time = 24
y = Volume of riser / Volume of Casting = 14
B = Relative Contraction on Freezing = 8
C = Constant = 10

L = x (y -B) – C
L = 24 (14 – 8) – 10
L = 24 (6) – 10
L = 144 – 10
L = 134

Therefore, the constant is 134.