1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MArishka [77]
3 years ago
13

) If the blood viscosity is 2.7x10-3 Pa.s, length of the blood vessel is 1 m, radius of the blood vessel is 1 mm, calculate the

tubular resistance of the blood vessel (in GPa.S/m3 ). If the blood pressure at the inlet of the above vessel is 43 mm Hg and if the blood pressure at the outlet of the above vessel is 38 mm Hg. Calculate the flow rate (in ml/min).
Engineering
1 answer:
Illusion [34]3 years ago
5 0

Answer:

a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b) the flow rate is 5.8 ml/min

Explanation:

Given the data in the question;

Length of blood vessel L = 1 m

radius r = 1 mm = 0.001 m

blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s

Now, we know that Resistance = 8μL / πr⁴

so we substitute

Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]

Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]

Resistance = 6.8755 ≈ 6.88 Gpa.s/m³

Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b)

blood pressure at the inlet of the vessel = 43 mm Hg

blood pressure at the outlet of the vessel = 38 mm Hg

flow rate = ?

we know that;

flow rate Q = ΔP / R

where ΔP is change in pressure and R is resistance.

ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg =  665 pa

R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹

so we substitute

Q =  665 pa / 114.5916 pa.m.ml⁻¹

Q = 5.8 ml/min

Therefore, the flow rate is 5.8 ml/min

You might be interested in
For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:
lana [24]

Answer:

a) The planar defect that exists is twin boundary defect.

b) The planar defect that exists is the stacking fault.

Explanation:      

I am using bold and underline instead of a vertical line.

a. A B C A B <u>C</u><u> </u>B A C B A

In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.

b. A B C A <u>B C  B C</u> A B C

In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.

6 0
4 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

NTU = 0.661

          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0
3 years ago
A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a stat
dalvyx [7]

Answer:

123.9 Btu

Explanation:

The energy balance on the air is:

∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W

ignore  ∆KE and ∆PE,

W = ∆U − Q = m(u2 − u1) − Q;                               (u2 − u1 = 51.94 Btu/lb)

ideal gas properties is attached

W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu

u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb

6 0
3 years ago
Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters
KonstantinChe [14]

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

Background image

I only know this

sorry for errors

4 0
3 years ago
Other questions:
  • A vacuum gauge indicates that the pressure of air in a closed chamber is 02 bar ( vacuum) The pressure of the surrounding atmosp
    10·1 answer
  • Who play 1v1 lol unblocked games 76
    8·2 answers
  • Cuáles son los sentidos comunes que se pueden utilizar, para identificar un metal, gases o líquido
    7·1 answer
  • Identify the fluid property responsible for the development of the velocity boundary layer and the kinds of fluids. For the idea
    14·1 answer
  • Cite five factors that lead to scatter in measured material properties. Upon what three criteria are factors of safety based? A
    6·1 answer
  • While discussing the diagnosis of an EI system in which the crankshaft and camshaft sensor tests are satisfactory but a spark te
    13·1 answer
  • Does anyone know how to fixes rope lights? Half of the rope is not on<br><br>Someone.. Anyone..
    9·1 answer
  • Which of the following identifies the beginning phase of the engineering design process?
    14·1 answer
  • Paper-like fiber clothing is intended as a one-use product. Is this statement TRUE or<br> FALSE?
    10·2 answers
  • Where can you go to find the system requirements?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!