At-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 btu/lbm-r). the temperatures are measured to be t1 = 50 f, t2 = 120 f at the inlets and t3 = 80 f at the exit. the pipe diameters are d1 = d3 = 2" sch 40 and d2 = 1¼" sch 40. if the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

m2=3.2722lbm/s

Explanation:

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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

heats surface water up

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