Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that 
For simplicity we work with 
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula 
the "normal" or "standard" weight of an object is given by 
when 
, so we need to find the value of 
that makes 
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides 
we cancel equal tems on both sides and obtain that 
Maybe it’s a vending machine, I’m confused too
Answer:
Q=486.49 KJ/kg
Explanation:
Given that
V= 0.2 m³
At initial condition
P= 2 MPa
T=320 °C
Final condition
P= 2 MPa
T=540°C
From steam table
At P= 2 MPa and T=320 °C
h₁=3070.15 KJ/kg
At P= 2 MPa and T=540°C
h₂=3556.64 KJ/kg
So the heat transfer ,Q=h₂ - h₁
Q= 3556.64 - 3070.15 KJ/kg
Q=486.49 KJ/kg
