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vodomira [7]
4 years ago
11

2. A fluid at 14.7 psi (lb-f per square inch) with kinematic viscosity (????????) 1.8 x10-4 ft2/sec and density(????????) 0.076

lb/ft3 enters a 10 inch diameter pipe with a uniform velocity and a Reynolds number 1000. Determine the decrease in pressure going from the entrance to 100 inch downstream the entrance. The entrance length, LLee is given by, LLee = 0.0288DD. RRRRDD. (Hint: calculate the pressure drop separately between 1 and 2 and between 2 and 3 because the region 1-2 shows developing flow and region 2-3 shows developed flow) region. The flow becomes fully developed after the entrance length, LLee. The thickness of the boundary layer 1 −� given as ????????(xx) = 5.0xxRRRR 2. Show that entrance length for this flow can be expressed LL = 0.01DD. RRRR . xxeeDD

Engineering
1 answer:
jenyasd209 [6]4 years ago
4 0

Answer:

See explaination

Explanation:

We are going to define Pressure drop as the difference in total pressure between two points of a fluid carrying network. A pressure drop usually occurs when frictional forces, caused by the resistance to flow, act on a fluid as it flows through the tube.

See attachment for the step by step solution of the given problem.

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"Carbon 14 (C-14), a radioactive isotope of carbon, has a half-life of 5730 ± 40 years. Measuring the amount of this isotope lef
igor_vitrenko [27]

Answer:

The age of the bones is approximately 14172 years.

Explanation:

The age of the bones can be determinated using the following decay equation:

N_{(t)} = N_{0}e^{-\lambda t}   (1)

<u>Where:</u>

N(t): is the quantity of C-14 at time t

No: is the initial quantity of C-14  

λ: is the decay rate      

t: is the time

First, we need to find λ:

\lambda = \frac{ln(2)}{t_{1/2}}

<u>Where:</u>

t(1/2): is the half-life of C-14 = 5730 y

\lambda = \frac{ln(2)}{5730 y} = 1.21 \cdot 10^{-04} y^{-1}

Now, we can calculate the age of the bones by solving equation (1) for t:

t = \frac{-ln(\frac{N_{(t)}}{N_{0}})}{\lambda}

We know that the bones have lost 82% of the C-14 they originally contained, so:

N_{t} = (1 - 0.82)N_{0} = 0.18N_{0}

t = \frac{-ln(0.18)}{1.21 \cdot 10^{-04} y^{-1}}

t = 14172 y

Therefore, the age of the bones is approximately 14172 years.

I hope it helps you!

3 0
3 years ago
Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per
Tomtit [17]

Answer:

Explanation :

The given  information to be listed can are Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per Gallon.

Check the attached document for the solution.

5 0
3 years ago
It is the tool used to measure the amount of electric current​
sergeinik [125]

Answer:

Ammeter

Explanation:

Instrument for measuring either direct or alternating electric current, in amperes. Ammeters vary in their operating principles and accuracies

7 0
3 years ago
Read 2 more answers
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
4 years ago
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
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