Water vapor initially at 3.0 MPa and 300°C (state 1) is contained within a piston- cylinder. The water is cooled at constant volume until its temperature is 200°C (state 2). The water is then compressed isothermally to a state where the pressure is 2.5 MPa (state 3).a. Locate states 1, 2, and 3 on a T-v and P-v diagram.b. Determine the specific volume at all three states.c. Calculate the compressibility factor Z at state 1 and comment.d. Find the quality (if applicable) at all three states.
Answers
Answer:
a. T-V and P-V diagram are included
b. State 1: Specific volume = 0.0811753 m³/kg
State 2: Specific volume = 0.0811753 m³/kg
State 3: Specific volume = 0.0804155 m³/kg
c. Z = 51.1
d. Quality for state 1 = 100%
Quality for state 2 = 63.47%
Quality for state 3 = 100%
Explanation:
a. T-V and P-V diagram are included
b. State 1: Water vapor
P₁ = 3.0 MPa = 30 bar
T₁ = 300°C = 573.15
Saturation temperature = 233.86°C Hence the steam is super heated
Specific volume = 0.0811753 m³/kg
State 2:
Constant volume formula is P₁/T₁ = P₂/T₂
Specific volume = 0.0811753 m³/kg
T₂ = 200°C = 473.15
Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa
At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg
[tex]v_g[/tex] = 0.127222 m³/kg
State 3:
P₃ = 2.5 MPa
T₃ = 200°C
Isothermal compression P₂V₂ = P₃V₃
V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg
Specific volume = 0.0804155 m³/kg
2) Compressibility factor is given by the relation;
[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]
Z = 51.1
3) Gas quality, x, is given by the relation
[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]
Quality at state 1 = Saturated quality = 100%
State 2 Vapor + liquid Quality
Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%
State 3: Saturated vapor, quality = 100%.
Related Questions
The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.The uniform sign has a weight of 1500 lb and is supported bythe pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in.. If the face of the sign issubjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at pointsC and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.
Answers
Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
[tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Answers
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answers
Answer:
Explanation:
Given that,
The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]
The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]
The area of the styrofoam [tex]A_s=11m^2[/tex]
The thickness of the styrofoam [tex]t_s=0.20m[/tex]
The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]
The thermal conductivity of the styrofoam [tex]k_s=0.010J(s.m.C^o)[/tex]
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]
The heat loss due to conduction in the wall is
[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]
The net heat loss of the wall and the window is
[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]
The percentage of heat lost by the window is
[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]
In contouring, it is necessary to measure position and not velocity for feedback.
a. True
b. False
In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
a. True
b. False
Job shop is another term for process layout.
a. True
b. False
Airplanes are normally produced using group technology or cellular layout.
a. True
b. False
In manufacturing, value-creating time is greater than takt time.
a. True
b. False
Answers
Answer:
(1). False, (2). True, (3). False, (4). False, (5). True.
Explanation:
The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.
(1). In contouring, it is necessary to measure position and not velocity for feedback.
ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.
(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
ANSWER: a=> True.
(3). Job shop is another term for process layout.
ANSWER: b => False
JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.
(4). Airplanes are normally produced using group technology or cellular layout.
ANSWER: b => False.
(5). In manufacturing, value-creating time is greater than takt time.
ANSWER: a => True.
WHAT IS A VACUOMETER?
Answers
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15 mm. The bar is 270-mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine Poisson’s ratio.
Answers
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5 5 m/s and the manometer reading is h 5 58 cm, estimate the total force resisted by the fl ange bolts.
Answers
Answer:
The total force resisted by the flange bolts is 163.98 N
Explanation:
Solution
The first step is to find the pipe cross section at the inlet section
Now,
A₁ = π /4 D₁²
D₁ = diameter of the pipe at the inlet section
Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²
=50.265 cm² * ( 1 m²/100² cm²)
= 5.0265 * 10^⁻³ m²
Secondly, we find cross section area of the pipe at the inlet section
A₂ = π /4 D₂²
D₂ = diameter of the pipe at the inlet section
Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²
= 19.63 cm² * ( 1 m²/100² cm²)
= 1.963 * 10^⁻³ m²
Now,
we write down the conversation mass relation which is stated as follows:
Q₁ = Q₂
Where Q₁ and Q₂ are both the flow rate at the exist and inlet.
We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂
So,
V₁ and V₂ are defined as the velocities at the inlet and exit
We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂
= 5.0265 * 5 = 1.963 * V₂
V₂ = 12.8 m/s
Note: Kindly find an attached copy of the part of the solution to the given question below
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.
