Ask question

Ask question

All categories

BaLLatris [955]

3 years ago

6

An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

mine: (a) the molar air-fuel ratio and air-fuel ratio by mass. (b) the lower heating value, in kJ per kmol of fuel and in kJ per kg of fuel.(c) the higher heating value, in kJ per kmol of fuel and in kJ per kg of fuel. (d) the dew point temperature of the combustion products at 1 atm, in °C. Consider a reference temperature and pressure of 25°C, 1 atm.

1 answer:

slava [35]3 years ago

4 0

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

a) determine the molar air-fuel ratio and air-fuel ratio by mass

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

You might be interested in

Air enters a 34 kW electrical heater at a rate of 0.8 kg/s with negligible velocity and a temperature of 60 °C. The air is disch

Flura [38]

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

This is specific volume.

The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

p1*v1 = 287 * 333 = 95.6 kJ/kg

p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.

4 0

4 years ago

What is one experiment that has been done in the manufacturing

Bingel [31]

Do you have a picture of any sort so I can see if there’s any options ;)

8 0

3 years ago

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical

Olin [163]

Answer:

The theoretical density for Niobium is $1.87 g/cm^3$ .

Explanation:

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density of the unit cell

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

$a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm$

$1 nm = 10^{-7} cm$

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}$

$\rho =1.87 g/cm^3$

The theoretical density for Niobium is $1.87 g/cm^3$ .

6 0

3 years ago

Read 2 more answers

Coastal regions are often affected by nearby ocean currents. The map shows how different ocean currents affect the United States

Effectus [21]

Effect of Ocean currents on the Coastal regions are described below.

Explanation:

1.Ocean currents effects the temperature of the region they travel through. Warm currents increase the temperature of the coasts along which they move. Sea breezes that blow over warm currents get warmed in turn and absorb moisture. Then they carry the warmth and moisture to neighboring coastal areas.

2.When land borders the ocean, the currents of the ocean warm or cool it, depending on the nature of the particular current that flows by that land. In cases where a warm current flows along a particular coast, that coastal area will generally be warmer than it would otherwise be if it were landlocked.

3.They produce high and low tides along coastal regions.They move vertically pushing warm water and nutrients to the surface along coastal regions.

4.The circumpolar current merges the waters of the Atlantic, Indian, and Pacific Oceans and carries up to 150 times the volume of water flowing in all of the world's rivers.

5 0

4 years ago

Which of the following is NOT a common problem or mistake with blueprints?

Nikitich [7]

Answer:

illegible.writng

Explanation:

am I right

3 0

3 years ago