Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = [tex]\int\limits^a_b P \, dV[/tex] = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button,
Answer:
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.
A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pressure changes with volume according to the relation P = aV + b, where a = –1200 kPa/m3 and b = 500 kPa. Calculate the work done during this process (a) by plotting the process on a P-V diagram and finding the area under the process curve and (b) by performing the necessary integrations.
Answer:
[tex]W=-52 800\ \text{J}=-52.8\ \text{kJ}[/tex]
Explanation:
First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.
To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:
[tex]W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}[/tex]
Which of the following would require the filing of an EIS? a. expansion of an airport in a national park b. allowing snowmobiles in a national park c. erecting a research telescope facility in a national park d. All of the above would require an EIS.
Answer:
d. All of the above would require an EIS.
Explanation:
A document prepared with the aim of describing the impacts of suggested operations on the environment is an Environmental Impact Statement (EIS). There was a mistake. An Environmental Impact Statement (EIS) is therefore a report describing the environmental effects resulting from a current action. All of the activities above would have an effect on the environment and therefore must fill an EIS
To use Allowable Stress Design to calculate required dimensions using a specified factor of safety. A structural element that carries a load must be designed so that it can support the load safely. There are several reasons why an element may fail at loads that are less than the theoretical limit. For example, material properties may not exactly equal the reference values used in the design. The actual loading may differ from the design loading. The exact dimensions of the member may be different from the nominal values. These scenarios, and others, make it important to design structural members so that the expected load is less than the expected load that would make the member fail. One method of doing this uses a factor of safety. The allowed load Fallow can be related to the load that causes failure, Ffail, using a constant called the factor of safety, F.S.=FfailFallow, which should be larger than 1. For preliminary analysis, the stresses that are developed are assumed to be constant, so that the load and the stress are related by N=σA or V=τA, where A is the area subjected to the load.
An anchor rod with a circular head supports a load Fallow = 11 kN by bearing on the surface of a plate and passing through a hole with diameter h = 2.6 cm. One way the anchor could break is by the rod failing in tension.
What is the minimum required diameter of the rod if the factor of safety for tension failure is F.S. = 1.6, given that the material fails in tension at σfail = 60 MPa? Assume a uniform stress distribution.
To determine the minimum required diameter of an anchor rod, we must calculate the allowable tensile stress using the factor of safety and the ultimate failure stress, then solve the area formula for a circular cross-section for the rod's diameter.
Explanation:To calculate the minimum required diameter of an anchor rod designed using Allowable Stress Design and considering a factor of safety (F.S.), we must ensure that the designed stress ( au_{allow}) does not exceed the allowable stress. The allowable stress is derived from the material's ultimate (failure) stress ( au_{fail}) divided by the factor of safety (F.S.). Given the allowable load (F_{allow}) the rod must support and the factor of safety (F.S.), the failure load (F_{fail}) is F_{fail}= F_{allow} \times F.S.
Since au_{allow} = F_{allow} / A, rearranging to solve for the cross-sectional area (A) needed gives us A = F_{allow} / au_{allow}. And for a circular cross-section rod, A = \pi (d/2)^{2}, where d is the diameter of the rod.
With the failure stress given as au_{fail} = 60 MPa and the factor of safety F.S. = 1.6, the allowable tensile stress ( au_{allow}) is au_{allow} = au_{fail} / F.S. = 60 MPa / 1.6.
Using the formula A = \pi (d/2)^{2} and the given load F_{allow} = 11 kN, we can solve for d. First, we convert the load to Newtons (since 1 kN = 1000 N), then substitute the values of F_{allow} and au_{allow} into the area equation and solve for the diameter (d).
Upon calculating d, we obtain the minimum required diameter of the rod that will ensure it supports the given load with the required factor of safety.
Factor of safety: The factor of safety is defined as the ratio of the load that causes failure to the load the member can safely support. In this case, the factor of safety for tension failure is given as 1.6. A factor of safety larger than 1 ensures structural stability.
