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WITCHER [35]
2 years ago
14

La Patrulla Fronteriza de los Estados Unidos analiza la compra de un helicóptero nuevo para la vigilancia aérea de la frontera d

e Nuevo México y Texas con la República Mexicana. Hace 4 años se adquirió un helicóptero similar con un costo de $ 140,000.00. con una tasa de interés de 7% anual. Calcula el factor de pago único y el factor valor presente con los datos anteriores con la tabla y fórmula. Dibuja el diagrama de flujo.
Engineering
1 answer:
patriot [66]2 years ago
7 0

Answer:

The Border Patrol of the United States analyzes the purchase of a new helicopter for the aerial surveillance of the border of New Mexico and Texas with the Mexican Republic. 4 years ago a similar helicopter was purchased at a cost of $ 140,000.00. with an interest rate of 7% per year. Calculate the single payment factor and the present value factor with the above data with the table and formula. Draw the flow chart.

Explanation:

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Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl
Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio \frac{v_1}{v_2}  = 10

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

\frac{p_1}{p_2}  = (\frac{v_2}{v_1} )^n Therefore p₂ = p₁ ÷ (\frac{v_2}{v_1} )^n = (1 bar) ÷ (\frac{1}{10} )^{1.3} = 19.953 bar

Similarly for a polytropic process we have

\frac{T_1}{T_2}  = (\frac{v_2}{v_1} )^{n-1} or T₂ = T₁ ÷ (\frac{v_2}{v_1} )^{n-1} = \frac{310}{0.1^{0.3}} = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = \frac{8.3145}{28.9628} = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R =  1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = \frac{R(T_2-T_1)}{n-1} = \frac{0.287(618.531-310)}{1.3 - 1}= 295.16 kJ/kg

Q =(\frac{n-\gamma}{\gamma - 1} )W = (\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = \frac{R(T_4-T_3)}{n-1} = Where T₄ is given by  \frac{T_4}{T_3}  = (\frac{v_3}{v_4} )^{n-1} or T₄ = T₃ ×0.1^{0.3}

= 2200 ×0.1^{0.3}  T₄ = 1102.611 K

W =  \frac{0.287(1102.611-2200)}{1.3 - 1}= -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

\eta = 1-\frac{T_4-T_1}{T_3-T_2} =1-\frac{1102.611-310}{2200-618.531} = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

p_{m}  = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}  where r = compression ratio and r_p = \frac{p_3}{p_2}

However p₃ = \frac{p_2T_3}{T_2} =\frac{(19.953)(2200)}{618.531} =70.97 atm

r_p = \frac{p_3}{p_2} = \frac{70.97}{19.953}  = 3.56

Therefore p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9} = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0
3 years ago
Along with refining craft skills another way to increase the odds for career advancement is to
Xelga [282]

The acquisition of additional certifications with a personal refined craft skills can increase the odds for career advancemen.

<h3>What is a career advancement?</h3>

An advancement is achieved in a career if a professional use their skill sets, determination or perserverance to achieve new career height.

An example of a career advancement is when an employee progresses from entry-level position to management and transits from an occupation to another.

Therefore, the Option A is correct.

Read more about career advancement

<em>brainly.com/question/7053706</em>

7 0
2 years ago
You work for a printing company, and you realize that your colleague sent incorrect price quotes to a client. You begin to write
xxTIMURxx [149]

Answer:

The sentence excerpted from the e-mail uses passive voice.  

Given the  purpose of your message, this voice is appropriate.

Explanation:

Because the objective is to remedy the situation a passive voice is great because it emphasizes the action and the object instead of the subject.

We want to emphasize the document and the incorrect information, not our colleague.

4 0
3 years ago
Given asphalt content test data:
VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 0
3 years ago
A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c
Eduardwww [97]

Answer:

h=1.99998\ W/m^2.C

k=33.333\ W/m.C

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0       Eq (1)

Where:

k is thermal Conductivity

\dot e_{gen} is uniform thermal generation

T(x) = a(L^2-x^2)+b

\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a

Putt in Eq (1):

-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C

Energy balance is given by:

Q_{convection}=Q_{conduction}

h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L     Eq  (2)

T(x) = a(L^2-x^2)+b

Putting x=L

T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC

\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2

From Eq (2)

h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C

7 0
3 years ago
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