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Sphinxa [80]
3 years ago
7

A liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3

, and a characteristic velocity of 1600 m/s. The nozzle has the following characteristics: throat area of 0.010 m2 and an expansion ratio of 10. Calculate the following:
1. Thrust coefficient at sea level
2. Specific impulse at sea level
3. Altitude at optimal expansion
4. Thrust coefficient at optimal expansion
5. Mass flux through the throat
Engineering
1 answer:
Cloud [144]3 years ago
8 0

Answer:

  1. 1.55
  2. 260 N.s
  3. 3370 m
  4. 1.6
  5. 43.75 kg/s

Explanation:

1) Thrust coefficient at sea level.

Cfsl = TSL / Pca

TSL = Mp * Vc  + ( Pc - Pa )Ac

Mp = mass flux = 43.75 kg/s

∴ Cfsl  = Mp Vc / Pca  + ( Pc - Pa )/Pc * ( Ac / A* )

           = 1.6 - 0.04923 = 1.55

<u>2) Specific impulse at sea </u>

Isp = Vc / g = 2549.75 / 9.81

                   = 260 N.s

3) Altitude at optimal expansion

H = 3370 m

<u>4) thrust coefficient at optimal expansion </u>

CF = 1.6

attached below is the detailed solution

<u>5) Mass flux through the throat </u>

Mass flux = P1 * At / Cc

                = ( 7*10^6 * 0.01 ) / 1600

                = 43.75 kg/s

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A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°c. determine
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By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

<u>Given the following data:</u>

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

<h3>How to calculate the average heat transfer coefficient.</h3>

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

  • Kinematic viscosity, v = 1.750 \times 10^{-5} m²/s.
  • Thermal conductivity, k = 0.02699 W/mk.
  • Thermal diffusivity, α = 2.416 \times 10^{-5} m²/s.
  • Prandtl number, Pr = 0.7241.

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10}  }\\\\R_{aL}=1.48 \times 10^{7}

Also take note, g(Pr) is given by this equation:

g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }

g(Pr) = 0.430

For GrL, we have:

G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times  [\frac{4}{3} \times  (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

N_{uL}=\frac{h_{L}L}{k} =0.68 +  \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19}   } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 +  \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19}   })\\\\h_{L}=4.87\;W/m^2k

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Answer:

Explanation:

Let us Begin with the first Policy;

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  • Reason/Purpose - The motivation behind the email strategy is to affirm the correct utilization of email address of the organization by the representatives.

  • Scope - The arrangement is covering the utilization of any email which is sent from an organization email address. It additionally applies to all related representatives, clients, and different partners who are working in the interest of the organization.

  • Policy

  1. The email address of the organization ought to be utilized by workers just for business reason. Representatives are restricted to unlawfully utilize the organization email address.
  2. The utilization of the email office must be steady with other security approaches of the organization. It gives the guarantee of the moral conduct inside the organization.
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  6. It isn't permitted to advance email to any outsider utilizing the organization email address.
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  • Policy Compliance - The security group of the organization will confirm the consistence of the approach utilizing suitable strategies, for example, stroll through, inward and outside reviews, business instrument reports, video checking, and input. The security group must need to educate and support any sort of special case to the arrangement ahead of time.

  • Related Standards, Policies, and Processes - Data assurance guidelines
  • Definitions and Terms - None

The Second Policy;

WIFI and Internet usage policy

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  • Purpose - The motivation behind WIFI and Internet use strategy is to affirm the best possible utilization of WIFI and Internet by the representatives inside the organization.

  • Scope - The arrangement applies to each web client including full-time representatives, low maintenance workers, colleagues, and so on of the organization.

  • Policy

  1. Asset utilization – Employees will get consent and endorsement to get to WIFI and Internet if sensible business needs are found. As per the present place of employment obligations, the WIFI and Internet administrations will be conceded to the workers.
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  3. Individual use – Employees can by and by utilize the WIFI and Internet assets by taking consent from the IT office. They should need to take care during individual use as the organization consistently screens their web use exercises.
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  • Policy Compliance - The security group of the organization will confirm the consistence of the arrangement utilizing proper techniques, for example, stroll through, interior and outer reviews, business apparatus reports, video checking, and input.

  • Related Standards, Policies, and Processes - The organization must need to utilize "Web utilization inclusion affirmation structure" for taking the affirmation from every representative in the wake of perusing the arrangement. The affirmation structure guarantees that each representative appropriately read the arrangement and get it.

Affirmation

I peruse and comprehend the approach. I am concurred with the arrangement.

Worker name and signature________________________

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Date_____________________

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3 0
3 years ago
Chaplets are used to support a sand core inside a sand mold cavity. The design of the
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Answer:

a) 2 chaplets

b) 9 chaplets

Explanation:

Before we can determine the minimum number of chaplets that should be placed beneath and above the core, we must know the mass of the sand used to make the surface of the mold cavity as well as the mass of the steel metal poured inside the mold.

Density is defined as the ratio of mass of a substance to its density.

Density = Mass/Volume

For the STEEL:

Density of steel = 7.82g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of steel = 7.82×5450

Mass of steel = 42619g

Mass of steel in kg = 42.619kg

For the SAND:

Density of sand = 1.6g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of sand = 1.6×5450

Mass of sand = 8720g

Mass of sand in kg = 8.72kg

a) Since the the chaplet support the sand from beneath the core, and each chaplet weighs 45N, we need to know the amount of force possessed by the sand.

Since the mass of the sand is 87.2kg

Weight = mass × acceleration due to gravity

Weight = 8.72×9.81

Weight of sand used to mold the core = 85.54N

Since 1 chaplet weighs 45N

This means that (85.54N/45N) i.e 1.9 which is approximately 2 chaplets must be placed beneath the core to sustain it before the steel metal is poured.

b) Since the metal poured in the core is steel, this means that the chaplet placed above the core must be able to withstand the strength of the steel.

Weight of steel = mass of steel × acceleration due to gravity

Weight of steel = 42.619×9.81

Weight of steel = 418.09N

Since one chaplet weigh 45N, the amount of chaplets that must be placed above the core is;

418.09/45

= 9.3

Therefore the minimum number of chaplets that should be placed above the core after pouring steel metal is 9chaplets.

4 0
3 years ago
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