Calculating required diameter: Using the allowable stress design approach, the minimum required diameter of the rod can be calculated to ensure it can support the load safely without failing in tension. Given the material failure stress and the factor of safety, the required diameter can be determined based on uniform stress distribution.
Engineering analysis: Engineers use factors of safety and stress calculations to design structural elements that can safely support loads. Understanding stress, strain, and material properties are essential in ensuring the structural integrity and safety of a design.
Describe briefly the main advantages offered by satellite communications. Explain what is meant by a distance-insensitive communications system.
Answer:
Advantage of satellite communication are as follow
- it help in mobile and wireless communications
- it can covers the wide area in single time
- installation of satellite communication is easy
- it can be used for any form of communication i.e. video, audio, etc.
- service can be available in uniformly
Explanation:
Advantage of satellite communication are as follow
- it helps in mobile and wireless communications
- it can cover the wide-area in a single time
- installation of satellite communication is easy
- it can be used for any form of communication i.e. video, audio, etc.
- service can be available in uniformly
Distance insensitive communication system is describe in terms of limitation discovered area. Example of this is mobile and satellite. The mobile discover area is less than satellite.
What are the effects of applying an additional layer of insulation to a cylindrical pipe or a spherical shell?
Answer:
When an additional layer of insulation is applied to a cylindrical pipe or a spherical shell, the insulation layer works to increase its conduction resistance but at the same time, lowers the convection resistance of the surface.
Explanation:
Generally, when more insulation is added to a wall, the resulting effect is that heat transfer decreases. With increasing thickness of the insulation, the heat transfer rate becomes lesser. This is because the thermal resistance of the wall becomes more with the added insulation, wherein the heat transfer area and convection resistance aren't affected.
However, a different scenario occurs when an additional layer of insulation is applied to a cylindrical pipe or spherical shell. The conduction resistance increases, but the surface convection resistance decreases since the outer surface area for convection also increases and does not remain constant.
Overall, the heat transfer from the cylindrical pipe or spherical shell may increase or decrease, which depends on the effect that dominates.
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and loosest state of this soil are 0.51 and 0.87. Find the relative density and degree of saturation.
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, [tex]e_{min}[/tex] = 0.51
Void ratio in the loosest state, [tex]e_{max}[/tex] = 0.87
Now,
Dry density, [tex]\gamma_d=\frac{\gamma_t}{1+w}[/tex]
[tex]=\frac{18}{1+0.05}[/tex]
= 17.14 kN/m³
Also,
[tex]\gamma_d=\frac{G\gamma_w}{1+e}[/tex]
here, G = Specific gravity = 2.7 for sand
[tex]17.14=\frac{2.7\times9.81}{1+e}[/tex]
or
e = 0.545
Relative density = [tex]\frac{e_{max}-e}{e_{max}-e_{min}}[/tex]
= [tex]\frac{0.87-0.545}{0.87-0.51}[/tex]
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
(a) find the current density at P(rho = 3, φ = 30◦, z = 2);
(b) determine the total current flowing outward through the circular band rho = 3, 0 < φ < 2π, 2 < z < 2.8
Answer:
(a) Current density at P is [tex]J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex].
(b) Total current I is 3.257 A
Explanation:
Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;
[tex]J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\[/tex]
where [tex]\textbf{a}_{\rho}[/tex] and [tex]\textbf{a}_{\phi}[/tex] unit vectors.
