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3 years ago

13

In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)

volume d. Isentropic Expansion

1 answer:

docker41 [41]3 years ago

3 0

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

Reversible adiabatic compression
Addition of heat at constant volume
Reversible adiabatic expansion
Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

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Consider an ideal gas undergoing a constant pressure process from state 1 to state

Radda [10]

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.

6 0

4 years ago

If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo

Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

$\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)}$ and considering that $E(s)=D(s)-G_c(s)C(s)$ then

$\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}$

$E(s)=\frac {D(s)}{1+G_c(s)G(s)}$

For ramp disturbance d(t)=at

$D(s)=\frac {a}{s^{2}}$ therefore, the steady state error is given by

$e(\infty)= \lim_{s \to 0} s E(s)$

$e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]$

Whenever $G_c(s)$ has a double intergrator, the error $e(\infty)$ becomes zero

3 0

3 years ago

The worst case signal-to-noise ratio at the output of an FM detector occurs when: ________

AlexFokin [52]

Answer:

a. the desired signal is 90 degrees out of phase with the intelligence signal.

Explanation:

The signal to noise ratio of FM detector is defined as function of modulation index for SSB FM signal plus narrow band Gaussian noise at input. The ratio is usually higher than 1:1 which indicates more signals than noise.

6 0

3 years ago

A closed, rigid, 0.45 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the

liberstina [14]

Answer:

initial quality = 0.3690

heat transfer = 979.63 kJ/kg

Explanation:

Given data:

volume of tank 0.45^3

weight of water 12 kg

Initial pressure 20 bar

final pressure 4 bar

Specific volume $v = \frac {0.45}{12} = 0.0375 m^3/kg$

At Pressure = 20 bar, from saturated water table

$v_f = 0.01177 m^/kg$

$v_g = 0.099587 m^3/kg$

$x = \frac{v -v_f}{v_g -v_f} = \frac{0.0375 - 0.001177}{0.099587 - 0.001177}$

inital quality is x =0.3690

Heat transfer is calculated as

$u_1 = h_f + x(h_g - h_f) = v_f + x( h_{fg})$

from saturated water table, for pressure 20 bar ,

$h_f = 908.79 kJ/kg, h_{fg} = 1890.7 kJ/kg$

=908.79 + 0.0357(1890.7)

= 979.63 kJ/kg

7 0

3 years ago

(a) Determine the temperature of the insulated walls. (b) Determine the net radiation heat rate from surface 2 per unit conduit

bazaltina [42]

Answer:

Explanation:

the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you

3 0

4 years ago