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Kruka [31]
3 years ago
9

You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210

centipoise [cP]. What is the kinematic viscosity of Fluid A, in units of stokes [St]
Engineering
1 answer:
Naily [24]3 years ago
4 0

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

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g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0
Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵  * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α  * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0
2 years ago
A wire is 0.92 m long and 1.2 mm2 in cross-sectional area. It carries a current of 5.0 A when a 2.2 V potential difference is ap
Flura [38]

The solution is in the attachment

3 0
3 years ago
Which of the following uses pressure and flow to transmit power from one location to another?
lord [1]

Answer:

fluid power

Explanation:

fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases

5 0
2 years ago
Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running
Nikolay [14]

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1   2   3   4   5   6   7   8   9   10   11   12   13   14   15   16   17   18   19   20   21   22   23   24   25   26   27   28   29    30

A  B  C   D  E  A  B   C* D  E    A    B    D   E    A   B   D*   E     A   B     E     A    B* E   A     E  A    E*  A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6       7-14        15-24      25-26        27-30

   

 B           E             A             C            D

     AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10      11-16      17-18      19-22      23-30

 

   A            B              C            D              E

   

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2        3-6         7-12         13-20      21-30

C           D            B               E                A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

3 0
3 years ago
A company has a stack that emits a hazardous air pollutant. The ground mass concentration directly downwind of the plume sometim
fredd [130]

Answer:

do the wam wam

Explanation:

3 0
3 years ago
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