(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation. Therefore
[tex]J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex]
(b) Total current flowing outward can be calculated by using the relation,
[tex]I=\int {\textbf{J} \, \textbf{ds}[/tex]
where integral is calculated through the circular band given in the question. We can write the integral as below,
[tex]I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\[/tex]
due to unit vector multiplication. Then,
[tex]I=10\int\(\rho^3z.dz.d\phi[/tex]
where [tex]\rho=3,\ 0<\phi<2\pi, \ 2<z<2.8[/tex]. Therefore
[tex]I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A[/tex]
Following are the solution to the given points:
Calculating the current density:
[tex]\bold{J = 10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi} \frac{mA}{m^2}}{}[/tex]
For point a:
Calculating the current density at point P:
[tex]\to p=2 \\\\ \to \phi =60^{\circ}\\\\ \to Z=3[/tex]
[tex]\to J= 10(2)^2 3 (3) a_{\rho} - (4\times 2 \times ( \cos 60^{\circ})^2) a_{\phi}\\\\[/tex]
[tex]=120 \ a_{\rho} - 4 \times 2 \times \frac{1}{2^2} a_{\phi} \\\\= 120 a_{\phi} - 2 a_{\phi} \frac{mA}{m^2}\\\\[/tex]
For point b:
Calculating the total current:
[tex]\to I=\int \int \vec{J} \cdot \vec{ds}\\\\[/tex]
[tex]= \int \int (10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi}) \cdot (\rho d \phi \cdot dz) a_{\rho}[/tex]
Note:
[tex]\text{p=constont then} \vec{ds} = \rho d \phi dz \hat{a}_{\rho} \\\\[/tex]
[tex]I= \int \int (10 \rho^2 z) \cdot ( d \phi dz)[/tex]
[tex]= (10 \rho^3 \int^{8}_{3} z dz \int^{2 \pi}_{\pi} 1 \cdot d \phi\\\\= (10 \times 4^3 [\frac{z^2}{2}]^{8}_{3} \times [\phi]^{2 \pi}_{\pi}\\\\= 10 \times 64 \times \frac{(64-9)}{2} \times (2\pi-\pi) MA \\\\= 55292 \ MA \\\\= 55.292\ A[/tex]
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A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining points A and D.
Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
BE / BC = 45 / 60 = 0.75
Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
[tex]M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right][/tex]
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in
The moment of the force about the line joining points A and D is; 617.949 lb.in
What is the moment of the force?We are given;
Force exerted by cable EF at E; T_EF = 46 lb.
From the diagram of the guy wire, we can draw a triangle and we will have the following coordinates;
A(0, 0, 0)
D(48, -12, 36)
E(E_x, 96, E_z)
Also, we can get that;
BC² = 48² + 36²
BC = √(48² + 36²)
BC = 60 in
Also, from similar triangles, we will have the coordinate of E as;
E(36, 96, 27)
Position of Vector of EF is;
EF = {(21 - 36)i + (-14 - 96)j + (57 - 27)k} in
EF = {-15i - 110j + 30k} in
Magnitude of EF from online calculation = 115 in
Force along cable EF is;
F_EF = 46{(-15i - 110j + 30k)/115}
F_EF = {-6i - 44j + 12k} lb
Position vector of AE is {36i + 96j + 27k} in
Position vector of AD is {48i - 12j + 36k} in
Magnitude of AD = 61.188 N
Unit vector of AD; λ_ad = {48i - 12j + 36k}/61.188
λ_ad = 0.7845i - 0.1961j + 0.5883k
M_ad = λ_ad × r_ea × T_EF
M_ad = [tex]\left[\begin{array}{ccc}0.7845&-0.1961&0.5883\\36&96&27\\6&-44&12\end{array}\right][/tex]
Solving this gives;
M_ad = 617.949 lb.in
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Water is flowing in a 3 cm diameter garden hose at a rate of 25 L/min. A 20 cm long nozzle is attached to the hose which decreases the diameter to 1.2 cm. The magnitude of the acceleration of a fluid particle moving down the centerline of the nozzle is
Answer:
[tex]a=5.908\ \text{m/s$^2$}[/tex]
Explanation:
To get the magnitude of the acceleration you need to find the change in velocity across the nozzle. To do that you can use the continuity equation which stated that the flow rate is the same in every point in the flow. So in this case you define the points at the nozzle inlet and at the nozzle outlet:
[tex]q_v_1=q_v_2\\A_1v_1=A_2v_2\\[/tex]
To get the velocity at point 1 you can just divide the flow rate by the surface:
[tex]v_1=\cfrac{q_v1}{A_1}=0.207\ \text{m/s}[/tex]
Now the expression for the velocity at point 2 is:
[tex]v_2=\cfrac{A_1}{A_2}v_1=\cfrac{d_1^2}{d_2^2}v_1=1.294\ \text{m/s}[/tex]
Now if you assume linear change of velocity across the nozzle you can use the following expression for the acceleration:
[tex]a=\cfrac{2v_1^2}{L}\left(\cfrac{2x}{L}+1\right)[/tex]
where L is your nozzle length (20 cm).
