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3 years ago

14

Consider a steady developing laminar flow of water in a constant-diameter horizontal discharge pipe attached to a tank. The flui

d enters the pipe with nearly uniform velocity V and pressure P1. The velocity profile becomes parabolic after a certain distance with a momentum correction factor of 2 while the pressure drops to P2. Identify the correct relation for the horizontal force acting on the bolts that hold the pipe attached to the tank.

1 answer:

dalvyx [7]3 years ago

6 0

Answer:

hello attached is the free body diagram of the missing figure

Fr = $\frac{\pi }{4} D^2 [ ( P1 - P2) - pV^2 ]$

Explanation:

Average velocity is constant i.e V1 = V2 = V

The momentum equation for the flow in the Z - direction can be expressed as

-Fr + P1 Ac - P2 Ac = mB2V2 - mB1V1 ------- equation 1

Fr = horizontal force on the bolts

P1 = pressure of fluid at entrance

V1 = velocity of fluid at entrance

Ac = cross section area of the pipe

P2 and V2 = pressure and velocity of fluid at some distance

m = mass flow rate of fluid

B1 = momentum flux at entrance , B2 = momentum flux correction factor

Note; average velocity is constant hence substitute V for V1 and V2

equation 1 becomes

Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )

Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2

equation for mass flow rate

m = <em>p</em>AcV

<em>p</em> = density of the fluid

insert this into equation 2 EQUATION 2 BECOMES

Fr = ( P1 - P2) Ac - <em>p</em>AcV^2

= Ac [ (P1 - P2) - pV^2 ] ---------- equation 3

Note Ac = $\frac{\pi }{4} D^2$

Equation 3 becomes

Fr = $\frac{\pi }{4} D^2$ [ (P1 -P2 ) - pV^2 ] ------- relation for the horizontal force acting on the bolts

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Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

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