Answers
Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]
Carrier frequency 100kHz
Upper side band
[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]
A) The frequency Components of the Detector Output are;
80 Hz, 120 Hz and 380 Hz
B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz
Message SignalsA) We are given the frequency components in the message signal as;
f1 = 100Hzf2 = 200Hzf3 = 400HzWe are told that the carrier signal has a frequency; fc = 100kHz
Thus, the frequency components generated are;
Lower side band:
100 kHz - 100 Hz = 99.9 kHz100 kHz - 200 Hz = 99.8 kHz100 kHz - 400 Hz = 99.6 kHzUpper side band:
100 kHz + 100 Hz = 100.1 kHz100 kHz + 200 Hz = 100.2 kHz100 kHz + 400 Hz = 100.4 kHzWe are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.
Thus, the output frequencies are;
100.02 kHz - 100.1 kHz = 80 Hz
100.02 kHz - 100.2 kHz = 180 Hz
100.02 kHz - 100.4 kHz = 380 Hz
B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;
100.02 kHz - 99.9 kHz = 120 Hz
100.02 kHz - 99.8 kHz = 220 Hz
100.02 kHz - 99.6 kHz = 420 Hz
Read more about Message Signals at; https://brainly.com/question/25904079
g A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answers
Answer:
the overall elongation δ of the bar is 1.2337 mm
Explanation:
From the information given :
According to the principle of superposition being applied to the axial load P of the system; we have:
[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]
where;
δ = overall elongation
[tex]\delta _{AB}[/tex] = elongation of bar AB
[tex]\delta _{BC}[/tex] = elongation of bar BC
[tex]\delta _{CD} =[/tex] elongation of bar CD]
If we replace; [tex]\dfrac{PL}{AE}[/tex] for δ and bt for area;
we have:
[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]
where ;
P = load
L = length of the bar
A = area of the cross-section
E = young modulus of elasticity
Let once again replace:
P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex] (since load in all member of AB, BC and CD will remain the same )
[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],
[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and
[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]
[tex]2\dfrac{b}{3}[/tex] for [tex]b_{BC}[/tex]
b for [tex]b_{CD}[/tex]
[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]
[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]
[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]
The stress in the central portion can be calculated as:
[tex]\sigma = \dfrac{P}{A}[/tex]
[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]
[tex]\sigma = \dfrac{3P}{2bt}[/tex]
So; Now:
[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]
[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]
δ = 1.2337 mm
Therefore, the overall elongation δ of the bar is 1.2337 mm
The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?
Answers
Find the given attachments for complete explanation
what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the uniform magnetic field and name it. How can you represent this output as the periodic function?
Answers
Answer:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
Explanation:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field
B ( wave flux density ) = Bm sinwt and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 4200 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assume uniformly accelerated motion. determine the number of revolutions that the motor executes
(a) in reaching its rated speed,
(b) in coating to rest.
Answers
Answer:
a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]
Explanation:
a) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]
[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 175\,rev[/tex]
b) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]
[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 2450\,rev[/tex]
You are standing at the edge of the roof of the engineering building, which is H meters high. You see Professor Murthy, who is h meters tall, jogging towards the building at a speed of v meters/second. You are holding an egg and want to release it so that it hits Prof Murthy squarely on top of his head. What formulas describes the distance from the building that Prof Murthy must be when you release the egg?
Answers
Answer:
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Explanation:
First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:
Vf = Vi + gt
where,
Vf = Velocity of egg at the time of hitting head of the Professor
Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)
g = acceleration due to gravity
t = time taken by egg to come down
Therefore,
Vf = 0 + gt
t = Vf/g ----- equation (1)
Now, we use 3rd euation of motion for Vf:
2gs = Vf² - Vi²
where,
s = height dropped = H - h
Therefore,
2g(H - h) = Vf²
Vf = √[2g(H - h)]
Therefore, equation (1) becomes:
t = √[2g(H - h)]/g
t = √[2(H - h)/g]
Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:
s = vt
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
A fully recrystallized sheet of metal with a thickness of 11 mm is to be cold worked by 40% in rolling. Estimate the necessary roll force if the sheet was 0.5 m wide and there was no lateral spreading during rolling. The strength coefficient is 200 MPa, the work hardening exponent of 0.1 and the roll contact length is 40 mm. Assume no friction.
Answers
Answer:
Roll force, F = 5.6 MN
Explanation:
Sheet width, b = 0.5 m
Roll contact length, [tex]l_{p} = 40 mm[/tex]
Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]
Thickness, h = 11 mm
Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:
Δh = 40% * 11 = 0.4 * 11 = 4.4 mm
Strain, e = (Δh)/h
e = 4.4/11 = 0.4
The roll force is calculated by the formula:
[tex]F = \sigma_{f} l_{p} b[/tex]
[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]
Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:
[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]
Solid spherical particles having a diameter of 0.090 mm and a density of 2002 kg/m3 are settling in a solution of water at 26.7C. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity and the Reynolds number.
