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2 years ago

11

stimate the maximum efficiency of an automobile engine that has a compression ratio of 5:1.0. Assume the engine operates accordi

ng to the Otto cycle and assume γ = 1.4.

1 answer:

Fed [463]2 years ago

5 0

Answer:

Efficiency based on Otto cycle.

Effotto = 47.47%

Explanation:

Efficiency based on Otto cycle.

effotto = 1 – (V2 / V1)^γ-1

effotto = 1 – (1 / 5)^1.4 - 1

effotto = 47.47%

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Which one of the following is not an economic want?

Nostrana [21]

The available options are:

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Answer:

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3 years ago

A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

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extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

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3 years ago

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mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

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