Answer:
The efficiency of this fuel cell is 80.69 percent.
Explanation:
From Physics we define the efficiency of the automotive fuel cell ([tex]\eta[/tex]), dimensionless, as:
[tex]\eta = \frac{\dot W_{out}}{\dot W_{in}}[/tex] (Eq. 1)
Where:
[tex]\dot W_{in}[/tex] - Maximum power possible from hydrogen flow, measured in kilowatts.
[tex]\dot W_{out}[/tex] - Output power of the automotive fuel cell, measured in kilowatts.
The maximum power possible from hydrogen flow is:
[tex]\dot W_{in} = \dot V\cdot \rho \cdot L_{c}[/tex] (Eq. 2)
Where:
[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]\rho[/tex] - Density of hydrogen, measured in kilograms per cubic meter.
[tex]L_{c}[/tex] - Heating value of hydrogen, measured in kilojoules per kilogram.
If we know that [tex]\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}[/tex], [tex]\rho = 0.0899\,\frac{kg}{m^{3}}[/tex], [tex]L_{c} = 141790\,\frac{kJ}{kg}[/tex] and [tex]\dot W_{out} = 80\,kW[/tex], then the efficiency of this fuel cell is:
(Eq. 1)
[tex]\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)[/tex]
[tex]\dot W_{in} = 99.143\,kW[/tex]
(Eq. 2)
[tex]\eta = \frac{80\,kW}{99.143\,kW}[/tex]
[tex]\eta = 0.807[/tex]
The efficiency of this fuel cell is 80.69 percent.
A ramjet operates by taking in air at the inlet, providing fuel for combustion, and exhausting the hot air through the exit. Th e mass fl ow at the inlet and outlet of the ramjet is 60 kg/s (the mass fl ow rate of fuel is negligible). Th e inlet velocity is 225 m/s. Th e density of the gases at the exit is 0.25 kg/m3 , and the exit area is 0.5 m2 . Calculate the thr
Answer:
15300 N
Explanation:
[tex]\rho_i[/tex] = Density of air at inlet
[tex]\dfrac{m}{t}[/tex] = Mass flow rate = 60 kg/s
[tex]v_i[/tex] = Inlet velocity = 225 m/s
[tex]\rho_o[/tex] = Density of gas at outlet = [tex]0.25\ \text{kg/m}^3[/tex]
[tex]A_i[/tex] = Inlet area
[tex]A_o[/tex] = Outlet area = [tex]0.5\ \text{m}^2[/tex]
Since mass flow rate is the same in the inlet and outlet we have
[tex]\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}[/tex]
Thrust is given by
[tex]F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}[/tex]
The thrust generated is 15300 N.
An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 300 mm, and the rolls rotate at 100 rpm. Calculate the roll force and the power required in this operation
Answer:
The roll force is 1.59 MN
The power required in this operation is 644.96 kW
Explanation:
Given;
width of the annealed copper, w = 228 m
thickness of the copper, h₀ = 25 mm
final thickness, hf = 20 mm
roll radius, R = 300 mm
The roll force is given by;
[tex]F = LwY_{avg}[/tex]
where;
w is the width of the annealed copper
[tex]Y_{avg}[/tex] is average true stress of the strip in the roll gap
L is length of arc in contact, and for frictionless situation it is given as;
[tex]L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm[/tex]
Now, determine the average true stress, [tex]Y_{avg}[/tex], for the annealed copper;
The absolute value of the true strain, ε = ln(25/20)
ε = 0.223
from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.
Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa
Finally determine the roll force;
[tex]F = LwY_{avg}[/tex]
[tex]F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN[/tex]
The power required in this operation is given by;
[tex]P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW[/tex]
A function is different from a procedure because a functiondoes not contain a set of instructions.can have only a limited number of steps.returns a value.is mathematical.
Answer: A function returns a value and a procedure just executes commands.
Explanation:
Answer:
c - returns a value.
