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Rama09 [41]
3 years ago
11

Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm have the form: u(r) = 0.15[1 − (r/ro ) 2 ] T(r

) = 314.8 + 75.0(r/ro ) 2 − 18.8(r/ro ) 4 with units of m/s and K, respectively. Determine the corresponding value of the mean (or bulk) temperature, Tm, at this axial position.

Engineering
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

Tm = 366.66k

Explanation:

check for the step by step explanation in the attachment

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A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above
adoni [48]

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = \frac{dV}{dt}

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,  

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

4 0
3 years ago
A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50
anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure  = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r =  120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0
3 years ago
Three things that i have learned about the topics in this learning activity sheet
AURORKA [14]
Where’s the sheet or the picture ?
3 0
2 years ago
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan
hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

R = \dfrac{u^2}{15(e+f_s)}

where;

R= radius

f_s = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

f_s = 0.20

So;

R = \dfrac{30^2}{15(0.08+0.20)}

R = \dfrac{900}{15(0.28)}

R = \dfrac{900}{4.2}

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0
3 years ago
A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; p
Slav-nsk [51]

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

8 0
3 years ago
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