Answers
Answer:
Settling Velocity (Up)= 2.048*10^-5 m/s
Reynolds number Re = 2.159*10^-3
Explanation:
We proceed as follows;
Diameter of Particle = 0.09 mm = 0.09*10^-3 m
Solid Particle Density = 2002 kg/m3
Solid Fraction, θ= 0.45
Temperature = 26.7°C
Viscosity of water = 0.8509*10^-3 kg/ms
Density of water at 26.7 °C = 996.67 kg/m3
The velocity between the interface, i.e between the suspension and clear water is given by,
U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]
D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)
D = 2.147
U = 0.0003m/s (n = 4.49)
Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s
Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3
Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel flow with a velocity of 12.5 m/sec. The plate has a length of 2.70 m (in the direction of the fluid flow), a width of 0.65 m, and is maintained at a constant temperature of 127.0°C. Determine the heat transfer rate from the top of the plate due to forced convection.
Answers
Answer:
Explanation:
Given that:
V = 12.5m/s
L= 2.70m
b= 0.65m
[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]
[tex]T_s= 127^0C = (127+273)= 400K[/tex]
P = 1atm
Film temperature
[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]
dynamic viscosity =
[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]
density = 0.9946kg/m³
Pr = 0.708564
K= 229.7984 * 10⁻³w/mk
Reynolds number,
[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]
[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]
we have,
[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]
we have,
heat transfer rate from top plate
[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]
You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answers
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity (m/s) to which water can be accelerated by the nozzles before striking the turbine blades.
Answers
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
[tex]V2 = 37.417ms^{-1}[/tex]
Explanation:
Given the following data;
Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.
Nozzle outlets = 100kPa = 100000pa.
Density of water = 1000kg/m³.
We would apply, the Bernoulli equation between the inlet and outlet;
[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]
Where, V1 is approximately equal to zero(0).
Z[tex]z_{1} = z_{2}[/tex]
Therefore, to find the maximum velocity, V2;
[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]
[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]
[tex]V2 = \sqrt{2(800-100)}[/tex]
[tex]V2 = \sqrt{2(700)}[/tex]
[tex]V2 = \sqrt{1400}[/tex]
[tex]V2 = 37.417ms^{-1}[/tex]
Hence, the maximum velocity, V2 is 37.417m/s
Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most transmissions today use compound (multiple) planetary gear sets. Which technician is correct?
Answers
Answer:
Both technician A and technician B are correct
Explanation:
A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.
One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.
So, both technician A and technician B are correct.
Two blocks of rubber (B) with a modulus of rigidity G = 14 MPa are bonded to rigid supports and to a rigid metal plate A. Knowing that c = 80 mm and P = 46 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 7 mm.
Answers
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]
Making a the subject formula;
[tex]a = \frac{Gd}{r}[/tex]
Substituting into the above formula;
[tex]a = \frac{14000000*0.007}{1400000}[/tex]
[tex]a = \frac{98000}{1400000}[/tex]
[tex]a = 0.07m or 70mm[/tex]
a = 0.07m or 70mm.
Also, shearing stress;
[tex]r = \frac{P}{2bc}[/tex]
Making b the subject formula;
[tex]b = \frac{P}{2cr}[/tex]
Substituting into the above equation;
[tex]b = \frac{46000}{2*0.08*1400000}[/tex]
[tex]b = \frac{46000}{224000}[/tex]
[tex]b = 0.205m or 205mm[/tex]
b = 0.205m or 205mm
there is usually a positive side and a negative side to each new technological improvement?
Answers
Answer:
positive sides:
low cost improves production speedless timeeducational improvementsnegative sides:
unemployment lot of space required increased pollution creates lots of ethical issuesA piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à 2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à 3: Constant volume cooling to p3=5 bar Process 3à 4: Constant pressure heating Process 4à 1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.
Answers
Answer:
[tex]W_{net} = - 1223 kJ[/tex]
Explanation:
State 1:
[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]
State 2:
[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]
State 3:
[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]
[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]
State 4:
[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]
Calculate the network done for the cycle
[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]
An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 750 kPa and 48°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. (Round the final answers to the nearest whole number.)
Answers
Answer:
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
Explanation:
An ideal gas is represented by the following model:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.
[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]
As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]
Now, final pressure is cleared:
[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]
[tex]P_{2} = 375\,kPa[/tex]
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
two opposite poles repel each other
Answers
Answer:
South Pole and South Pole or North Pole and North Pole.
While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answers
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
A float valve, regulating the flow of water into a reservoir, is shown in the figure. The spherical float (half of the sphere is submerged) is 0.1553 m in diameter. AOB is the weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which flows into the reservoir. The link is mounted on a frictionless hinge at O, and the angle AOB is 135o. The length of OA is 253 mm and the distance between the center of the float and the hinge is 553 mm. When the flow is stopped AO will be vertical. The valve is to be pressed on to the seat with a force of 10,53 N to be completely stop the flow in the reservoir. It was observed that the flow of water is stopped, when the free surface of water in the reservoir is 353 mm below the hinge at O. Determine the weight of the float sphere.
