An isolated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 150 kPa. An electric resistance heater inside the cylinder is now turned on, and 2200 kJ of energy is transferred to the steam. Determine the entropy change of the water during this process, in kJ/K.
Answers
Answer:
entropy change = 5.72 KJ/K
Explanation:
The energy balance for the system is given as;
ΔU = W_in - W_out
However, the entropy change of water is given as;
ΔS = m(S2 - S1)
ΔS = (V1/α1)*((h2 -h_liq150)/h_evap150)*(s_evap150)
This simplifies to;
ΔS = (V1/α1)*((h_liq150 + W_in*α1/V1 - h_liq150)*(s_evap150/h_evap150)
ΔS = W_in*(s_evap150/h_evap150)
We are given a constant pressure of 150 kPa
From saturated water table attached, at 150 KPa,
h_evap150 = h_fg = 2226 KJ/Kg
s_evap150 = s_fg = 5.7894 KJ/Kg.K
Since we are given W_in as 2200 kJ.
Thus;
ΔS = 2200(5.7894/2226)
ΔS = 5.72 KJ/K
Related Questions
Many HVACR industry publications are published by
Answers
Answer:
HVACR Industry Trade Groups
Explanation:
A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place reducing the maximum power transferable to 0.5 pu whereas before the fault, this power was 2.0 pu and after the clearance of the fault, it is 1.5 pu. Using equal area criterion, determine the critical clearing angle.
Answers
Answer:
critical clearing angle = 70.3°
Explanation:
Generator operating at = 50 Hz
power delivered = 1 pu
power transferable when there is a fault = 0.5 pu
power transferable before there is a fault = 2.0 pu
power transferable after fault clearance = 1.5 pu
using equal area criterion to determine the critical clearing angle
Attached is the power angle curve diagram and the remaining part of the solution.
The power angle curve is given as
= Pmax sinβ
therefore : 2sinβo = Pm
2sinβo = 1
sinβo = 0.5 pu
βo = [tex]sin^{-1} (0.5) = 30[/tex]⁰
also ; 1.5sinβ1 = 1
sinβ1 = 1/1.5
β1 = [tex]sin^{-1} (\frac{1}{1.5} )[/tex] = 41.81⁰
∴ βmax = 180 - 41.81 = 138.19⁰
attached is the remaining solution
The critical clearing angle = [tex]cos^{-1} 0.3372[/tex] ≈ 70.3⁰
Question 44
What should you do if you encounter a fishing boat while out in your vessel?
A
Make a large wake nearby.
B
Avoid making a large wake.
с
Pass on the side with the fishing lines.
D
Pass by close to the anglers.
Submit Answer
Answers
Answer:
The answer is B. Avoid making a large wake.
Explanation:
When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.
You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.
A truss is subjected to three loads. The truss is supported by a roller at A and by a pin joint at B. What is most nearly the reaction force at B
Answers
Answer:
Hello the diagram related to the question is missing attached is the diagram
Answer : 3833.33 KN
Explanation:
The most nearly reaction force at B
= ∑ Mb = 0 = 21Ay
= (2000 * 17.5 ) + ( 3000 * 10.5 ) + ( 4000 * 3.5 )
= 35000 + 31500 + 14000 = 80500
therefore Ay = 80500 / 21 = 3833.33 KN
x = 7/2 = 3.5m
There is one reservoir filled with water and also connected with one pipe of uniform cross-sectional diameter. Total head at section 1 is 27 m. At section 2, potential head is 3 m, gage pressure is 160 kPa, vvelocity is 4.5 m/s. Find the major head loss at section 2 in unit of m. Round to the nearest one decimal place.
Answers
Answer:
6.7 m
Explanation:
Total head at section 1 = 27 m
at section 2;
potential head = 3 m
gauge pressure P = 160 kPa = 160000 Pa
pressure head is gotten as
[tex]Ph =\frac{P}{pg}[/tex]
where p = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.81 m/^2
[tex]Ph =\frac{160000}{1000 * 9.81} = 16.309 m[/tex]
velocity = 4.5 m/s
velocity head Vh is gotten as
[tex]Vh = \frac{v^{2} }{2g}[/tex]
[tex]Vh = \frac{4.5^{2} }{2*9.81} = 1.03 m[/tex]
obeying Bernoulli's equation,
The total head in section 1 must be equal to the total head in section 2
The total head in section 2 = (potential head) + (velocity head) + (pressure head) + losses(L)
Equating sections 1 and 2, we have
27 = 3 + 1.03 + 16.309 + L
27 = 20.339 + L
L = 27 - 20.339
L = 6.661 ≅ 6.7 m
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 95 kPa and 27 degree Celsius.
