Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
[tex]density=\frac{P}{RT}[/tex]
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)= gas constant ideal for air
[tex]density=\frac{120}{(0.287)(293)}=1.43kg/m^3[/tex]
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
The required work is "281.4 Kj/kg" and the power is "4 Kw".
According to the question,
As we know,
→ [tex]Density = \frac{P}{RT}[/tex]
[tex]= \frac{120}{0.287\times 293}[/tex]
[tex]= 1.43 \ kg/m^3[/tex]
and,
→ [tex]Mass = \frac{Density}{Flow \ rate}[/tex]
[tex]= \frac{1.43}{0.01}[/tex]
[tex]= 0.0143 \ kg/s[/tex]
hence,
The work will be:
→ [tex]w = C_p (T_1-T_2)[/tex]
By substituting the values,
[tex]= 1.005(300-20)[/tex]
[tex]= 281.4 \ KJ/kg[/tex]
and,
The power be:
→ [tex]P = mw[/tex]
[tex]= 0.0143\times 281.4[/tex]
[tex]= 4 \ kw[/tex]
Thus the answer above is correct.
Find our more information about work done here:
https://brainly.com/question/1634438
