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.
An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a heating value of 141,790 kJ/kg and a density of 0.0899 kg/m3, determine the efficiency of this fuel cell.
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.
How do I answer all the questions on this page?
Answer:
Create a google docs copy everything and paste hope this helps! :)
Explanation:
Assume that Randy’s photocopying Service charges $.10 per photocopy. If fixed costs are
$27000 a year and variable costs are $0.04 per copy.
1. How Randy can compute his breakeven point? Show the result in graph.
2. How many photocopies are required to earn $ 500 profit?
3. Identify the safety margin at breakeven point.
Answer:
1) 739 copies (739.7260273972603).
2) 5000 copies.
Explanation:
27000/365 = 73.97260273972603 (Their daily wins). How much times $.10 is equal to 73.97260273972603?
I used a two step equation to get it.
0.10 * x = 73.97260273972603/0.10 = /0.10 x = 739.7260273972603They need to sell 739.7260273972603 copies a day to get 27000$ a year.
an equation (y = mx+b) can be
y = 739.7260273972603x
To earn a 500$ profit we need to do the same two-step equation but instead of 73.97260273972603 use 500.
0.10 * x = 500/0.10 = /0.10 x = 50005000 copies needs to be sold to get a 500$ profit.
The last question I don't understand but I hope those 2 questions helped
Create a story with nuclear engineer (not so long and not so short)
Answer:
I can't write stories well, but I can give character advice!
Explanation:
A really important thing is to make sure your characters act human. In a lot of stories, humans act... well, like aliens or superheroes.Come up with unique traits and descriptions (ex: 2 different colored eyes, spunky, bright orange hair)POV really matters! Choosing the wrong POV might make the character seem to flawed, or too perfect, so POV is one of the most important things in a story!Don't be afraid to make your main character stray from the usual, but make sure there's other characters that stand out, because it gets quite boring if there's only 1 unique character!1. Drill press size is determined by the largest__
Expert Review is done by end users.
Answer:nononononono
Explanation:
