Answer: the average velocity decreases
Explanation:
From the provided data we have:
Vessel avg. diameter[mm] number
Aorta 25.0 1
Arteries 4.0 159
Arteioles 0.06 1.4*10^7
Capillaries 0.012 2.9*10^9
from the information, let 
be the mass flow rate, 
is density, n number of vessels, and A is the cross-section area for each vessel
the flow rate is constant so it is equal for all vessels,
The average velocity is related to the flow rate by:
we clear the side where v is in:
area is π*R^2 where R is the average radius of the vessel (diameter/2)
we get:
you can directly see in the last equation that if we go from the aorta to the capillaries, the number of vessels is going to increase ( n will increase and R is going to decrease ) . From the table, R is significantly smaller in magnitude orders than n, therefore, it wont impact the results as much as n. On the other hand, n will change from 1 to 2.9 giga vessels which will dramatically reduce the average blood velocity
Answer:
0.16 micron per day
Explanation:
Given:
The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m
Initial tensile stress, σ₁ = 120 MPa
Final stress = 30 MPa
now from Griffith's equation, we have
![\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}](https://tex.z-dn.net/?f=%5Csigma%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20a%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
where,
Gc and E are the material constants
now,
for the initial stage
![120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}](https://tex.z-dn.net/?f=120%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20%280.1%5Ctimes10%5E%7B-6%7D%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
........{1}
and for the final case
![30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}](https://tex.z-dn.net/?f=30%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20a_2%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
............{2}
on dividing 1 by 2, we get
![\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B120%7D%7B30%7D%3D%5B%5Cfrac%7Ba_2%7D%7B0.1%5Ctimes10%5E%7B-6%7D%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
or
a₂ = 4² × 0.1 × 10⁻⁶ m
or
a₂ = 1.6 micron
Now,
the change from 0.1 micron to 1.6 micron took place in 10 days
therefore, the rate at which the crack is growing = 
or
average rate of change of crack = 0.16 micron per day
Answer:
W = 12.8 KW
Explanation:
given data:
mass flow rate = 0.2 kg/s
Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)
we know that maximum possible efficiency is given as
rate of heat transfer is given as
Maximuim power is given as
W = 0.2038 * 62.7
W = 12.8 KW
Answer:
See explanation
Explanation:
Given:
Initial pressure,
p
1
=
15
psia
Initial temperature,
T
1
=
80
∘
F
Final temperature,
T
2
=
200
∘
F
Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.
R
=
0.04513
Btu/lbm.R
C
v
=
0.158
Btu/lbm.R
Find the work done during the isobaric process.
w
1
−
2
=
p
(
v
2
−
v
1
)
=
R
(
T
2
−
T
1
)
=
0.04513
(
200
−
80
)
w
1
−
2
=
5.4156
Btu/lbm
Find the change in internal energy during process.
Δ
u
1
−
2
=
C
v
(
T
2
−
T
1
)
=
0.158
(
200
−
80
)
=
18.96
Btu/lbm
Find the heat transfer during the process using the first law of thermodynamics.
q
1
−
2
=
w
1
−
2
+
Δ
u
1
−
2
=
5.4156
+
18.96
q
1
−
2
=
24.38
Btu/lbm
Answer:
Q = 424523.22 kw
Explanation:
k = 48.9 W/m - K
c = 0.115 KJ/kg- K
T_∞ = 35 degree celcius
velocity of air stream = 15 m/s
D = 40 cm
L = 200 cm
mass flow rate
solving for h
h = 675.6 kw/m^2K
Q = 675.6*2.513*(285-35)
Q = 424523.22 kw
