The pressure difference across the sensor housing will be "95 kPa".
According to the question, the values are:
Altitude,
Speed,
Pressure,
The temperature will be:
→ ![T = 15.04-[0.00649(9874)]](https://tex.z-dn.net/?f=T%20%3D%2015.04-%5B0.00649%289874%29%5D)
→ 
→ 
now,
→ ![P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}](https://tex.z-dn.net/?f=P_o%20%3D%20101.29%5B%5Cfrac%7B%28-49.042%2B273.1%29%7D%7B288.08%7D%20%5D%5E%7B%285.256%29%7D)
→
hence,
→ The pressure differential will be:
= 
= 
Thus the above solution is correct.
Learn more about pressure difference here:
brainly.com/question/15732832
Answer:
awnsers should be added to know to show additional
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:
In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.
Answer:
Explanation:
The schedule using shortest remaining time, non-preemptive priority and round Robin with quantum number 30 is shown in the attached file, please kindly go through it to access the answer.