Answer:
At the moment there are three astronauts working in the international space station.
Explanation:
The space station is as large as a U.S football field and is typically occupied by at least three astronauts and a maximum of six astronauts.
Answer:
The rate of work output = -396.17 kJ/s
Explanation:
Here we have the given parameters
Initial temperature, T₁ = 355°C = 628.15 K
Initial pressure, P₁ = 350 kPa
h₁ = 763.088 kJ/kg
s₁ = 4.287 kJ/(kg·K)
Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately
h₂ = 79.572 kJ/kg
The saturation temperature at the given
T₂ = 79°C
The rate of work output
=
×
×(T₂ - T₁)
Where;
= The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
= The mass flow rate = 2.0 kg/s
Substituting the values, we have;
= 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
= -396.17 kJ/s
True
(I’m not sure if that’s right a picture below may help with the answer too)
Answer:
The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F
Explanation:
I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.
Answer:
The right solution is "28.45%".
Explanation:
The given values are:



and,





At 45,
⇒ 

At
,

or,

then,
⇒ 

hence,
The isentropic efficiency of turbine will be:
⇒ 

(%)
The thermal efficiency of cycle will be:
⇒ 

(%)