Answer:
to filter out any impurities such as metal shavings in the oil
Explanation:
are bhai brainly aap english me questions kar ne ko hai
ye kon si bhasha hai ??????
Answer:
I couldn't find options for your question online, but I can give you an explanation so you can choose the correct option.
Explanation:
A spark knock is a form of unpredictable behavior that occurs in combustion, that is, in the chemical reaction that occurs between oxygen and an oxidizable material. Such combustion is usually manifested by incandescence or flame.
The spark knock is a detonation that occurs when there is a lot of pressure in the fuel.
<u>Some situations in which this can happen are:
</u>
- Engine overloaded.
- Maximum pressure in the cylinders.
- Engine overheated.
- Overheated air.
- Long and excessive engine ignition timing.
- Spark plug at high temperatures.
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
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