Answers
Answer:
9.29 N . . . . weight of 0.948 kg sphere
Explanation:
The sum of torques on the link BOA is zero, so we have ...
(right force at A)(OA) = (up force at B)(OB·sin(135°))
Solving for the force at B, we have ...
up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N
This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.
∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg
__
The center of the sphere of diameter 0.1553 m is below the waterline by ...
(553 mm)cos(45°) -(353 mm) = 38.0300 mm
The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...
V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L
So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...
1.64336 kg -0.695205 kg ≈ 0.948 kg
The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N
The weight of the 0.948 kg float sphere is about 9.29 N.
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice
Answers
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70% of the carnot effieciency. if 2kj of work are produced, how much heat is rejected to the low temperature reservior
Answers
Answer:
Heat rejected to cold body = 3.81 kJ
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]
eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]
eff = 1 - 0.25 = 0.75
efficiency of the heat engine = 70% of 0.75 = 0.525
work done by heat engine = 2 kJ
eff. of heat engine is gotten as = W/Q
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = 3.81 kJ
Given in the following v(t) signal.
a. Find the first 7 harmonics of the Fourier series function in cosine form.
b. Plot one side spectrum
c. Find the first 7 harmonics of the Fourier series function in exponential form.
d. Plot two side spectrum Given in the following v(t) signal.
Answers
Answer:
Check the v(t) signal referred to in the question and the solution to each part in the files attached
Explanation:
The detailed solutions of parts a to d are clearly expressed in the second file attached.
A shell-and tube heat exchanger (two shells, four tube passes) is used to heat 10,000 kg/h of pressurized water from 35 to 120 oC with 5000 kg/h pressurized water entering the exchanger at 300 oC. If the overall heat transfer coefficient is 1500 W/m^2-K, determine the required heat exchanger area.
Answers
Answer:
4.75m^2
Explanation:
Given:-
- Temperature of hot fluid at inlet: [tex]T_h_i = 300[/tex] °C
- Temperature of cold fluid at outlet: [tex]T_c_o = 120[/tex] °C
- Temperature of cold fluid at inlet: [tex]T_c_i = 35[/tex] °C
- The overall heat transfer coefficient: U = 1500 W / m^2 K
- The flow rate of cold fluid: m_c = 10,00 kg/ h
- The flow rate of hot fluid: m_h = 5,000 kg/h
Solution:-
- We will evaluate water properties at median temperatures of each fluid using table A-4.
Cold fluid: Tci = 35°C , Tco = 35°C
Tcm = 77.5 °C ≈ 350 K --- > [tex]C_p_c = 4195 \frac{J}{kg.K}[/tex]
Hot fluid: Thi = 300°C , Tho = 150°C ( assumed )
Thm = 225 °C ≈ 500 K --- > [tex]C_p_h = 4660 \frac{J}{kg.K}[/tex]
- We will use logarithmic - mean temperature rate equation as follows:
[tex]A_s = \frac{q}{U*dT_l_m}[/tex]
Where,
A_s : The surface area of heat exchange
ΔT_lm: the logarithmic differential mean temperature
q: The rate of heat transfer
- Apply the energy balance on cold fluid as follows:
[tex]q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = \frac{10,000}{3600} * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W[/tex]
- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :
[tex]T_h_o = T_h_i - \frac{q}{m_h * C_p_h} \\\\T_h_o = 300 - \frac{9.905*10^5}{\frac{5000}{3600} * 4660} \\\\T_h_o = 147 C[/tex]
- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).
- So the relations from the figure 11.11 are:
[tex]P = \frac{T_c_o - T_c_i}{T_h_i - T_c_i} \\\\P = \frac{120 - 35}{300 - 35} \\\\P = 0.32[/tex]
[tex]R = \frac{T_h_i - T_h_o}{T_c_o - T_c_i} \\\\R = \frac{300 - 147}{120 - 35} \\\\R = 1.8[/tex]
Therefore, P = 0.32 , R = 1.8 ---- > F ≈ 0.97
- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:
[tex]dT_l_m = \frac{( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) }{Ln ( \frac{( T_h_i - T_c_o )}{( T_h_o - T_c_i )} ) } \\\\dT_l_m = \frac{( 300 - 120 ) - ( 147 - 35 ) }{Ln ( \frac{( 300-120 )}{( 147-35)} ) } \\\\dT_l_m = 143.3 K[/tex]
- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:
[tex]dT_l = F*dT_l_m = 0.97*143.3 = 139 K[/tex]
- The required heat exchange area ( A_s ) can now be calculated:
[tex]A_s = \frac{9.905*10^5 }{1500*139} \\\\A_s = 4.75 m^2[/tex]
