In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answer: hello your question lacks some data below is the missing data
Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol
H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.
H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
Values are gotten from the table named: blackbody radiation functions
a) Calculate the band emission fractions for the visible region
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
b)calculate wavelength corresponding to the maximum spectral intensity
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)
) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address
Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as an ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%).
Answer:
a) specific work output of the actual turbine is 73.14 Btu/lbm
b) the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c) Isentropic efficiency of the turbine is 70.76%
Explanation:
Given the data in the question;
For an adiabatic turbine; heat loss Q = 0
For Initial State;
p₁ = 120 psia
T₁ = 500°F = 959.67°R
from table; { Gas Properties of Air }
At T₁ = 959.67°R
[tex]s_1^0[/tex] = 0.74102 Btu/lbm°R
[tex]h_1[/tex] = 230.98 Btu/lbm
For Finial state;
p₂ = 15 psia
T₂ = 200°F = 659.67°R
[tex]s^0_{2a[/tex] = 0.64889 Btu/lbm°R
[tex]h_{2a[/tex] = 157.84 Btu/lbm
we know that R for air is 0.06855 Btu/lbm.R
a)
The specific work output of the actual turbine Wₐ is;
W[tex]_a[/tex] = [tex]h_1[/tex] - [tex]h_{2a[/tex]
we substitute
W[tex]_a[/tex] = 230.98 - 157.84
W[tex]_a[/tex] = 73.14 Btu/lbm
Therefore, specific work output of the actual turbine is 73.14 Btu/lbm
b)
amount of specific entropy generation during the irreversible process.
To determine the entropy generation [tex]S_{gen[/tex];
[tex]S_{gen[/tex] = ΔS = [tex]s_{2a[/tex] - [tex]s_1[/tex] = [tex]s^0_{2a[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute in our values
[tex]S_{gen[/tex] = 0.64889 - 0.74102 - 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]S_{gen[/tex] = 0.64889 - 0.74102 + 0.1425457
[tex]S_{gen[/tex] = 0.050416 Btu/lbm°R
Therefore, the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c)
Isentropic efficiency of turbine η[tex]_{is[/tex]
η[tex]_{is[/tex] = {actual work output] / [ ideal work output ] = ([tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )
Now, for an ideal turbine;
ΔS = 0 = [tex]s_{2s[/tex] - [tex]s_1[/tex]
so, [tex]s_{2s[/tex] - s₁ = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
0 = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
[tex]s^0_{2s[/tex] = [tex]s_1^0[/tex] + R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute
[tex]s^0_{2s[/tex] = 0.74102 + 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]s^0_{2s[/tex] = 0.74102 - 0.1425457
[tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R
Now, from table; { Gas Properties of Air }
At [tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R; [tex]h_{2s[/tex] = 127.614 Btu/lbm
η[tex]_{is[/tex] = [( [tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )] × 100%
we substitute
η[tex]_{is[/tex] = [( 230.98 - 157.84 ) / ( 230.98 - 127.614 )] × 100%
η[tex]_{is[/tex] = [ 73.14 / 103.366] × 100%
η[tex]_{is[/tex] = 0.70758 × 100%
η[tex]_{is[/tex] = 70.76%
Therefore, Isentropic efficiency of the turbine is 70.76%
A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system, indicate whether the following statements are true or false. (Note the the volume MUST CHANGE upon the addition of water.)
A. The concentration of HCN will increase.
B. The concentration of CN" will decrease.
C. The equilibrium concentration of Hy0 will remain the same 4
D. The pH will remain the same.
E. The ratio of [HCN]/[CN] will decrease.
banana with an average mass of 0.15 kg and average specific heat of 3.35 kJ/kg · °C is cooled from 20°C to 5°C. The amount of heat transferred from the banana is
a.
62.1 kJ
b.
7.5 kJ
c.
None of these
d.
6.5 kJ
e.
0.85 kJ
f.
17.7 kJ
Answer:
The amount of heat transferred from the banana is (-)7.54 KJ
Explanation:
As we know,
[tex]Q = m*c*\Delta T[/tex]
Q = Amount of heat transferred
m = mass of banana
[tex]T_2 = 5[/tex] degree Celsius
[tex]T_1 = 20[/tex] degree Celsius
The amount of heat transferred from the banana =
[tex]0.15 * 3.35 * (5 -20)\\-7.54[/tex]KJ (negative sign represents reduction in heat energy)
The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.
Answer:
the liquid woulriekwvhrnsshsnekwb ndrhwmoadi
Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4
Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
Deflection angle at B
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
slope at B
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
Deflection angle at C
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
Slope at C
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be
Answer:
the surface temperature of the plates when they come out of the oven is approximately 445 °C
Explanation:
Given the data in the question;
thickness t = 3 cm = 0.03 m
so half of the thickness L = 0.015 m
thermal conductivity of brass k = 110 W/m°C
Density p = 8530 kg/m³
specific heat [tex]C_p[/tex] = 380 J/kg°C
thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s
Temperature of oven T₀₀ = 700°C
initial temperature T[tex]_i[/tex] = 25°C
time t = 10 min = 600 s
convection heat transfer coefficient h = 80 W/m².K
Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.
So, using analytical one-term approximation method, the Fourier number > 0.2.
now, we determine the Biot number for the process
we know that; Biot number Bi = hL / k
so we substitute
Bi = hL / k
Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109
Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )
The interpolation method used to find the
λ₁ = 0.1039 and A₁ = 1.0018
so
The Fourier number т = ∝t/L²
we substitute
Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²
т = 0.02034 / 0.000225
т = 90.4
As we can see; 90.4 > 0.2
So, analytical one-term approximation can be used.
∴ Temperature at the surface will be;
θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation
θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )
so we substitute
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )
θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998
θ(L,t)[tex]_{wall[/tex] = 0.3775
so we substitute into equation 1
θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)
0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )
0.3775 = ( T(L,t) - 700 ) / ( - 675 )
0.3775 × ( - 675 ) = ( T(L,t) - 700 )
- 254.8125 = T(L,t) - 700
T(L,t) = 700 - 254.8125
T(L,t) = 445.1875 °C ≈ 445 °C
Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C
You will be hiking to a lake with some of your friends by following the trails indicated on a map at the trailhead. The map says that you will travel 1.7 mi directly north, then 2.7 mi in a direction 36° east of north, then finally 1.7 mi in a direction 15° north of east. At the end of this hike, how far will you be from where you started, and what direction will you be from your starting point?
Explanation:
