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3 years ago

How does the clearance volume affect the efficiency of the Otto cycle?

Engineering

1 answer:

eduard3 years ago

7 0

Answer:

Explanation:

A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.

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Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$

Evaporation rate = $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

6 0

3 years ago

Is a street the same as a avenue

-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0

3 years ago

Read 2 more answers

9. To forward bias an NPN transistor, the

murzikaleks [220]

Answer:

a. the base must be more positive than the emitter

Explanation:

A transistor can be defined as a semiconductor component that is used to control the flow of voltage or current and as a gate (switch) for electronic signals. Thus, a transistor allows for the amplification, control and generation of electronic signals in a circuit. The three (3) basic parts of a transistor are; base, emitter and collector.

Basically, there are two (2) main types of transistor and these are;

1. PNP transistor.

2. NPN transistor.

Biasing of a transistor can be defined as the process of providing the controlled amount of direct current (DC) voltage or current conditions so as to enable the transistor amplify the alternating current (AC) input signal correctly.

Hence, to forward bias an NPN transistor, the base must be more positive than the emitter because the majority carriers are electrons which are moved from the n-type region to the p-type region while the minority carriers are holes.

This ultimately implies that, for an NPN transistor to conduct current in milliamps, its base-emitter junction must be forward biased.

4 0

3 years ago

A fluid has a viscosity of 13 P and a specific gravity of 0.94. Determine the kinematic viscosity of this fluid in units of ft²/

OlgaM077 [116]

Answer:

kinematic viscosity is 0.0149 ft²/s

Explanation:

given data

specific gravity S = 0.94

density ρ = 0.94 × 1000

viscosity μ = 13 Poise = 1.3 Pa-sec

we know 1 poise = 0.1 pas

to find out

kinematic viscosity

solution

we will apply here Kinematic viscosity formula that is

kinematic viscosity = $\frac{\mu}{\rho}$ ..................1

put here value in equation 1

and here ρ is density and μ is viscosity

kinematic viscosity = $\frac{1.3}{0.94*1000}$

kinematic viscosity = 1.382978 × $10^{-3}$ m³/s

so kinematic viscosity is 0.0149 ft²/s

3 0

3 years ago

What types of issues MAY occur to slow or prevent the best outcome?

german

Answer:

im sorry but i cant find any studies about this and im 3 days late

4 0

3 years ago