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3 years ago

What is a quasi-equilibrium process? What is its importance in engineering?

Engineering

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.

Explanation:

Step1

Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.

Step2

All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.

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An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205

lara [203]

Answer:

$0.845\ \text{N}$

Explanation:

g = Acceleration due to gravity at sea level = $9.81\ \text{m/s}^2$

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

$g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2$

Weight at sea level

$W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}$

Weight at the given height

$W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}$

Change in weight $W_h-W=636.805-637.65=-0.845\ \text{N}$

Her weight reduces by $0.845\ \text{N}$ .

8 0

3 years ago

6) Describe the differences between the troposphere and stratosphere.

vodka [1.7K]

Answer:

frejfru8758u37

Explanation:

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2 years ago

A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the

ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0

2 years ago

Differences between acidic and basic Bessemer process

dybincka [34]

Answer:

In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.

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2 years ago

Identify the right components for gsm architecture that consists of the hardware or physical equipment such as digital signal pr

sergiy2304 [10]

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<h3>What are the 4 main components?</h3>

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Under the GSM framework, a cell tele cell smartphone is called a Mobile Station and is partitioned into wonderful additives: the Subscriber Identity Module (SIM) and the Mobile Equipment (ME).