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MakcuM [25]
3 years ago
8

A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu

gh a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 6.5 kg/m3 . Determine the mass of air that has entered the tank.
Engineering
1 answer:
Mumz [18]3 years ago
6 0

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

Density = \frac{mass}{volume}

Therefore, mass is:

mass = denisty *volume

Density of air = 2 kg/m^3; Volume of the tank =  3.5 m^3

=> Mass = 3.5 * 2 = 7 kg

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = 6.5 kg/m^3

The new mass will be:

Mass = 6.5 * 3.5 = 22.75 kg

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

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b. equal to the specific entropy of the gas at the inlet.

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If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

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When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th
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Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

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Differentiating the above equation:

\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\

\to V_A=  \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\

        = \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}

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2 years ago
You can change lanes during a turn long as there’s no traffic and you driving slowly
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5 0
2 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

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attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
2 years ago
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Semenov [28]

Answer:

all exhaust gases from all gasoline engines

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