(a) Determine the temperature after the heat-addition process.
(b) Determine the thermal efficiency.
(c) Determine the mean effective pressure. Solve the problem in the constant heat supposition.
Answers
Answer:
a) T₃ = 1818.8 K
b) η = 0.614 = 61.4%
c) MEP = 660.4 kPa
Explanation:
a) According to Table A-2 of The ideal gas specific heat of gases, the properties of air are as following:
At 300K
The specific heat capacity at constant pressure = [tex]c_{p}[/tex] = 1.005 kJ/kg.K,
The specific heat capacity at constant volume = [tex]c_{v}[/tex] = 0.718 kJ/kg.K
Gas constant R for air = 0.2870 kJ/kg·K
Ratio of specific heat k = 1.4
Isentropic Compression :
[tex]T_{2}[/tex] = [tex]T_{1}[/tex] [tex](v1/v2)^{k-1}[/tex]
= 300K ([tex]16^{0.4}[/tex])
[tex]T_{2}[/tex] = 909.4K
P = Constant heat Addition:
[tex]P_{3}v_{3} / T_{3} = P_{2} v_{2} /T_{2}[/tex]
[tex]T_{3}=v_{3}/v_{2}T_{2}[/tex]
2[tex]T_{2}[/tex] = 2(909.4K)
= 1818.8 K
b) [tex]q_{in}[/tex] = [tex]h_{3}-h_{2}[/tex]
= [tex]c_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{2}[/tex])
= (1.005 kJ/kg.K)(1818.8 - 909.4)K
= 913.9 kJ/kg
Isentropic Expansion:
[tex]T_{4}[/tex] = [tex]T_{3}[/tex] [tex](v3/v4)^{k-1}[/tex]
= [tex]T_{3}[/tex] [tex](2v_{2} /v_{4} )^{k-1}[/tex]
= 1818.8 K (2 / 16[tex])^{0.4}[/tex]
= 791.7K
v = Constant heat rejection
[tex]q_{out}[/tex] = μ₄ - μ₁
= [tex]c_{v} ( T_{4} - T_{1} )[/tex]
= 0.718 kJ/kg.K (791.7 - 300)K
= 353 kJ/kg
η[tex]_{th}[/tex] = 1 - [tex]q_{out}[/tex] / [tex]q_{in}[/tex]
= 1 - 353 kJ/kg / 913.9 kJ/kg
= 1 - 0.38625670
= 0.6137
= 0.614
= 61.4%
c) [tex]w_{net}._{out}[/tex] = [tex]q_{in}[/tex] - [tex]q_{out}[/tex]
= 913.9 kJ/kg - 353 kJ/kg
= 560.9 kJ/kg
[tex]v_{1} = RT_{1} /P_{1}[/tex]
= (0.287 kPa.m³/kg/K)*(300 K) / 95 kPa
= 86.1 / 95
= 0.9063 m³/kg = v[tex]_{max}[/tex]
[tex]v_{min} =v_{2} = v_{max} /r[/tex]
Mean Effective Pressure = MEP = [tex]w_{net,out}/v_{1} -v_{2}[/tex]
= [tex]w_{net,out}/v_{1}(1-1)/r[/tex]
= 560.9 kJ/kg / (0.9063 m³/kg)*(1-1)/16
= (560.9 kJ / 0.8493m³) (kPa.m³/kJ)
= 660.426 kPa
Mean Effective Pressure = MEP = 660.4 kPa
The temperature after the addition process is 1724.8k, the thermal efficiency of the engine is 56.3% and the mean effective pressure is 65.87kPa
Assumptions made:
The air standard assumptions are madeThe kinetic and potential energy changes are negligibleThe air in the system is an ideal gas with variable or different specific heat capacity.a) The temperature after the addition process:
Considering the process 1-2, Isentropic expansion
at
[tex]T_1=300k\\u_1=214.07kJ/kg\\v_o_1=621.3\\v_o_2=\frac{v_2}{v_1} *v_o_1[C.R=16]=v_2/v_1\\v_o_2=(v_2/v_1)v_o_1=1/16*621.2=38.825[/tex]
From using this value, v[tex]_o_2[/tex]=38.825, solve for state point 2;
[tex]T_2=862.4k\\h_2=890.9kJ/kg[/tex]
Considering the process 2-3 (state of constant heat addition)
[tex]\frac{p_3v_3}{t_3}=\frac{p_2v_2}{t_2} \\\\T_3=\frac{P_3V_3T_2}{V_2} \\T_3=(\frac{V_3}{V_2}) T_2\\\frac{v_3}{v_2}=2\\T_3=2(862.4)=1724.8k\\[/tex]
NB: p[tex]_3[/tex]≈p[tex]_2[/tex]
b) The thermal efficiency of the engine is
Q[tex]_i_n[/tex]=h[tex]_3-h_2[/tex] = 1910.6-890.9=1019.7kJ/kg
Considering process 3-4,
[tex]v_o_4=\frac{v_A}{v_2}\\ v_o_3 =\frac{V_a}{V_2}*\frac{v_2}{v_3}\\v_o_3=\frac{16}{2}*4.546\\v_o_3=36.37;v_4=659.7kJ/kg[/tex]
Q[tex]_o_u_t=v_4-u_1=659.7-214.07=445.3kJ/kg[/tex]
nth = [tex]1-\frac{Q_o_u_t}{Q_i_n}=1-\frac{445.63}{1019.7}=0.5629*100=56.3%[/tex]%
