The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
cycle performance?
1 answer:
Answer:
Suction and exhaust processes do not affect the performance of Otto cycle.
Explanation:
Step1
Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.
Step2
Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.
Step3
The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:
Process 0-1 is suction process and process 1-0 is exhaust process.
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<u>Answer</u>:
a. r(t) = 6.40 cos (ωt + 38.66°) units
b. r(t) = 6.40 cos (ωt - 38.66°) units
c. r(t) = 6.40 cos (ωt - 38.66°) units
d. r(t) = 6.40 cos (ωt + 38.66°) units
<u>Explanation</u>:
To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:
(i) Convert the phasor to polar form. The polar form is written as;
r∠Ф
Where;
r = magnitude of the phasor =
Ф = direction = tan⁻¹ ()
(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:
r(t) = r cos (ωt + Φ)
Where;
ω = angular frequency of the sinusoid
Φ = phase angle of the sinusoid
(a) 5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
5 + j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
(b) 5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
5 - j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(c) -5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
-5 + j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(d) -5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
-5 - j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
<em>first step : calculate the value of h2 using the equation below </em>
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
Answer:
Applying Kirchhoff's current law at inverting terminal
Explanation:
Detailed explanation is given in the attached document.
