5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h
1 answer:
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Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ =
/ t₂
t₂ = t₁
t₂ = t₁ /
t₂ = ( /
)t₁
t₂ =
/
× t₁
so we substitute
t₂ = 0.0049 / 0.0018
× 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Assumptions:
- Steady state.
- Air as working fluid.
- Ideal gas.
- Reversible process.
- Ideal Otto Cycle.
Explanation:
Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):
- Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
- Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).
- Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
- Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.
- Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
- Exhaust 1-0: the working fluid is vented to the atmosphere.
If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:
where:
Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.
Answer:
See image attached.
Answer:
a) the amount of energy produced in kJ/K is 0.73145 kJ/K
b) the amount of energy produced in kJ/K is 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Explanation:
Draw the T-s diagram.
a)
= 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280
R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar
Δs =
Substitute all parameters in the equation
Δs =
Δs = 5 kg × 0.14629 kJ/kg.K
= 0.73145 kJ/K
b)
Δs =
Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K
T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K
R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar
Δs =
= 5 kg (0.13795 kJ/kg.K)
= 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.