Answer:
They both have the same efficiency.
Explanation:
The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.
Therefore in comparison, the efficiency is the same for both.
Answer:
474.59 mg/L
Explanation:
Given that
BOD = 30 mg/L
Original BOD = 30 mg/L × dilution factor
Original BOD = 30 mg/L × 10 = 300 mg/L
here is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day
Answer:
a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s
Explanation:
Diffusion is governed by Arrhenius equation
I will be using R in the equation instead of k_b as the problem asks for molar activation energy
I will be using
and
°C + 273 = K
here, adjust your precision as neccessary
Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm
So:
and
You might notice that these equations have the form of
You can solve this equation system easily using calculator, and you will eventually get
After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation
And you should get D = 2.76*10^-16 m^/s as an answer for c)
Answer:
The circulation around the cylinder is 0.163
Explanation:
Given :
Velocity of spinning cylinder
Sea level density
Sea level span
Lift per unit circulation is given by,
Where circulation around cylinder
Therefore, the circulation around the cylinder is 0.163
Answer:
u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)
Explanation:
Given:
- Electric Field strength near earth's surface E = 145 V / m
- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg
Find:
- How much energy is stored per cubic meter in this field?
Solution:
- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:
u_e = 0.5*e_o * E^2
- Plug in the values given:
u_e = 0.5*8.854 *10^-12 *145^2
u_e = 9.30777 * 10^-8 J/m^3