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Mrac [35]
3 years ago
6

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

ir conditioning unit is
Engineering
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds  

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

                          = 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

                              = 2622.6 / 25  = 104.904 ft^3/sec

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A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.
stira [4]

Answer:

c. less than 60 mi/h

Explanation:

To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.

Total Distance Traveled = S = 100 mi + 100 mi

S = 200 mi

Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.

Total Time = t = Time from A to B + Time from B to C

t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)

t = 2 h + 1.43 h

t = 3.43 h

Now, the average speed of bus will be given as:

Average Speed = V = S/t

V = 200 mi/3.43 h

<u>V = 58.33 mi/h</u>

It is clear from this answer that the correct option is:

<u>c. less than 60 mi/h</u>

7 0
3 years ago
Calculate and plot the radial and circumferential stress distribution in the left ventricle at the end of systole (p 5 80 mmHg;
alexandr402 [8]

Answer:

62990.08 N/M

Explanation:

Circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a stress in directions coplanar with but perpendicular to the symmetry axis.

See attachment for the step by step solution of the given problem..

8 0
3 years ago
Hi, any kind of help on these questions will be appreciated.
Zielflug [23.3K]

Answer:

IDK

Explanation:

8 0
3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
why is the thermal conductivity of super insolation order of magnitude lower than the thermal conductivity of ordinary insulatio
Anni [7]

Answer:

Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.

Explanation:

3 0
3 years ago
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