Answer:
The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )
Explanation:
Liquids stored at : 1 atm , 25⁰c and they are vented with air
Determining whether the equilibrum vapor above the liquid will be flammable
We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable
A) For acetone
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 16.6513
B = 2940.46
C = -35.93
T = 298 k input values into Antoine equation
therefore ; [tex]p^{out}[/tex] = 228.4 mg
calculate volume percentage using Dalton's law
= V% = (saturation vapor pressure / pressure ) *100
= (228.4 mmHg / 760 mmHg) * 100 = 30.1%
The liquid is not flammable because its UFL = 12.8%
B) For Benzene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A= 15.9008
B = 2788.52
C = -52.36
T = 298 k input values into the above equation
[tex]p^{out}[/tex] = 94.5 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (94.5 / 760 ) * 100 = 12.4%
Benzene is not flammable under the given conditions because its UFL =7.1%
C) For cyclohexane
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 15.7527
B = 2766.63
c = -50.50
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 96.9 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= ( 96.9 mmHg /760 mmHg) * 100 = 12.7%
cyclohexane not flammable under the given conditions because its UFL= 8%
D) For Toluene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten from appendix E ( chemical process safety (3rd edition) )
A = 16.0137
B = 3096.52
C = -53.67
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 28.2 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (28.2 mmHg / 760 mmHg) * 100 = 3.7%
Toluene is flammable under the given conditions because its UFL= 7.1%
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wire required to create a secondary impedance of 25 ohms
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.
Answer:
[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,
[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]
For the maximum value of internal torque
[tex]\dfrac{dt(x)}{dx}=0[/tex]
Therefore
[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]
Thus the maximum internal torque will be at x= 0.25 L
[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select from a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Answer:
59°C
Explanation:
Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)
and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)
Therefore the minimum and maximum heat capacities are:
Cmin = Cc = 5803.2(W/K)
Cmax = Ch = 11634.3(W/K)
The capacity ratio is:
Cr = Cmin / Cmax = 0.499 = 0.5
The maximum possible heat transfer rate is:
Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)
And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99
Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)
Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C