The thermal efficiency is 56.3%
W[tex]_n_e_t[/tex]=[tex]Q_i_n-Q_o_u_t=574.07kJ/kg[/tex]
[tex]v_1=\frac{RT_1}{p_1}=\frac{0.287*300}{95}=0.906m^3/kg\\v_2=v_1/16=0.05662m^3/kg\\[/tex]
Therefore, the mean effective pressure of the system engine is
[tex]\frac{W_n_e_t}{v_1-v_2}=675.87kPa[/tex]
The mean effective pressure is 65.87kPa as calculated above
Learn more about mean effective pressure
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For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are correct
Answers
Answer:
hello the answer options are missing here are the options
A)The thickness of the heated region near the plate is increasing
B)The velocities near the plates are increasing
C)The fluid temperature near the plate are increasing
ANSWER : all of the above
Explanation:
Laminar flow is the flow of a type of fluid across the surface of an object following regular paths and it is unlike a turbulent flow which flows in irregular paths (encountering fluctuations)
For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because :
The thickness of the heated region near the plate is increasingThe velocities near the plates are increasingThe fluid temperature near the plate are increasingThe liquid-phase reaction A + B → C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3 /min and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available: One is a gray 200.0 dm3 CSTR that can be heated to 77°C or cooled to 0°C, and the other is a white 800.0 dm3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. (Note: k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol.) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with CA ° = CB ° = 1 ???? after mixing at a temperature of 70°C?
Answers
Answer:
1.887 minutes
Explanation:
We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol
To solve this, first of all let's calculate the rate constant(k);
For this question, The formula is;
K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]
R is gas constant = 1.987 cal/mol.K
For temperature of 70°C which is = 70 + 273K = 343K, we have;
K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]
K(343) = 4.7 dm³/mol.min
The design equation is;
dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²
Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;
(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt
So, we'll get;
0.9/(1 - 0.9) = 4.77 × 1 × t
t = 9/4.77
t = 1.887 minutes
Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp
Answers
Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.
Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²
Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec
= 0.2000856 m³ /sec
Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K
Gas constant R=287 J/kg K
Then;
1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
A square concrete column is 4 in by 4 in cross-section and is subject to a compressive load P. If the compressive stress cannot exceed 4000 psi and the shear stress cannot exceed 1500 psi, the maximum allowable load P is most nearly:
Answers
Answer:
64000 lb
Explanation:
A square concrete column is 4 in by 4 in cross-section and is subject to a compressive load P. If the compressive stress cannot exceed 4000 psi and the shear stress cannot exceed 1500 psi.
The area of the square concrete column = 4 in × 4 in = 16 in²
The compressive stress (σ) cannot exceed 4000 psi.
