A T-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)?


Answer 1





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


3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out



divide both sides of the equation by 48.03


4. Subtract the equations obtained in steps 3 and 4

m1            +      m2       =  m3


0.376m1   +  1.832(m2) =m3



solving for m2



5. Mass flow is the product of density by velocity across the cross-sectional area


internal Diameter for  2" Sch 40=2.067in=0.17225ft

A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2

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




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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.



115 ⁰C


Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_(1) +q_(2) =-q_(3) -----eqution 1


q_(1) is the heat absorbed by the solid at 0⁰C

q_(2) is the heat absorbed by the liquid at 0⁰C

q_(3) the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature


q=n*\delta {_f_u_s} -------equation 3


q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_(2) = -q_(3)

Step 4: calculate the final temperature of the water

79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)

Substitute in the values; we will have,

79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})

79.13 kJ + 990.66J* (T_(f)-218}) = -1463J*(T_(f)-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_(f)-218}) = -1.463KJ*(T_(f)-100})

79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3

collect like terms,

2.45366T_(f) = 283.133

T_(f) = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.


Given Information:  

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:  

Change in length = ?  

Change in diameter = ?  


Change in length = 0.01257 mm

Change in diameter = -0.003772 mm


Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

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g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball



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Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t =0.5s

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As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None



B-  Velocity


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Where is the density of the material greater, at point B or point C?
Explain why.


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