Answer:
Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road
Explanation:
A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus of Elasticity [GPa] Tensile Strength [MPa] Aramid fiber 131 3600 Polycarbonate 2.4 65
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa. For this composite, compute the following:
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity
Answer:
1. 1111.5MPa
2. 56.1GPa
Explanation:
1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;
σcl = σm (1 - Vf) + σfVf
Substituting the figures, we will have;
45(1 - 0.30) + 3600(0.30)
45(0.70) + 1080
31.5 + 1080
= 1111.5MPa
2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;
Ecl = EmVm + EfVf
Substituting the formula gives;
= 2.4 (1 - 0.30) + 131 (0.30)
= 2.4(0.70) + 39.3
= 16.8 + 39.3
= 56.1GPa
Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
Longitudinal tensile stress can be obtained using the relation :
σcl = σm (1 - Vf) + σfVfSubstituting the values into the relation:
45(1 - 0.30) + 3600(0.30)
45 × 0.70 + 1080
31.5 + 1080
= 1111.50 MPa
2.)
Longitudinal modulus of elasticity is obtained using the relation :
Ecl = EmVm + EfVfSubstituting the values thus :
2.4 (1 - 0.30) + 131 (0.30)
= 2.4 × 0.70 + 39.3
= 16.8 + 39.3
= 56.10 GPa
Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
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The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 3^106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a hand-held flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 hours a year intermittently. Taking the unit cost of energy to be $4.35/10^6 Btu, determine the annual energy and cost savings as a result of tuning up the boiler.
Answer:
Energy Saved = 6.93 x 10⁹ Btu
Cost Saved = $ 30145.5
Explanation:
The energy generated by each boiler can be given by the following formula:
[tex]Annual\ Energy = (Heat\ In)(Combustion\ Efficiency)(Operating\ Hours)[/tex]
Now, the energy saved by the increase of efficiency through tuning will be the difference between the energy produced before and after tuning:
[tex]Energy\ Saved = (Heat\ In)(Efficiency\ After\ Tune - Efficiency\ Before\ Tune)(Hours)[/tex][tex]Energy\ Saved = (5.5\ x\ 3\ x\ 10^{6}\ Btu/h)(0.8-0.7)(4200\ h)[/tex]
Energy Saved = 6.93 x 10⁹ Btu
Now, for the saved cost:
[tex]Cost\ Saved = (Energy\ Saved)(Unit\ Cost)\\Cost\ Saved = (6.93\ x\ 10^{9}\ Btu)(\$4.35/10^{6}Btu)\\[/tex]
Cost Saved = $ 30145.5
a) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure and temperature at the beginning of compression process of 1 bar, 300 k. Determine:
(i) Maximum pressure and temperature in the cycle
(ii) Thermal efficiency
(iii) Mean effective pressure.
Assume for air Cp = 1.005 kJ/kg K, Cp = 0.718 kJ/kg K and R = 287 kJ/kg K.
(b) Explain any four types of classification of an Internal combustion engines.
:
Answer:
a) i) The maximum pressure is approximately 122.37 bar
ii) The thermal efficiency is approximately 56.47%
iii) The mean effective pressure is approximately 20.974 bar
b) (b) Four types of internal combustion engine includes;
1) The diesel engine
2) The Otto engine
3) The Brayton engine
4) The Wankel engine
Explanation:
The parameters of the Otto cycle are;
The heat added, [tex]Q_{in}[/tex] = 2,800 kJ/kg
The compression ratio, r = 8
The beginning compression pressure, P₁ = 1 bar
The beginning compression temperature, T₁ = 300 K
Cp = 1.005 kJ/kg·K
Cv = 0.718 kJ/kg·K
R = 287 kJ/kg·K
K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4
T₂ = T₁×r^(k - 1)
∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K
[tex]\dfrac{P_1\cdot V_1}{T_1} = \dfrac{P_2\cdot V_2}{T_2}[/tex]
[tex]P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2} = \dfrac{V_1}{V_2} \cdot \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot \dfrac{P_1 \cdot T_2}{T_1 }[/tex]
∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar
[tex]Q_{in}[/tex] = m·Cv·(T₃ - T₂)
∴ [tex]Q_{in}[/tex] = 2,800 ≈ 0.718 × (T₃ - 689.219)
T₃ = 2,800/0.718 + 689.219 = 4588.94 K
P₃ = P₂ × (T₃/T₂)
P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar
The maximum pressure = P₃ ≈ 122.37 bar
(ii) The thermal efficiency, [tex]\eta_{Otto}[/tex], is given as follows;
[tex]\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}[/tex]
Therefore, we have;
[tex]\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647[/tex]
The thermal efficiency, [tex]\eta_{Otto}[/tex] ≈ 0.5647
Therefore, the thermal efficiency ≈ 56.47%
(iii) The mean effective pressure, MEP is given as follows;
[tex]MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}[/tex]
Therefore, we get;
[tex]MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974[/tex]
The mean effective pressure, MEP ≈ 20.974 bar
(b) Four types of internal combustion engine includes;
1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion
2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel
The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust
3) The Brayton engine works on the principle of the steam turbine
4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes
To convert a measurement in centimeters to
meters, you simply move the decimal point
a. two places to the left
b. three places to the right
c. three places to the left
do four places to the right
Answer:
a
Explanation:
To convert a measurement in centimeters to meters, simply move the decimal point two places to the left. Thus, the correct option is A). two places to the left.
What is the way to convert centimeters to meters and decimals?There are 100 centimeters in every meter, which means that dividing the measurement in centimeters by 100 will convert it to meters. This conversion is very quick and easy by simply moving the decimal point in the measurement 2 spaces to the left.
The centimeter to meter conversion (cm to m) is basically, the conversion from centimeters to meters. One centimeter is approximately equal to 0.01 meter or we can say that one meter equals to 100 centimeters.
Simply, in order to convert cm to m, multiply the given centimeter value by 0.01 m. For example:- 5 cm = 5 x 0.01 m
5 cm = 0.05 m
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