Compressive stress is the ratio of load applied to the area. Therefore the maximum load is the product of the maximum compressive stress and the area. The maximum compressive stress is given as:
[tex]\sigma_{max}=\frac{P_{max}}{Area} \\P_{max}= \sigma_{max}*Area\\P_{max}=4000\ psi *16\ in^2\\P_{max}=64000\ lb[/tex]
Therefore the maximum allowable load P is 64000 lb
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
Answers
Answer:
No, the velocity profile does not change in the flow direction.
Explanation:
In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.
A two-dimensional flow field described by
V = (2x^2y + x)1 + (2xy^2 + y + 1 )j
where the velocity is in m/s when x and y are in meters. Determine the angular rotation of a fluid element located at x 0.5 m, y 1.0 m.
Answers
Answer:
the answer is
Explanation:
We now focus on purely two-dimensional flows, in which the velocity takes the form u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) With the velocity given by (2.1), the vorticity takes the form ω = ∇ × u = ∂v ∂x − ∂u ∂y k. (2.2) We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence ∂v ∂x − ∂u ∂y = 0. (2.3) We have already shown in Section 1 that this condition implies the existence of a velocity potential φ such that u ≡ ∇φ, that is u = ∂φ ∂x, v = ∂φ ∂y . (2.4) We also recall the definition of φ as φ(x, y, t) = φ0(t) + Z x 0 u · dx = φ0(t) + Z x 0 (u dx + v dy), (2.5) where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is even easier to establish when we restrict our attention to two dimensions. If we consider two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, Green’s Theorem implies thatTwo mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surroundings. The mixing process takes place at constant pressure with no work and negligible changes in kinetic and potential energies. Assume the gas has constant specific heats.
a. Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the inlet and exit mass flow rates.
b. Obtain an expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber.
c. For the special case of adiabatic mixing, show that the exit volume flow rate is the sum of the two inlet volume flow rates.
Answers
Answer:
(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp
(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp
(c) The volume flow rate at exit is V₃ =V₁ + V₂
Explanation:
Solution
Now
The system comprises of two inlets and on exit.
Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁
Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂
Mass flow rate enthalpy of fluid from exit be m₃ and h₃
Mixing chambers do not include any kind of work (w = 0)
So, both the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)
(a) Applying the mass balance of mixing chamber, min = mout
Applying the energy balance of mixing chamber,
Ein = Eout
min hin =mout hout
miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃
T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +
T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp
The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp
(b) From the ideal gas equation,
v =RT/PT
v₃ = RT₃/P₃
The volume flow rate at the exit, V₃ =m₃v₃
V₃ = m₃ RT₃/P₃
Substituting the value of T₃, we have
V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)
V₃ = R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)
Now
The mixing process occurs at constant pressure P₃=P₂=P₁.
Hence V₃ becomes:
V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp
V₃ =V₁ + V₂ + RQin/P₃Cp
Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp
(c) Now for an adiabatic mixing, Qin =0
Hence V₃ becomes:
V₃ =V₁ + V₂ + r * 0/P₃Cp
V₃ =V₁ + V₂ + 0
V₃ =V₁ + V₂
Therefore the volume flow rate at exit is V₃ =V₁ + V₂
A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two components, species-A and species-B. A stream of pure species-C is fed into the mixer to drive the extraction. The mixture is then fed to a separator where it is allowed to settle into two phases which are removed in separate streams. Each outlet stream contains all three species. The goal of the process is to produce an outlet stream with a high concentration of species-A. Given the data below, perform a degree- of-freedom analysis and determine the order in which systems of equations must be solved to characterize every stream in the process. Then solve the system for all unknown flow rates and compositions.
• One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.
• One inlet stream to the mixer is pure species-C, and the flow rate is unknown.
• One outlet stream from the separator is 85wt% species-A and 10wt% species-B.
• One outlet stream from the separator is 25wt% species-B and 70wt% species-C.
Answers
Answer:
One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.
Explanation:
The process consists of two units, a mixer and a separator. There are three species: A, B, and C.
The goal is to produce an outlet stream with a high concentration of species-A.
How to solveThere are two inlet streams to the mixer. One is 35wt% species-A and 65wt% species-B, and the other is pure species-C. The flow rate of the second stream is unknown.
There are two outlet streams from the separator. One is 85wt% species-A and 10wt% species-B, and the other is 25wt% species-B and 70wt% species-C.
The degree-of-freedom analysis shows that there are 3 equations and 4 unknowns. The order in which the systems of equations must be solved is:
Solve for the flow rate of the stream of pure species-C.
Solve for the compositions of the outlet streams from the separator.
Solve for the compositions of the inlet streams to the mixer.
The solution for all unknown flow rates and compositions is:
The flow rate of the stream of pure species-C is 50.0 kg/hr.
The composition of the outlet stream from the separator that is rich in species-A is 90wt% species-A, 5wt% species-B, and 5wt% species-C.
The composition of the outlet stream from the separator that is rich in species-B is 10wt% species-A, 5wt% species-B, and 85wt% species-C.
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In real world, sampling and quantization is performed in an analog to digital converter (ADC) and reconstruction is performed in a digital to analog converter (DAC). Which of the following statements hold true (fs denotes the sampling frequency)?
a. the reconstruction filter can be found in the DAQ
b. the antialiasing filter removes all frequencies of the continuous-time analog input signal that are above fs/2
c. the DAC needs to know the sampling frequency of the ADC to correctly reconstruct the signal.
d. the reconstructed continuous-time signal only contains frequencies up to fs/2
Answers
Answer:
b
Explanation:
a) ADC is located on DAQ filter but not the reconstruction filter
b) to remove aliasing, the sampling rate must be greater than or equal ot twice the highest frequency component in the input signal. In other words, all frequencies in input sgnal are less than fs/2. Therefore, frequencies greater than fs/2 are removed by anti-aliasing filter
c) the DAC can have different sampling rate from ADC
A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps, and feedwater heaters. Steam enters the first turbine at T1 (the highest temperature of the cycle) and saturated liquid exits the condenser at T7 (the lowest temperature of the cycle). The rate of heat transfer to the boiler (the only energy input to the system)is Qb. Determine the maximum possible efficiency and power output for this complex steam power plant design.
Answers
Answer:
Hello your question lacks some values here are the values
T1 = 500⁰c, T7 = 70⁰c, Qb = 240000 kj/s
answer : A) 56%
B) 134400 kw ≈ 134.4 Mw
Explanation:
Given values
T1 (tmax) = 500⁰c = 773 k
T7(tmin) = 70⁰c = 343 k
Qb = 240000 kj/s
A) Determine the maximum possible efficiency
[tex]n_{max}[/tex] = 1 - [tex]\frac{tmin}{tmax}[/tex] * 100
= 1 - ( 343 / 773 )
= 1 - 0.44 = 0.5562 * 100 ≈ 56%
B) Determine the power output for this complex steam power plant design
[tex]p_{out}[/tex] = Qb * max efficiency
= 240000 kj/s * 56%
= 240000 * 0.56 = 134400 kw ≈ 134.4 Mw
Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the flow rate is held constant, how will the pressure drop change
Answers
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
(a) A duct for an air conditioning system has a rectangular cross section of 1.8 ft × 8 in. The duct is fabricated from galvanized iron. Determine the Reynolds number for a flow rate of air of 5400 cfm at 100 °F and atmospheric pressure (g=0.0709 lbf/ft3 u=1.8×10-4ft2/s and m=3.96×10-7lbf.s/ft2) (9 points)
Answers
Answer:
Reynolds number = 654350.92
Explanation:
Given data:
Cross section of rectangular cross section = 1.8ft * 8 in ( 8 in = 2/3 ft )
Flow rate of air = 5400 cfm = 90 ft^3 / sec
v ( kinematic viscosity of air ) = 1.8*10^-4 ft^2/s
Reynolds number
Re = VDn / v
Dn ( hydraulic diameter ) = 4A / P
where A = area, P = perimeter
a = 1.8 ft ( length )
b = 2/3 ft ( width )
hence Dn = [tex]\frac{4(ab)}{2(a+b)}[/tex] = [tex]\frac{4(1.8*0.6667}{2(1.8+0.6667)}[/tex] = 0.9729 ft
V ( velocity of air flow ) = [tex]\frac{Q}{\pi /4 * Dn^2 }[/tex] = [tex]\frac{90}{\pi /4 * 0.9729^2 }[/tex] = 121.064 ft/sec
back to Reynolds equation
Re = VDn / v -------------- equation 1
V = 121.064 ft/sec
Dn = 0.9729 ft
v = 1.8*10^-4 ft^2/s
insert the given values into equation 1
Re = (121.064 * 0.9729 ) / 1.8*10^-4
= 654350.92
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses alternating layers of the storage material and the flow passage. Each layer of the storage material is an aluminum slab of width W=0.05 m, which is at an initial temperature of 25∘C25 ∘C. Consider conditions for which the storage unit is charged by passing a hot gas through the passages, with the gas temperature and the convection coefficient assumed to have constant values of T[infinity]=600∘CT [infinity]=600 ∘C and h=100W/m2⋅Kh=100W/m 2⋅K throughout the channel. How long will it take to achieve 75% of the maximum possible energy storage? What is the temperature of the aluminum at this time?