This equation depends on the position of your particle moving down the centerline of the nozzle. I sketched it:
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is heated, and the gage pressure at the nal state is 367 kPa. Determine the nal temperature, in °C. The local atmospheric pressure is 1 atm.
Answer:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]{T_2}=T_1\times \dfrac{P_2}{P_1}[/tex]
Now by putting the values in the above equation we get
[tex]{T_2}=300\times \dfrac{368}{301}\ K[/tex]
The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C
A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s= 1.5 ft.
Answer:
attached below
Explanation:
The concept of equilibrium which is useful when dealing with forces is states that a body on which exactly balanced forces act has no net force acting on it and the body is said to be in equilibrium
The weight of the suspended block E is approximately 18.33-lb
The reason why the value of the weight is correct is as follows:
The given parameters are;
The length of the chord = 4-ft.
The weight of the block D = 10-lb
The horizontal distance between pin A and C = 1 ft.
The length of the segment of the string s when the system is in equilibrium = 1.5 ft.
Required:
To determine the weight of the suspended block E, if the system is in equilibrium
Solution:
The tension in the cord, T = The weight of block D = 10-lb
By equilibrium of forces, we have;
E = 2 × T × cos(θ)
Where:
E = The weight of the block located at E
θ = The angle between segment CB and the vertical
[tex]sin(\theta) = \dfrac{Opposite}{Hypotenuse}[/tex]
The length of the opposite side to θ = 1 ft./2 = 0.5 ft.
The length of the hypotenuse side = [tex]\dfrac{4 - 1.5}{2} = 1.25[/tex]
Therefore;
[tex]sin(\theta) = \dfrac{0.5}{1.25} = 0.4[/tex]
θ ≈ 23.58°
E = 2 × 10 × cos(23.58°) ≈ 18.33 lb.
The weight of the suspended block E ≈ 18.33 lb.
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an ideal otto cycle has a compression ratio of 8. at the begining of the compression process, air is at 95 kpa and 27 c, and 750kj/kg of heat is transferred to air during the constant volume heat addition process Taking into account the variation of specific heats with temperature, determine
(a) the pressure and temperature at the end of the heataddition process,
(b) the net work output,
c) the thermal efficiency, and
(d) the mean effective pressure for the cycle.
Answer:
(a) the pressure and temperature at the end of the heat addition process is 1733.79 K and 4392.26 Kpa respectively
(b) the net work output is 423.54 KJ/Kg
(c) the thermal efficiency is 56.5%
(d) the mean effective pressure for the cycle cannot be determined without initial volume of the process
Explanation:
Assumptions:
changes in kinetic and potential energy is negligibleair is an ideal gas with constant specific heatsThe properties of air at room temperature;
Cp = 1.005 KJ/kg.K, Cv = 0.718KJ/kg.K, R = 0.287KJ/kg.K and K = 1.4
Part a:
For isentropic compression:
[tex]T_2=T_1[\frac{V_1}{V_2}]^{K-1}[/tex]
Where;
T₁ = (27+273)K =300K
V₁/V₂ = 8
[tex]T_2=300[8]^{1.4-1} = 300[8]^{0.4} = 689.22K[/tex]
Based on the assumption above;
Q₁ₙ = U₃ -U₂
For an ideal gas with constant specific heats, the change in internal energy in terms of the change in temperature is shown below
U₃ -U₂ = Cv(T₃-T₂)
Thus; Q₁ₙ = Cv(T₃-T₂)
