A device called oacillatwr is used to swnd waves along a streatched string to send waves along a streatched string. The string i
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Answer:
m2=3.2722lbm/s
Explanation:
Hello!
To solve this problem follow the steps below
1. Find water densities and entlapies in all states using thermodynamic tables.
note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)
through prior knowledge of two other properties, such as pressure and temperature.
D1=Density(Water;T=50;x=0) =62.41 lbm/ft^3
D2=Density(Water;T=120;x=0) =61.71 lbm/ft^3
D3=Density(Water;T=80;x=0) =62.21 lbm/ft^3
h1=Enthalpy(Water;T=50;x=0) =18.05 BTU/lbm
h2=Enthalpy(Water;T=120;x=0) =88 BTU/lbm
h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm
2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit
m1+m2=m3
3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out
m1h1+m2h2=m3h3
18.05(m1)+88(m2)=48.03(m3)
divide both sides of the equation by 48.03
0.376(m1)+1.832(m2)=m3
4. Subtract the equations obtained in steps 3 and 4
m1 + m2 = m3
-
0.376m1 + 1.832(m2) =m3
--------------------------------------------
0.624m1-0.832m2=0
solving for m2
(0.624/0.832)m1=m2
0.75m1=m2
5. Mass flow is the product of density by velocity across the cross-sectional area
m1=(D1)(A)(v1)
internal Diameter for 2" Sch 40=2.067in=0.17225ft
m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s
6. use the equation from step 4 to find the mass flow in 2
0.75m1=m2
0.75(4.3629)=m2
m2=3.2722lbm/s