Ask question

Ask question

All categories

3 years ago

7

Fluid originally flows through a tube at a rate of 100 cm^3/s. To illustrate the sensitivity of the Poiseuille flow rate to vari

ous factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions:
a. Another tube is used with a radius 0.100 times the original.
b. Pressure difference increases by a factor of 1.60.
c. A new fluid with 3.00 times greater viscosity is substituted.
d. The tube is replaced by one having 4.00 times the length.

1 answer:

ankoles [38]3 years ago

5 0

Answer:

a) 0.01

b) 150 cm^3/s

c) 300 cm^3/s

d) 25 cm^3/s

Explanation:

a) We know that :

Q=ΔP/R

R=8ηl/π*r^4

Givens:

r^2 = 0.1 r_1

Plugging known information to get :

Q=ΔP/R

=ΔP*π*r^4/8*η*l

Q_2/r_2^4 =Q_1/r_1^4

Q_2=Q_1/r_1^4*r_2^4

=Q_1/r_1^4*r*0.0001*r_1^4

Q_2 = 0.01

b) From the rate flow of the fluid we know that :

Q=ΔP/R (1)

F=η*Av/l (2)

R=8*ηl/π*r^4 (3)

<em>Where: </em>

ΔP is the change in the pressure .

r is the raduis of the tube .

l is the length of the tube .

η is the coefficient of the vescosity of the fluid .

R is the resistance of the fluid .

Givens: Q1 = 100 cm^3/s , ΔP= 1.5

Plugging known information into EQ.1 :

Q=ΔP/R

Q_2/ΔP2=Q_1/ΔP

Q_2=150 cm^3/s

c) we know that :

F = η*Av/l

can be written as :

ΔP = F/A = η*v/l

Givens: η_2 = 3η_1

Q=ΔP/R

Q=η*v/l*R

Q_2/η_2=Q_1/η_1

Q_2=300 cm^3/s

d) We know that :

Q=ΔP/R

R=8*ηl/π*r^4

Givens: l_2 = 4*l_1

Plugging known information to get :

Q=ΔP/R

Q=ΔP*π*r^4/8*ηl

Q_2/l_2=Q_1/l_1

Q_2 = 25 cm^3/s

You might be interested in

If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with

SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

$\sigma_b = \frac{M}{I} \times y$

putting all value to get bending stress

$\sigma_b = \frac{50.112}{144} \times 3.96$

$\sigma_b = 1.37 psi$

Bending stress at point 3.96 is $\sigma_b$ = 1.37 psi

3 0

3 years ago

Which of the following is true about ideal standards? a.Ideal standards provide allowance for normal breakdowns and interruption

marysya [2.9K]

Answer: D. All of the choice A, B and C are correct.

8 0

3 years ago

The ignition temperature of 87-octane gasoline vapor is about 430 ∘C and, assuming that the working gas is diatomic and enters t

Flura [38]

Answer:

I have solved the problem below. I hope it will let you clear the concept.

For any inquiries ask me in the comments.

Explanation:

6 0

3 years ago

Which of the following is most characteristic of OSHA's relationship with professional trade organizations in residential constr

geniusboy [140]

Answer:

its beneficial

Explanation:

took the test

3 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago