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3 years ago

12

How does a turbo charger work

2 answers:

yKpoI14uk [10]3 years ago

8 0

Answer:

The turbocharger on a car applies a very similar principle to a piston engine. It uses the exhaust gas to drive a turbine. This spins an air compressor that pushes extra air (and oxygen) into the cylinders, allowing them to burn more fuel each second.

Explanation:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

The turbocharger on a car applies a very similar principle to a piston engine. It uses the exhaust gas to drive a turbine. This spins an air compressor that pushes extra air (and oxygen) into the cylinders, allowing them to burn more fuel each second

Explanation:

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Select the correct answer. Felix aspires to be an engineer working for the government. What credentials will Felix require to ap

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3 years ago

Add the following vector given in rectangular form and illustrated the process graphically A = 16+j12, B= 6+j10.4

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Answer:

A=16+j12…'B=6+j10.4

Explanation:

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2 years ago

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

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2 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

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