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3 years ago

The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False

Engineering

1 answer:

Alex3 years ago

4 0

Answer:

(B) FALSE

Explanation:

view factor $F_{ij}$ depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by $F_{ij}$

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How many grams is one molecule of calcium oxide?

MakcuM [25]

Answer:

Molar mass -56.0774 g/mol

There is a way to calculate this

atomic number for calcium is 20

atomic number for oxygen is 8

The molar mass for calcium is 40

The molar mass for oxygen is 16

CALCIUM OXIDE

40. +. 16

=56g/mol

1 molecule = 1×56=56g/mol

2 molecules=2×56 =112g/mol

5 0

2 years ago

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.

kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0

2 years ago

A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

S = $\frac{\pi D}{4}$ ....................1

and

ratio of flow rate through the square and circle is here

$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

4 0

4 years ago

A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of th

fenix001 [56]

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\$

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft$

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

6 0

3 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 24 C and exits at 1 ba

Katen [24]

Answer:

$h_{1} = h_2} = 293.45 KJ/kg$ .

The quality of the refrigerant exiting the expansion valve is

$x_{2}=0.193596$ .

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature $T_{1}$ = $24^{o} C$ .

Pressure $P_{1}$ = 10 bar.

Exit 2:-

Pressure $P_{2}$ = 1 bar.

Solution:-

6 0

3 years ago