What are three automotive safety systems

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

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

4 0

3 years ago

By using a book of the OHS Act, Act 85 of 1993, find the act or regulation where the following extraction comes from "every empl

AnnyKZ [126]

Answer:

Is very important to employ safety and care

For every staff

Is very necessary is protected

Her is going very high stage

Because of every one say that this company is best for others company

Is reason safety is very important in our life

3 0

3 years ago

Explain, with reasons, whether the LTIC systems described by the following differential equations are (i) stable or unstable in

taurus [48]

Answer:

Explanation:

The LTIC system is the Linear Time Invariant Theory, also known as LTI system theory, investigates the response of a linear and time-invariant system to an arbitrary input signal.

For the step by step Solution to the question you asked, go through the attached documents.

8 0

3 years ago

Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS t

Anna11 [10]

Answer:

Proof is as follows

Proof:

Given that , $V = V_{ac} + V_{dc}$

for any function f with period T, RMS is given by

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }$

In our case, function is $V = V_{ac} + V_{dc}$

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }$

Now open the square term as follows

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }$

Rearranging terms

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac} } \, dt }$

You can see that

second term is square of RMS value of Vac

Third terms is average of VdcVac and given is that average of $V_{ac}V_{dc} = 0$

so

$RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

$RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

So it has been proved that given expression for root mean square (RMS) is valid

7 0

3 years ago

g Write a program that takes in a positive integer as input, and outputs a string of 1's and 0's representing the integer in bin

Phoenix [80]

Answer:

The solution is written in Python

binary = ""
decimal = 13
quotient = int(decimal / 2)
remainder = decimal % 2
binary = str(remainder) + binary
while(quotient >0):
decimal = int(decimal / 2)
quotient = int(decimal / 2)
remainder = decimal % 2
binary = str(remainder) + binary
print(binary)

Explanation:

Firstly, we declare a variable binary and initialize it with an empty string (Line 1). This binary is to hold the binary string.

Next, we declare variable decimal, quotient and remainder (Line 2-4). We assign a test value 13 to decimal variable and then get the first quotient by dividing decimal with 2 (Line 3). Then we get the remainder by using modulus operator, % (Line 4). The first remainder will be the first digit joined with the binary string (Line 5).

We need to repeat the process from Line 3-5 to get the following binary digits. Therefore create a while loop (Line 7) and set a condition that if quotient is bigger than 0 we keep dividing decimal by 2 and calculate the quotient and remainder and use the remainder as a binary digit and join it with binary string from the front (Line 9-11).

At last, we print the binary to terminal (Line 13).

4 0

4 years ago