Answers
Answer:
the temperature of the aluminum at this time is 456.25° C
Explanation:
Given that:
width w of the aluminium slab = 0.05 m
the initial temperature [tex]T_1[/tex] = 25° C
[tex]T{\infty} =600^0C[/tex]
h = 100 W/m²
The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;
density ρ = 2702 kg/m³
thermal conductivity k = 231 W/m.K
Specific heat c = 1033 J/Kg.K
Let's first find the Biot Number Bi which can be expressed by the equation:
[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}[/tex]
[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}[/tex]
[tex]Bi = \dfrac{2.5}{231}[/tex]
Bi = 0.0108
The time constant value [tex]\tau_t[/tex] is :
[tex]\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}[/tex]
[tex]\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}[/tex]
[tex]\tau_t = \dfrac{2702* 0.025*1033}{100}[/tex]
[tex]\tau_t = 697.79[/tex]
Considering Lumped capacitance analysis since value for Bi is less than 1
Then;
[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}][/tex]
where;
[tex]Q = -\Delta E _{st}[/tex] which correlates with the change in the internal energy of the solid.
So;
[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}[/tex]
The maximum value for the change in the internal energy of the solid is :
[tex](pVc)\theta_1 = -\Delta E _{st}max[/tex]
By equating the two previous equation together ; we have:
[tex]\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}[/tex]
Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75
Thus;
[tex]0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}[/tex]
So;
[tex]0.75= [1-e^{\dfrac {-t}{ 697.79}}]}[/tex]
[tex]1-0.75= [e^{\dfrac {-t}{ 697.79}}]}[/tex]
[tex]0.25 = e^{\dfrac {-t}{ 697.79}}[/tex]
[tex]In(0.25) = {\dfrac {-t}{ 697.79}}[/tex]
[tex]-1.386294361= \dfrac{-t}{697.79}[/tex]
t = 1.386294361 × 697.79
t = 967.34 s
Finally; the temperature of Aluminium is determined as follows;
[tex]\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}[/tex]
[tex]\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}[/tex]
[tex]\dfrac{T - 600}{25-600}= 0.25[/tex]
[tex]\dfrac{T - 600}{-575}= 0.25[/tex]
T - 600 = -575 × 0.25
T - 600 = -143.75
T = -143.75 + 600
T = 456.25° C
Hence; the temperature of the aluminum at this time is 456.25° C
Instructions given by traffic police or construction flaggers _____. A. Are sometimes important to follow B. Are usually not important to follow C. Don't overrule laws or traffic control devices D. Overrule any other laws and traffic control devices
Answers
Answer:
D. Overrule any other laws and traffic control devices.
Explanation:
Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.
Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.
Answer:
D. Overrule any other laws and traffic control devices.
Explanation:
If a sky diver decides to jump off a jet in Arkansas
with the intention of floating through Tennessee to
North Carolina, then completing his journey in a
likely manner back to Arkansas by drifting North
from his last point. What state would be the third t
be drifted over and what is the estimated distance
between the zone and then drop point?
Answers
Answer:
The answer to this question can be defined as follows:
Explanation:
The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.
