Tech A says that LED brake lights illuminate faster than incandescent bulbs. Tech B says that LED brake lights have

Optimum engine oil pressure at operating temperature and moderate engine load should be __________ ps

kobusy [5.1K]

Answer:

What Is Normal Oil Pressure Temperature? Oil temperatures typically range from 200 degrees to 200 degrees Fahrenheit, depending on the weather. The oil pressure ranges from 100psi to 10-15psi depending on the weather at start up.

Explanation:

3 0

2 years ago

Is a water proof material used around tubs and

Leviafan [203]

I think it’s hard board

4 0

3 years ago

Read 2 more answers

The products of combustion from burner are routed to an industrial application through a thin-walled metallic duct of diameter D

slamgirl [31]

Answer:

Start by calculating the heat lost falling from Tm,i to Tm,o

As a first approximation use an average Cp of (Tm,i + Tm,o)/2 and an average density at (Tm,i + Tm,o)/2

Cp air at (Tm,i + Tm,o)/2 ----(1)

Density of air at (Tm,i + Tm,o)/2 : PV=nRT, V=T1/T2, air density at 293K = 1.204kg/m^3

Air density at (Tm,i + Tm,o)/2 1.204×293/(Tm,i + Tm,o)/2

Energy lost per kg for (Tm,i - Tm,o)K drop: (Tm,i - Tm,o)K × (1) =

Time of travel: t

Energy lost per kg drop/t seconds = 24.3 kW

Volume occupied by 1kg air at at mean tempera ture = 1/Air density at mean temperature

Length of pipe ( Di m diameter) needed to hold Volume occupied by 1 kg of air at mean temperature :

Cross section = π/4 Di^2, Volume occupied by 1 Kg of air at mean temp. ÷ π/4Di^2

Surface area of pipe Di m diameter by L m long = Length of pipe to hold Volume of air in m × π*Di

Q/A=k (delta T)/ thickness,

Thickness of insulation = Area × k ×dT / Q

Explanation:

6 0

3 years ago

The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in t

vovikov84 [41]

Answer:

Explanation:

$r_2=$ ∞

$q=4\pi kT_1(T_2-T_1)\\$

$q=2\pi kD.$ ΔT--------(1)

$q=hA$ ΔT $=4\pi r_1^2(T_2_s-T_1_s)\\$

$N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)$

By equation (1) and (2)

$2\pi kD.$ ΔT=h.4 $\pi r_1^2$ ΔT

$2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\$ -------(4)

From equation (3) and (4)

So for sphere $R_a$ →0

6 0

3 years ago

For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:

lana [24]

Answer:

a) The planar defect that exists is twin boundary defect.

b) The planar defect that exists is the stacking fault.

Explanation:

I am using bold and underline instead of a vertical line.

a. A B C A B C B A C B A

In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.

b. A B C A B C B C A B C

In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.

6 0

4 years ago