Answer:
19.0476 kN/m^2
Explanation:
Given data:
Dimension of rectangular area = 2.5m x 5.0m
external load = 150 KN/m^2
load depth = 6.25 m
calculate the vertical stress increase due to load at depth 6.25
we will use the approximate method which is
[tex]V_{s} = \frac{qBL}{(B +Z)(L+Z)}[/tex] ------- (1)
q = 150 kN/m^2
B = 2.5 m
L = 5 m
Z = 6.25 m
substitute the given values into the equation (1) above
hence the Vs ( vertical stress ) = 19.0476 kN/m^2
True or false Osha engineering controls are the last strategy of control an employer should use for job hazards
Answer:
false
Explanation:
3 mA are flowing through an 18 V circuit, how much resistance (in kΩ) is in that circuit?
Answer:
6kΩ
Explanation:
If we assume that the entire circuit's current is 3 mA, then we can compute the resistance within the circuit with the application of Ohm's law:
V = IR
V/I = R
R = V / I
R = 18V / 3 mA
R = 6kΩ
Hence, the resistance of the circuit is 6kΩ.
Cheers.
Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:
a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.
Answer:
44.59°c
Explanation:
Given data :
Total pressure = 105 kpa
complete combustion
A) Determine air-fuel ratio
A-F = [tex]\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }[/tex]
N = number of mole
m = molar mass
A-F = [tex]\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}[/tex] = 22.2 kg air/fuel
hence the ratio of Fuel-air = 1 : 22.2
B) Determine the temperature at which water vapor in the products start condensing
First we determine the partial pressure of water vapor before using the steam table to determine the corresponding saturation temp
partial pressure of water vapor
Pv = [tex]\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )[/tex]
N watervapor ( number of mole of water vapor ) = 3
N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol
Pro = 105
hence Pv = ( 3/33.53 ) * 105 = 9.39kPa
from the steam pressure table the corresponding saturation temperature to 9.39kPa = 44.59° c
Temperature at which condensing will start = 44.59°c
An equation showing the products of propylene with their mole numbers is attached below