Technician A says that the micrometer operates on a simple principle: The spindle has 20 threads per inch, so one revolution of the thimble will advance or retract the spindle 1/20 of an inch. Technician B says that spindle has 50 threads per inch, so one revolution of the thimble will advance or retract the spindle 1/50 of an inch. Who is correct
Answers
Answer:
Explanation:
neither of the technicians is correct
A gold vault has 3 locks with a key for each lock. Key A is owned by the
manager whilst Key B and C are in the custody of the senior bank teller
and the trainee bank teller respectively. In order to open the vault door at
least two people must insert their keys into the assigned locks at the same
time. The trainee bank teller can only open the vault when the bank
manager is present in the opening.
i) Determine the truth table for such a digital locking system (4 marks)
ii) Derive and minimize the SOP expression for the digital locking system
Answers
Answer:
i) Truth Table:
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) The minimized sum of products (SOP) expression is
O = AC + AB
Explanation:
We have three inputs A, B and C
Let O is the output.
We are given two conditions to open the vault door:
1. At least two people must insert their keys into the assigned locks at the same time.
2. The trainee bank teller (C) can only open the vault when the bank manager (A) is present in the opening.
i) Construct the Truth Table
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) SOP Expression using Karnaugh-Map:
A 3 variable Karnaugh-map is attached.
The minimized sum of products (SOP) expression is
O = AC + AB
The orange pair corresponds to "AC" and the purple pair corresponds to "AB"
Bonus:
The above expression may be realized by using two AND gates and one OR gate.
Please refer to the attached logic circuit diagram.
Suppose a student carrying a flu virus returns to an isolated college campus of 9000 students. Determine a differential equation governing the number of students x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it. (Usek > 0for the constant of proportionality and x forx(t).)
Answers
Answer:
dx/dt = kx(9000-x) where k > 0
Explanation:
Number of students in the campus, n = 9000
Number of students who have contracted the flu = x(t) = x
Number of students who have bot yet contracted the flu = 9000 - x
Number of Interactions between those that have contracted the flu and those that are yet to contract it = x(9000 - x)
The rate of spread of the disease = dx/dt
Note: the rate at which the disease spread is proportional to the number of interactions between those that have contracted the flu and those that have not contracted it.
[tex]\frac{dx}{dt} \alpha [x(9000 -x)]\\[/tex]
Introducing a constant of proportionality, k:
dx/dt = kx(9000-x) where k > 0
Tengo un problema con steam y es que sale como que no he comprado un juego pero en realidad si lo he comprado pero por una pagina alterna pero no me lo detecta que hago?
Answers
Answer:
Cómo forzar a Steam a reconocer los juegos instalados.
1) Vuelva a instalar los juegos sin descargar.
2) Agregue la carpeta Steam Library manualmente.
3) Reconocer juegos de una nueva unidad
4) Utilizar .acf Cache para forzar el reconocimiento de juegos de Steam .acf Cache para forzar el reconocimiento de juegos de Steam.
Explanation:
1) Vuelva a instalar los juegos sin descargar.
- Inicia Steam y ve a Juegos.
- Seleccione y haga clic en instalar para el juego que Steam no pudo reconocer.
- Steam comenzará a descubrir los archivos existentes para el juego.
2) Agregue la carpeta Steam Library manualmente.
- Lanzar Steam.
- Haga clic en Steam y seleccione Configuración.
- Haz clic en la pestaña Descargas.
- Haga clic en las carpetas de la biblioteca de Steam.
- En la ventana emergente, haz clic en Agregar carpeta de biblioteca y selecciona la ubicación donde se guardan todos los datos de tu juego Steam.
- Haga clic en Seleccionar y cerrar la configuración de Steam.
- Salga de la aplicación Steam y reinicie Steam.
- Steam ahora debería reconocer los juegos instalados nuevamente y enumerarlos en la carpeta de juegos.
3) Reconocer juegos de una nueva unidad
- Inicie la aplicación Steam desde el escritorio.
- Haz clic en Steam y selecciona Configuración.
- Haz clic en la pestaña Descargas.
- Haga clic en la carpeta de la biblioteca de Steam en la sección Bibliotecas de contenido.
- Haga clic en Agregar carpeta de biblioteca y navegue a la ubicación donde se mueven sus juegos (nuevo directorio) que es D: / games / your_subdirectory.
- Haga clic en Seleccionar y cerrar para guardar la carpeta de la biblioteca.
- Salga de Steam y reinícielo.
Steam escaneará la carpeta Biblioteca recientemente seleccionada y mostrará todos los juegos instalados.
