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choli [55]
2 years ago
7

Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No

Engineering
1 answer:
Degger [83]2 years ago
5 0

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is 1.557\times10^{-7}m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

Re=\frac{vd}{\nu}

Here, v is velocity, \nu is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

Re=\frac{vd}{\nu}

Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

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Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

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2 years ago
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A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou
Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor (f_{p})can be estimated using the equation below:

f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}

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6 0
2 years ago
A steam power plant operates on the reheat Rankine cycle. Steam enters the highpressure turbine at 12.5 MPa and 550°C at a rate
gayaneshka [121]

Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg,  S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = \frac{h3-h4}{h3-h4s} = \frac{3476.5 - h4}{3476.5 - 2948.1}

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg,  s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95  and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

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8 0
2 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

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Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

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The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

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By adding the above equation

2  σ₁ = 530.4

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2 years ago
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katen-ka-za [31]

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8 0
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