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Serga [27]
3 years ago
6

A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed

of 320 km/h in air at 15 C. The student decided to build a 1/50 scale model and test it in a water tunnel at the same temperature.
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype

Engineering
1 answer:
yan [13]3 years ago
3 0

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.

See attachment for the remaining.

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Outline the structure of an input-output model (including assumptions about supply and demand). What is an inverse matrix? Why i
pishuonlain [190]

Answer:

Explanation:

C.1 Input-Output Model

It is a formal model that divides the economy into 2 sectors and traces the flow of inter-industry purchases and sales. This model was developed by Wassily Leontief in 1951. In simpler terms, the inter-industry model is a quantitative economic model that defines how the output of one industry becomes the input of another industrial sector. It is an interdependent economic model where the output of one becomes the input of another. For Eg: The Agriculture sector produces output using the inputs from the manufacturing sector.

The 3 main elements are:

Concentrates on an economy which is in equilibrium

Deals with technical aspects of production

Based on empirical investigations and assumptions

Assumptions

2 sectors - " Inter industry sector" and "final sector"

Output of one industry is the input for another

No 2 goods are produced jointly. i.e each industry produces homogenous goods

Prices, factor suppliers and consumer demands are given

No external economies or diseconomies of production

Constant returns to scale

The combinations of inputs are employed in rigidly fixed proportions.

Structure of IO model

See image 1

Quadrant 1: Flow of products which are both produced and consumed in the process of production

Quadrant 2: Final demand for products of each producing industry.

Quadrant 3: Primary inputs to industries (raw materials)

Quadrant 4: Primary inputs to direct consumption (Eg: electricity)

The model can be used in the analysis of the labor market, forecast economic development of a nation and analyze economic developments of various regions.

Leontief inverse matrix shows the output rises in each sector due to a unit increase in final demand. Inverting the matrix is significant since it is a linear system of equations with unique solutions. Thus, the final demand vector for the required output can be found.

C.2 Linear programming problems

Linear programming problems are optimization problems in which objective function and the constraints are all linear. It is most useful in making the best use of scarce resources during complex decision makings.

Primal LP, Dual LP, and Interpretations

Primal linear programming: They can be viewed as a resource allocation model that seeks to maximize revenue under limited resources. Every linear program has associated with it a related linear program called dual program. The original problem in relation to its dual is termed as a primal problem. The objective function is a linear combination of n variables. There are m constraints that place an upper bound on a linear combination of the n variables The goal is to maximize the value of objective functions that are subject to the constraints. If the primal linear programming has finite optimal value, then the dual has finite optimal value, and the primal and dual have the same optimal value. If the optimal solution to the primal problem makes a constraint into a strict inequality, it implies that the corresponding dual variable must be 0. The revenue-maximizing problem is an example of a primal problem.

Dual Linear Programming: They represent the worth per unit of resource. The objective function is a linear combination of m values that are the limits in the m constraints from the primal problem. There are n dual constraints that place a lower bound on a linear combination of m dual variables. The optimal dual solution implies fair prices for associated resources. Stri=ong duality implies the Company’s maximum revenue from selling furniture = Entrepreneur’s minimum cost of purchasing resources, i.e company makes no profit. Cost minimizing problem is an example of dual problems

See image 2

n - economic activities

m - resources

cj - revenue per unit of activity j

4 0
3 years ago
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The thermal efficiency of two reversible power cycles operating between the same thermal reservoirs will a)- depend on the mecha
mestny [16]
C ,, i’m pretty sure .
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3 years ago
The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement
scoundrel [369]

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

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8 0
2 years ago
The separation and purification process is used for which of the following purposes?
Svetach [21]

The separation and purification process is used for DEVELOPING VACCINES. The development of vaccines must follow strict guidelines.

A vaccine is a biological preparation used to protect against harmful diseases (e.g., viruses) in a simple and safe manner.

Vaccine purification is a strict process consisting of clarification, concentration and chromatography in order to separate the extract that contains the active principle of the vaccine.

The techniques used for vaccine filtration include membrane filtration, depth filtration and centrifugation.

Learn more about vaccines here:

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6 0
3 years ago
A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons
antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

5 0
3 years ago
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