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
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Learn lore about problem solving here:
brainly.com/question/24528689
Answer:
EH buddy use a sparkplug use a drill through a hose im from da bronx
Explanation:
Answer:
Exit velocity 
m/s.
Explanation:
Given:
At inlet:
Properties of steam at 100 bar and 600°C
At exit:Lets take exit velocity 
We know that if we know only one property inside the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.
Properties of saturated steam at 80 bar
So the enthalpy of steam at the exit of turbine
Now from first law for open system
In the case of adiabatic nozzle Q=0,W=0
m/s
So Exit velocity m/s.
