Answer:
isentropic efficiency = 0.818
Explanation:
given data
pressure P1 = 95 kPa
temperature = 27°C
pressure P2 = 600 kPa
temperature = 277°C
to find out
isentropic efficiency of the compressor and exit temperature of the air
solution
we know from ideal gas of properties of air is
Pr1 at 27°C = 1.3860
and h1 at 300 K = 300.19 kJ/kg
and h2 at 550 K = 555.74 kJ/kg
and
we know equation for isentropic process that is
.........................1
put here value we get
solve we get Pr2
Pr2 = 8.75
by ideal gas of properties of air will be at Pr2
h2s = 508.66
T2s = 505.5 K
'so
isentropic efficiency will be here as
isentropic efficiency =
isentropic efficiency =
isentropic efficiency = 0.818
Answer:
The appropriate solution is "1481.76 N".
Explanation:
According to the question,
Mass,
m = 540 kg
Coefficient of static friction,
= 0.28
Now,
The applied force will be:
⇒
By substituting the values, we get
Answer:
-effective technical skills.
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Explanation:
is this what ur looking for? if so there ya go lol
Answer:
The molecular weight of the gas mixture is 35.38 g/mol.
Explanation:
The molecular weight of the gas can be found using the following equation:
Where:
m: is the mass = 230 g
n: is the number of moles
First, we need to find the number of moles using Ideal Gas Law:
Where:
P: is the pressure = 135 psi
V: is the volume = 15 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 465 °R (K = R*5/9)
Finally, the molecular weight of the gas is:
Therefore, the molecular weight of the gas mixture is 35.38 g/mol.
I hope it helps you!