All categories

3 years ago

1. Why is outside air mixed with return air?

Engineering

1 answer:

Dmitrij [34]3 years ago

3 0

Answer:

Explanation:

outside air must be mixed with return air at some point in time because here is an example a warm current of air comes in contact with a cold current of air the warm current has to cool down so it could mix with the cold current or if they don´t mix they would continue to bump into each other and probably cause a tornado. probably.

You might be interested in

What does CPU stand for? computer processing unit central programming unit central processing unit computer programming unit

Harman [31]

Central processing unit

6 0

3 years ago

Read 2 more answers

Define schema & Instance...<br><br><br>don't spam..<br>

zubka84 [21]

Answer:

<h3>Hey there ! </h3><h3>Here is your Answer Buddy | </h3>

<h3>• Define Instance :- </h3>

The data stored in database at a particular moment of time is called instance of database.

<h3>• Define Schema :- </h3>

Database schema defines the variable declarations in tables that belong to a particular database

<h3><u>The value of these variables at a moment of time is called the instance of that database.</u></h3><h3 />

Explanation:

<h3>Hope this helps !! </h3>

4 0

3 years ago

Read 2 more answers

A commonly used strategy to improve the efficiency of a power plant is to withdraw steam from the turbine before it fully expand

algol [13]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

6 0

3 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

Molodets [167]

Answer:

Explanation:

Mean temperature is given by

$T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}$

Tmean = (Ti + T∞)/2

$T_mean = 107.5^{0}$

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

$\mu$ = 221.6 × 10⁻⁷N.s/m²

$\kappa$ = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/ $\mu$

Substituting for Pv

Re = 4m/(πD $\mu$ )

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0

3 years ago

1. Why is outside air mixed with return air?​

1. Why is outside air mixed with return air?