All categories

3 years ago

Sketch the velocity profile for laminar and turbulent flow.

Engineering

1 answer:

Margarita [4]3 years ago

4 0

Answer:

The laminar flow is generally given in high viscosity fluids such as honey or oil, it has the characteristic of flowing in an orderly manner, the walls of the tube have a zero speed while in the center it has a maximum speed.

turbulent flow is characterized by fluid velocity vectors presenting themselves in a disorderly manner and in all directions.

I attached the drawings for the velocity profile in laminar and turbulent flow.

You might be interested in

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

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:

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

4 0

3 years ago

Fill in the blank to correctly complete the statement below.

frutty [35]

Answer:

The invention of the pendulum-driven ___clocks___ in the 1600s paved the way for a new industrial era.

4 0

3 years ago

50.38

klemol [59]

Answer:

International Building Code (IBC)

Explanation:

6 0

2 years ago

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

Given

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

Required

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

Solution

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0

3 years ago

Water flows through a pipe and enters a section where the cross sectional area is larger. Viscosity, friction, and gravitational

BaLLatris [955]

Answer:

(A) and (D)

Explanation:

1) P2 is less than P1, that is when P1 increases in pressure, the velocity V1 of the water also increases. Therefore, on the other hand, since P2 is directly proportional to V1, P2 and V2 will be less than P1 and V1 respectively.

2) For P2 greater than P1 and V2 also is greater than V1. Since P2 is directly proportional to V2, hence, when P2 increases in pressure, P1 reduces in pressure. Similarly, velocity, V2 also increases and V1 reduces.

3 0

3 years ago