Answer:
Net discharge per hour will be 3.5325
Explanation:
We have given internal diameter d = 25 mm
Time = 1 hour = 3600 sec
So radius
We know that area is given by
We know that discharge is given by , here A is area and V is velocity
So
So net discharge in 1 hour =
Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)= gas constant ideal for air
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
Answer:
a) The stress in the bar when F is 32,000 is approximately 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed is approximately 110,000 lb
Explanation:
The question topic relates to stresses in structures;
The given parameters of the steel bar are;
The width of the steel bar, W = 4.0 in.
The thickness of the steel bar, t = 1.125 in.
The formula for stress in a bar is given as follows;
The cross sectional area of bar, A = W × t = 4.0 in. × 1.125 in. = 4.5 in.²
∴ The cross sectional area of bar, A = 4.5 in.²
a) The stress in the bar for F = 32,000 lb, is given as follows;
The stress in the bar when F is 32,000 is σ = 7,111. psi ≈ 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed, σ = 25,000 psi is given as follows;
Therefore;
P = σ × A = 25,000 psi × 4.5 in² = 112,500 lb
For the axial stress of 25,000 psi not to be exceeded, the maximum load that can be supported by the bar, P = 112,500 lb ≈ 110,000 lb.
Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e = =
Specific gravity =
% Saturation S = ×
=
×
water content w = =
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w =
Recall; mass = Density × volume
w =
From eqn. (2) =
∴
putting eqn. (6) into (5)
w =
Again, from eqn (1)
substituting into eqn. (7)
∴
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given,
∴
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil ( )
Given, , S = 50% = 0.5
%S = ×
=
×
0.50 =
mass of water =