One kilogram of "as received" yard trimmings is made up of approximately 620 g moisture, 330 g of decomposable organic matter (r
1 answer:
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Answer:
Part a: The volume of vessel is 4.7680 and total internal energy is 3680 kJ.
Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.
Explanation:
Part a:
As per given data
m=2 kg
T=80 °C =80+273=353 K
Dryness=70% vapour =0.7
<em>From the steam tables at 80 °C</em>
Specific volume of saturated vapours=v_g=3.40527
Specific volume of saturated liquid=v_f=0.00102
Now the relation of total specific volume for a specific dryness value is given as
Substituting the values give
Now the volume of vessel is given as
So the volume of vessel is 4.7680.
Similarly for T=80 and dryness ratio of 0.7 from the table of steam
Pressure=P=47.4 kPa
Specific internal energy is given as u=1840 kJ/kg
So the total internal energy is given as
The total internal energy is 3680 kJ.
So the volume of vessel is 4.7680 and total internal energy is 3680 kJ.
Part b
Volume of vessel is given as 1.6
mass is given as 2 kg
Pressure is given as 0.2 MPa or 200 kPa
Now the specific volume is given as
So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives
Temperature=T=120 °C
Quality=x=0.903 ≈ 90.3%
Specific internal energy =u=2330 kJ/kg
The total internal energy is given as
So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05
Also given
Therefore, = 1
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given :
Therefore, =
= 0.1182 MPa
a). Work transfer, δW =
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
=
=
=
= 197.7 kJ