Explanation:
on edge pls mark brainiest
How do I answer all the questions on this page?
Answer:
Create a google docs copy everything and paste hope this helps! :)
Explanation:
Expert Review is done by end users.
Answer:nononononono
Explanation:
Phosphorous is added to make an n-type silicon semiconductor with an electrical conductivity of 1.75 (Ωm)-1 . Calculate the necessary number of charge carriers required
Answer:
The necessary number of electron charge carriers required is:
8.1019 × 10¹⁹ electrons/m³
Explanation:
The necessary number of charge carriers required can be determined from the resistivity. Given that, the phosphorus make an n-type of silicon semiconductor;
Resistivity [tex]\rho = \dfrac{1}{\sigma}[/tex]
[tex]\rho = \dfrac{1}{q \mu _n n_n}[/tex]
where;
The number of electron on the charge carriers [tex]n_n[/tex] is unknown??
The charge of the electron q = [tex]1.6 \times 10^{-19} \ C[/tex]
The electron mobility [tex]\mu_n[/tex] = 0.135 m²/V.s
The electrical conductivity [tex]\sigma[/tex] = 1.75 (Ωm)⁻¹
Making [tex]n_n[/tex] the subject from the above equation:
Then;
[tex]n_n = \dfrac{\sigma }{q \mu_n}[/tex]
[tex]n_n = \dfrac{1.75 \ \Omega .m^{-1} }{1.6 \times 10^{-19} \times 0.135 \ m^2/V.s}[/tex]
[tex]n_n =8.1019 \times 10^{19}[/tex] electrons/m³
Thus; the necessary number of electron charge carriers required is:
8.1019 × 10¹⁹ electrons/m³
For three control stations,there should be how many start buttons in parallel with the suxiliary contact
Do you think we would have a use for a technology that would mimic the flight of a hummingbird?
Answer:
I think we would.
Explanation:
Only if u invent it.
answer
i don't think so
but it will take a lot of thinking if it is
possible
1. Drill press size is determined by the largest__
CAD(computer-aided design) software and is used in__________and __________that show how to construct an object. Technical drawings show in detail how the pieces of something relate to each other.
Answer:
Plans; blueprints.
Explanation:
In Engineering, it is a common and standard practice to use drawings and models in the design and development of various tools or systems that are being used for proffering solutions to specific problems in different fields such as engineering, medicine, telecommunications and industries.
Hence, a design engineer make use of drawings such as pictorial drawings, sketches, or technical drawing to communicate ideas about a design to others, to record and retain informations (ideas) so that they're not forgotten and to analyze how different components of a design work together.
Technical drawing is mainly implemented with CAD (computer-aided design) software and is typically used in plans and blueprints that show how to construct an object.
Additionally, technical drawings show in detail how the pieces of something (object) relate to each other, as well as accurately illustrating the actual (true) shape and size of an object in the design and development process.
Pyramid is a type of ___________ structure.
Answer:
Massive or linteled
Explanation:
Pyramid is a type of massive or linteled structure.
These structures do no have not much internal spaces and they are huge edifices.
A pyramid is a solid body with outer triangular faces that converges on top. To construct a pyramid, large amounts of materials are usually involved. Pyramids were more prominent in times past before this present civilization.
PLEASE ANSWER SOON
In a science lab, Cash mixes two clear liquids together in a beaker. Bubbles are produced, and a white solid forms and settles to the bottom. Which statement below describes what happened?
a
A physical change occurred, a gas and precipitate was produced
b
A physical change occurred, only a gas was produced
c
A chemical change occurred, only a gas was produced
d
A chemical change occurred, a gas and precipitate was produced
Answer:
I think it is D
Explanation:
I did a couple mins of research on this topic but there is no clear line that separates a physical and chemical reaction, so do with that as you will. i hope I somewhat helped.
Convert 273.15 mL at 166.0 mm of Hg to its new volume at standard pressure.
Answer:
(166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)