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Bond [772]
3 years ago
5

The function of air conditioning system is a)-Cooling and de humidification b)-Cooling and humidification c)-all of these d)-Hea

ting and humidification
Engineering
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

a). Cooling and dehumidification

Explanation:

Air conditioning means conditioning the air around. Air conditioners are meant for cooling and dehumidification of air. By dehumidification, we mean to remove water particles or moister from the surrounding air.

      This is a basic process by which a typical air conditioner undergoes.

When the moist air is brought near cold surface, it gets cooled below its dew point temperature. Thus some water vapour condenses into liquid and leaves as liquid from the air through the air conditioning process, thereby decreasing both the moisture content and temperature of the surrounding air.

Thus, the function of air conditioning system is to cool and dehumidify the air.

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without using the routh hurwitz criterion, determing if the followign systems are asymptotically stable, marginally stable, or u
lesantik [10]

Answer:

The system is marginally stable.

Explanation:

Transfer function, M(s) = [10(s+2)]/(s³ + 3s² + 5s)

In control the stability properties of a system can be obtained from just the characteristic equation of its closed loop transfer function.

- The condition for stability is that all the roots of the characteristic equation be negative and real.

- The condition for asymptotic stability is that all the real parts of the roots must all be negative, since there'll be complex roots.

- The condition for marginal stability is that the real part of all the complex roots are negative, the roots without real parts must have distinct imaginary parts.

- The condition for instability is for at least one of the roots to be positive. Or if there are complex roots, the real part of the roots being positive indicates instability.

The characteristic equation for this transfer function is (s³ + 3s² + 5s)

Solving this polynomial

s = 0

s = [-3 - √(11i)]/2

s = [-3 + √(11i)]/2

These roots have all their real parts to be negative, and the zero root has a distinct imaginary part, hence the system is marginally stable

4 0
4 years ago
A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

To know more about process click here:

brainly.com/question/29310303

#SPJ4

5 0
1 year ago
I will rate 5 stars/brainliest if you help me!!! A company produces A, B, and C and can sell these products in unlimited quantit
Serhud [2]

Answer: 0

Explanation: i need to add it up to get what u want

3 0
3 years ago
Assume you have created a class named MyClass and that is contains a private field named
marysya [2.9K]

Answer:

a. myMethod() has access to and can use myField.

Explanation:

Logic programming is a kind of programming which is largely based on formal logic.  The statement are written in logical forms which express rules about the domain. In the given scenario the my method will have access to my field which is private field. My method non static public field can also use my field class.

7 0
4 years ago
Calculate the link parameter and channel utilization efficiency (in error-free channels) for a system with the following paramet
larisa86 [58]

Answer: a) 0.77 and 0.39 b) 3.9 and 0.20

Explanation:

We have to find two things here i.e link parameter and channel utilization efficiency.

a)

We are given

Bitrate= 12 Mbps,

Distance= 6 miles

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 500 bits

We know that The link parameter is related to propagation time and frame time

Where Propagation time= Bitrate x Distance= 12 Mbps x 6 miles

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 500

So,

Link parameter= \frac{12*10^6*6*1609.34}{3*10^8*500}

Link parameter= 0.77

Channel utilization efficiency= \frac{1}{1+2* link parameter}

Channel utilization efficiency= 0.39

b)

We are given

Bitrate= 10 Mbps,

Distance= 15 km

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 256 bits

We know that The link parameter is related to propagation time and frame time, Here QPSK is used so bitrate is multiplied by 2,

Where Propagation time= Bitrate x Distance= 2 x 10 Mbps x 15 km

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 256

So,

Link parameter= \frac{2*10*10^6*15*1000}{3*10^8*256}

Link parameter= 3.9

Channel utilization efficiency= \frac{1}{1+2* link parameter}

Channel utilization efficiency= 0.20

7 0
3 years ago
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