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Svetllana [295]
3 years ago
6

A classroom that normally contains 40 people is to be air-conditioned with window air-conditioning units of 5 kW cooling capacit

y. A person at rest may be assumed to dissi- pate heat at a rate of about 80 Watts. There are 10 light bulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h.
a. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window air-conditioning units required.
b. What is the source of heat being generated for the light bulbs and for the students?
Engineering
1 answer:
kap26 [50]3 years ago
4 0

Answer:

Explanation:

a ) Heat energy given out by 40 people

= 40 x 80 = 3200 J in one second

Heat energy given out by 10 light bulbs

= 10 x 100 = 1000 J in one second

Heat transfer by wall = 15000 x 1000 J in one hour

= 15000 x 1000 / 60 x 60

= 4166.67 J in one second

Total energy

= 3200 + 1000 + 4166 .67 J in one second

= 8366.67 J

To drive this much of heat , we need 2 air -conditioner of 5000 W each.

b ) Source of heat for bulb is electrical energy of bulb

For the students , source of heat by student is the bio- chemical reactions taking place in the body of the student which is exothermic.

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blondinia [14]

Answer:

The MATLAB Code for this PI Controller will be:

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

Explanation:

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

Obtain an open-loop response and determine what needs to be improved

Add a proportional control to improve the rise time

Add a derivative control to reduce the overshoot

Add an integral control to reduce the steady-state error

Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.

The further explanation is attached in the Word File.

Download docx
5 0
3 years ago
A master stud pattern is laid out somewhat<br> like a?
Svetradugi [14.3K]

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

3 0
2 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
3 years ago
Read 2 more answers
Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th
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Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

   # value_celsius is initialized to 0.0

   value_celsius = 0.0

   

   # value_celsius is calculated by

   # subtracting 273.15 from value_kelvin

   value_celsius = value_kelvin - 273.15

   # value_celsius is returned

   return value_celsius

   

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

   # value_kelvin is initialized to 0.0

   value_kelvin = 0.0

   

   # value_kelvin is calculated by

   # adding 273.15 to value_celsius

   value_kelvin = value_celsius + 273.15

   # value_kelvin is returned

   return value_kelvin

   

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 0
3 years ago
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mice21 [21]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

6 0
3 years ago
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