Answer:
a) Cr = Co - Fx / D
b) ΔC / Δx = ( CR - Cr ) / ( xR - xRo )
Explanation:
A) Derive an expression for the profile c(r) inside the tissue
F = DΔC / X = D ( Co - Cr ) / X ------ 1
where : F = flux , D = drug diffusion coefficient
X = radial distance between Ro and R
Hence : Cr = Co - Fx / D
B) Express the diffusive flux at outer surface of the balloon
Diffusive flux at outer surface = ΔC / Δx = CR - Cr / xR - xRo
Answer:
hello your question has some missing information attached to the answer is the missing component
Answer : αaxial,p = -6.034 ksi ( compressive )
αbend,p = 19.648 ksi ( tensile )
Explanation:
αaxial, p = equation 1
αbend, p = equation 2
P = load = 35 kips
A = area of column = 5.8
d = column cross section depth = 9.5 in
= 55.0
Hence equation 1 becomes
αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )
equation 2 becomes
αbend, p = = + 19.648 ksi ( tensile )
Answer:
(a) attached below
(b)
(c)
(d) Ω
(e) and
Explanation:
Given data:
(a) Draw the power triangle for each load and for the combined load.
°
°
≅
≅
The negative sign means that the load 2 is providing reactive power rather than consuming
Then the combined load will be
(b) Determine the power factor of the combined load and state whether lagging or leading.
or in the polar form
°
The relationship between Apparent power S and Current I is
Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.
(c) Determine the magnitude of the line current from the source.
Current of the combined load can be found by
(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω
Ω
(e) Compute the magnitude of the current in each capacitor and the line current from the source.
Current flowing in the capacitor is
Line current flowing from the source is
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice
Latent heat for ice H=336 KJ/kg
Specific heat for ice
We know that sensible heat given as
Heat for -15F to 32 F:
Q=858.69 KJ
Heat for 32 Fto 200 F:
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.