1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vadim26 [7]
3 years ago
11

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon

tal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

Engineering
1 answer:
Ulleksa [173]3 years ago
4 0

The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

What is the Rayleigh number for free convection on the outer sides of the duct?

What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?

What is the Rayleigh number for free convection on the top of the duct?  

What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?

What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

What is the total heat gain to the duct per unit length, in W/m?

Answers:

- 7709251  or 7.709 ×10⁶

- 4.87

- 965073

- 5.931 W/m² K

- 2.868 W/m² K

- 69.498 W/m

Explanation:

Find the given attachments for complete explanation

You might be interested in
How may a Professional Engineer provide notice of licensure to clients?
azamat

Find full question attached

Answer:

(b) By including a statement that he or she is licensed by the Board for Professional Engineers and Land Surveyors immediately above the signature line in at least 12 point type on all contracts for services

Explanation:

A PE(professional engineer) licensee must show that he is licensed in order to show and ensure public safety as he is qualified for the job he is handling. The California regulations on professional engineers holds that all professional engineers must be licensed by the board of professional engineers and Land surveyors in order to operate legally as an engineer. The engineer may show licensure through the following options:

The engineer might provide statement to each client to show he is licensed which would then be signed by the client

The engineer may choose to post a wall certificate in his work premises to show he is licensed

The engineer may choose to include a statement of license in a letterhead or contract document which must be above the client's signature line and not less than 12 point type

4 0
3 years ago
The formation of faults in Earth's crust is an effect. What causes faults to form in the crust? Global Positioning System sensor
allsm [11]

Answer: The movement of tectonic plates

Explanation:

Tectonic plates are the part of the earth's crust that both the ocean and land rest on. These plates are constantly moving as a result of currents in the mantle.

These movements cause stress on the surface which has the effect of fracturing rocks and thereby creating/ forming faults in the earth's crust. Sometimes faults form when these plates move away from each other and sometimes they are formed when they push into each other.

7 0
2 years ago
At a festival, spherical balloons with a radius of 140.cm are to be inflated with hot air and released. The air at the festival
Tpy6a [65]

Answer:

find attached

Explanation:

5 0
3 years ago
If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B
Sergio039 [100]

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

5 0
3 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
Other questions:
  • Where the velocity is highest in the radial direction? Why?
    9·1 answer
  • As you discovered in lab last week, the advantage of CMOS logic is that no drain current flows through the MOSFETs when the outp
    14·1 answer
  • Hey, can anyone tell me if Igneous rock is good to build on? Cheers!
    6·1 answer
  • An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of air and c
    15·1 answer
  • A completely mixed activated-sludge process is being designed for a wastewater flow of 10,000 m3/d (2.64 mgd) using the kinetics
    6·1 answer
  • The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards
    9·1 answer
  • Place these events in chronological order. Put the numbers 1-6 next to the<br> events.
    15·1 answer
  • Método de Programación lineal utilizado para resolver problemas en teoría de redes?
    15·1 answer
  • What Are 2 Properties electromagnets have that permanent magnets do not?
    8·2 answers
  • what do you expect the future trends of an operating system in terms of (a) cost (b) size (c) multitasking (d) portability (e) s
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!