Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:
a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.
Answers
Answer:
44.59°c
Explanation:
Given data :
Total pressure = 105 kpa
complete combustion
A) Determine air-fuel ratio
A-F = [tex]\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }[/tex]
N = number of mole
m = molar mass
A-F = [tex]\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}[/tex] = 22.2 kg air/fuel
hence the ratio of Fuel-air = 1 : 22.2
B) Determine the temperature at which water vapor in the products start condensing
First we determine the partial pressure of water vapor before using the steam table to determine the corresponding saturation temp
partial pressure of water vapor
Pv = [tex]\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )[/tex]
N watervapor ( number of mole of water vapor ) = 3
N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol
Pro = 105
hence Pv = ( 3/33.53 ) * 105 = 9.39kPa
from the steam pressure table the corresponding saturation temperature to 9.39kPa = 44.59° c
Temperature at which condensing will start = 44.59°c
An equation showing the products of propylene with their mole numbers is attached below
Related Questions
When water levels on the exterior of a building exceed water levels on the interior, hydrostatic loads become:_______
Answers
Answer: more dense
Explanation:
1:
Determine the dynamic pressure of water received at the site described below.
Water Tower holds water at an elevation of 265 feet
Site is at an elevation of 145 feet
The water supply system uses cast iron pipes.
The water travels through 3.2 miles of pipes before reaching the site
The pipe has a diameter of 8 inches
The water travels through 9 90-degree bends, 4 Branch tees, and 1 Swing Check Valve
The water has a flow rate of 105 gpm
Round to the hundredths place (2 places after the decimal)
2:
What is the static head of a water supply system if the water tower holds water at and elevation of 462 feet and the site that uses the water is at an elevation of 294 feet?
Answer value
Answers
Answer:
20.87 Pa
Explanation:
The formula for dynamic pressure is given as;
q= 1/2*ρ*v²
where ;
q=dynamic pressure
ρ = density of fluid
v = velocity of fluid
First find v by applying the formula for flow rate as;
Q = v*A where ;
Q= fluid flow rate
v = flow velocity
A= cross-sectional area.
A= cross-sectional vector area of the pipe given by the formula;
A= πr² = 3.14 * 4² = 50.27 in² where r=radius of pipe obtained from the diameter given divided by 2.
Q = fluid flow rate = 105 gpm----change to m³/s as
1 gpm = 0.00006309
105 gpm = 105 * 0.00006309 = 0.006624 m³/s
A= cross-sectional vector area = 50.27 in² -------change to m² as:
1 in² = 0.0006452 m²
50.27 in² = 50.27 * 0.0006452 = 0.03243 m²
Now calculate flow velocity as;
Q =v * A
Q/A = v
0.006624 m³/s / 0.03243 m² =v
0.2043 m/s = v
Now find the dynamic pressure q given as;
q= 1/2 * ρ*v²
q= 1/2 * 1000 * 0.2043² = 20.87 Pa
If a person runs a distance of 0.7 km in 3 min, what is his average speed in kilometres/hour
Answers
Answer:
14 km/hour
Explanation:
A well-insulated heat exchanger has one line with 2 kg/s of air at 125 kPa and 1000 K entering, and leaving at 100 kPa and 400 K. The other line has 0.5 kg/s water entering at 200 kPa and 20 °C, and leaving at 200 kPa. Calculate the exit temperature of the water and the total rate of entropy generation?
Answers
Answer:
120°CExplanation:
Step one:
given data
T_{wi} = 20^{\circ}C
T_{Ai}=1000K
T_{Ae}= 400kPa
P_{Wi}=200kPa
P_{Ai}=125kPa
P_{We}=200kPa
P_{Ae}=100kPa
m_A=2kg/s
m_W=0.5kg/s
We know that the energy equation is
[tex]m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}[/tex]
making [tex]h_{We}[/tex] the subject of formula we have
[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]
from the saturated water table B.1.1 , corresponding to [tex]T_{wi}= 20c[/tex]
[tex]h_{Wi}=83.94kJ/kg[/tex]
from the ideal gas properties of air table B.7.1 , corresponding to T=1000K
the enthalpy is:
[tex]h_{Ai}=1046.22kJ/kg[/tex]
from the ideal gas properties of air table B.7.1 corresponding to T=400K
[tex]h_{Ae}=401.30kJ/kg[/tex]
Step two:
substituting into the equation we have
[tex]h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})[/tex]
[tex]h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg[/tex]
from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex] we can obtain the specific enthalpy:
[tex]h_g=2706.63kJ/kg[/tex]
we can see that [tex]h_g>h_{Wi}[/tex], hence there are two phases
from saturated water table B.1.2 at [tex]P_{We}=200kPa[/tex]
[tex]T_{We}=120 ^{\circ} C[/tex]
You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the circuit?
Answers
Answer:
4mA
Explanation:
For this problem, we will simply apply Ohm's law:
V = IR
V/R = I
I = V / R
I = 12 volt / 3kΩ
I = 4mA
Hence, the current in the circuit is 4mA.
Cheers.
A flexible rectangular area measures 2.5 m X 5.0 m in plan. It supports an external load of 150 kN/m^2. Determine the vertical stress increase due to the load at a depth of 6.25 m below the corner of the footing.
Answers
Answer:
19.0476 kN/m^2
Explanation:
Given data:
Dimension of rectangular area = 2.5m x 5.0m
external load = 150 KN/m^2
load depth = 6.25 m
calculate the vertical stress increase due to load at depth 6.25
we will use the approximate method which is
[tex]V_{s} = \frac{qBL}{(B +Z)(L+Z)}[/tex] ------- (1)
q = 150 kN/m^2
B = 2.5 m
L = 5 m
Z = 6.25 m
substitute the given values into the equation (1) above
hence the Vs ( vertical stress ) = 19.0476 kN/m^2
You have a 20 Volt power source attached to a light bulb that you've measured has a resistance of 8 Ohms, what is the power output of this light bulb (in Watts)?
Answers
Answer:
50 Watts
Explanation:
For this problem, we simply apply Ohm's law:
V = IR
V / R = I
P = IV
P = ( V / R ) V
P = ( V^2 / R )
P = (20 V)^2 / 8 Ω
P = 400 V^2 / 8 Ω
P = 50 Watts
Hence, the power consumed by the lightbulb is 50 Watts.
Cheers.
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Answers
Answer:
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Explanation:
thats all you said
Answer:
hii my name is RAGHAV what is your name
Explanation:
this question is which chapter
True or false Osha engineering controls are the last strategy of control an employer should use for job hazards
Answers
Answer:
false
Explanation:
What prevented this weld from becoming ropey?
A lower ampera
A higher voltage
The position of the weld
The stepping motion of the weld
Answers
Answer:
If I am not mistaken I believe it is a higher voltage.
Explanation:
Hope this helps
You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?
Answers
Answer:
18 Watts
Explanation:
For this problem, we simply need to understand the relationship of power to voltage and current. This relationship is derived from Ohm's law:
Power = Voltage * Current
Given this equation, we can say the following to find the power consumption of the motor:
Power = 12volts * 1.5amps
Power = 18 Watts
Hence, the motor is consuming 18 Watts of power.
Cheers.
3 mA are flowing through an 18 V circuit, how much resistance (in kΩ) is in that circuit?
Answers
Answer:
6kΩ
Explanation:
If we assume that the entire circuit's current is 3 mA, then we can compute the resistance within the circuit with the application of Ohm's law:
V = IR
V/I = R
R = V / I
R = 18V / 3 mA
R = 6kΩ
Hence, the resistance of the circuit is 6kΩ.
Cheers.