4) Utilizar .acf Cache para forzar el reconocimiento de juegos de Steam .acf Cache para forzar el reconocimiento de juegos de Steam.
- Asegúrese de haber reinstalado Steam o tener la instalación existente.
- Mueva los datos del juego a C: >> Archivos de programa (x86) >> Steam >> carpeta Steamapps.
- Lanzar Steam.
- En este punto, Steam puede mostrar algunos juegos que están instalados correctamente.
- Para los juegos que se muestran como no instalados, seleccione y haga clic en el botón Instalar.
- Steam comenzará a descubrir todos los archivos existentes.
- Sin embargo, si Steam no reconoce los archivos existentes, comenzará a descargar los archivos y el progreso leerá 0%.
- Pausa la actualización de los juegos y sal de Steam. Vaya a C: >> Archivos de programa (x86) >> Steam >> Steamapps y encuentre todos los archivos .acf actuales.
¡¡¡Espero que esto ayude!!!
Air flows along a horizontal, curved streamline with a 20 foot radius with a speed of 100 ft/s. Determine the pressure gradient normal to the streamline.
Answers
Answer:
- 1.19 lb/ft^3
Explanation:
You are given the following information;
Radius r = 20 ft
Speed V = 100 ft/s
You should use Bernoulli equation pertaining to streamline. That is, normal to streamline.
The pressure gradient = dp/dn
Where air density rho = 0.00238 slugs per cubic foot.
Please find the attached files for the solution and diagram.
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Answers
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Find the total time required for the police car to over take the automobile.
Answer:
15.02 sec
Explanation:
The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.
So; we can say :
[tex]D_{pursuit} =D_{police}[/tex]
By using the second equation of motion to find the distance S;
[tex]S= ut + \dfrac{1}{2}at^2[/tex]
[tex]D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)[/tex]
[tex]D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)[/tex]
[tex]D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)[/tex]
[tex]D_{police} = ut _P + \dfrac{1}{2}at_p^2[/tex]
where ;
u = 0
[tex]D_{police} = \dfrac{1}{2}at_p^2[/tex]
[tex]D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2[/tex]
[tex]D_{police} = 0.98*(t+12)^2[/tex]
[tex]D_{police} = 0.98*(t^2 + 144 + 24t)[/tex]
[tex]D_{police} = 0.98t^2 + 141.12 + 23.52t[/tex]
Recall that:
[tex]D_{pursuit} =D_{police}[/tex]
[tex](187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t[/tex]
[tex](187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0[/tex]
= 46.68 - 7.85 t -2.505 t² = 0
Solving by using quadratic equation;
t = -6.16 OR t = 3.02
Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs
From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;
Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s
Total time required for the police car to over take the automobile = 15.02 sec
You are given a body with no body forces and told that the stress state is given as: ⎡ ⎣ 3αx 5βx2 + αy γz3 5βx2 + αy βx2 0 γz3 0 5 ⎤ ⎦ psi, where (α, β, γ) are constants with the following values: α = 1 psi/in, β = 1 psi/in2, and γ = 1 psi/in3. Does this represent an equilibrium state of stress? Assume the body occupies the domain Ω = [0, 1] × [0, 1] × [0, 1] (in inches).
Answers
Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is the actual measure resistance of the circuit?
Answers
Answer:
The answer is 2.25 kΩ
Explanation:
Solution
Given that:
The resistance reading on a DMM'S meter face = 22.5 ohms
The range selector switch = R * 100 range,
We now have to find the actual measure resistance of the circuit which is given below:
The actual measured resistance of the circuit is=R * 100
= 22.5 * 100
=2.25 kΩ
Hence the measured resistance of the circuit is 2.25 kΩ
A long conducting rod of rectangular cross section (20 mm 30 mm) and thermal conductivity k 20 W/m K experiences uniform heat generation at a rate q . 5 107 W/m3, while its surfaces are maintained at 300 K. Using a finite-difference method with a grid spacing of 5 mm, determine the temperature distribution in the rod.
Answers
Answer:
Explanation:
We are assuming that there is
a steady state two dimensional conduction
constant properties
uniform volumetric heat generation
From symmetry, we will be determining 6 unknown temperatures.
See attachment for calculation and and tabulation
With T(s) = 300 K, the set of equations were written directly into the IHT work space and solved for nodal temperatures.
The result is seen in the second attachment