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.
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.
How do I answer all the questions on this page?
Answer:
Create a google docs copy everything and paste hope this helps! :)
Explanation:
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.
