As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
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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:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Answer:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
