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3 years ago

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Which of the following is an acronym used in mechanical plans?

1 answer:

brilliants [131]3 years ago

8 0

Complete Question:

Which of the following is an acronym used in mechanical plans?

Group of answer choices

A. BTR

B. CAT

C. HVAC

D. TAN

Answer:

C. HVAC.

Explanation:

Mechanical plans can be defined as a construction document which typically comprises of drawings that illustrate, describe or show the placement, size, type and routes of various conduit and surface duct works and components.

An example of an acronym used in mechanical plans is HVAC.

HVAC is simply an acronym for heating, ventilation and air conditioning.

Basically, a HVAC system is a technology designed and developed to provide cooling and heating comfort (features) to vehicular and indoor environments such as submarines, cars, aeroplane, ships, restaurants, hotels, residential facilities or buildings etc.

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Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

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3 years ago

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp

cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

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Line(s) indicates passing is allowed if there are no oncoming cars.

anygoal [31]

Broken yellow b/c you can’t pass on a double solid yellow

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4 years ago

Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

mixas84 [53]

Answer:

work done = 48.88 × $10^{9}$ J

Explanation:

given data

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velocity = 310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J

work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J

work done = 48.88 × $10^{9}$ J

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SVETLANKA909090 [29]

Answer:

a scale model each size is a certain amount smaller

Explanation:

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