T₃ = (Q₁ₙ/Cv) + T₂
T₃ = (750/0.718) + 689.22 K = 1733.79 K
From general gas equation, we find the second stage pressure
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]P_2 =P_1[\frac{T_2}{T_1}][\frac{V_1}{V_2}] = 95 Kpa[\frac{689.22}{300}](8)[/tex]
P₂ = 1746.024Kpa
To obtain the pressure at stage 3
[tex]P_3 =P_2[\frac{T_3}{T_2}][\frac{V_2}{V_3}] = 1746.024 Kpa[\frac{1733.79}{689.22}](1)[/tex]
P₃ = 4392.26 Kpa
Part b:
To obtain net work output we consider overall energy balance on the cycle
[tex]Q_{out} = U_4-U_1 = C_v(T_4-T_1)[/tex]
For is isentropic expansion
[tex]T_4=T_3[\frac{V_3}{V_4}]^{K-1} = 1733.79 [\frac{1}{8}]^{0.4} = 754.68K[/tex]
[tex]Q_{out} = 0.718(754.68-300) = 326.46 KJ/Kg[/tex]
To solve for net work output:
[tex]Q_{net} = Q{in}- Q_{out}[/tex] = (750 - 326.46) KJ/Kg = 423.54 KJ/Kg
part c:
To calculate the thermal efficiency, we use net work output per input work
η = 423.54/750
η = 0.565 = 56.5%
part d:
Mean effective pressure for the cycle (MEP)
[tex]MEP = \frac{Q_{net}}{V_1-V_2}[/tex] = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(1-\frac{1}{8}) }[/tex]
MEP = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(0.875)}[/tex]
Thus mean effective pressure cannot be determined without initial volume of the process.
Consider a Venturi with a throat-to-inlet area ratio of 0.8, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level.
If the static pressure at the throat is 2100 lb/ft³, calculate the velocity of the airplane. Note that standard sea-level density and pressure are 1.23 kg/m³ (0.002377 slug/ft³) and 1.01 x 10⁵ N/m2 (2116lb/ft³), respectively.
Defining the radius of area throat to inlet we would have that the proportional relationship would be [tex]\frac{A_2}{A_1}.[/tex]
This equation or relationship is obtained from continuity where
[tex]A_1V_1 = A_2V_2[/tex]
[tex]\frac{V_2}{V_1} = \frac{A_2}{A_1}= 0.8[/tex]
Now applying the Bernoulli equation between inlet and throat section we have,
[tex]\frac{p_1}{\rho g}+ \frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+ \frac{v_2^2}{2g}+z_2[/tex]
Here,
[tex]z_1 = z_2[/tex]
Then for a Venturi duct, the velocity of the airplane [tex]V_1[/tex] will be
[tex]V = \sqrt{\frac{2(p_1-p_2)}{\rho[(\frac{A_1}{A_2})^2-1]}}[/tex]
Our values are,
[tex]\frac{A_2}{A_1} = 0.8[/tex]
[tex]\rho = 0.002377slug/ft^3[/tex]
[tex]p_1 = 2116lb/ft^2[/tex]
[tex]p_2 = 2100lb/ft^2[/tex]
Replacing,
[tex]V= \sqrt{\frac{2(2116-2100)}{(0.002377)[(\frac{1}{0.8})^2-1]}}[/tex]
[tex]V = 154.7ft/s[/tex]
Therefore the velocity of the airplane is 154.7ft/s
A clay sample has a wet mass of 417 g, a volume of 276 cm3, and a specific gravity of 2.70. When oven dried, the mass becomes 225 g. Calculate: a) The water content, b) the void ratio, and c) the degree of saturation.
a) The water content is 85.3%.
b) The void ratio is approximately 2.31.
c) The degree of saturation is 99.6%.
How do we determine the water content?
a) We shall calculate the water content (w) using the formula:
[tex]\frac{M_{w} }{Ms} * 100[/tex]
where:
[tex]M_{w}[/tex] = mass of water
[tex]M_{s}[/tex] = mass of solids (oven-dried mass)
Given:
[tex]M_{w}[/tex] = [tex]417g -225g = 192g[/tex]
[tex]M_{s}[/tex] = [tex]225g[/tex]
We shall now calculate water content (w):
[tex]w = (192/225) * 100[/tex]
[tex]w = 0.853 * 100[/tex]
[tex]w = 85.3[/tex]%
Thus, the water content is 85.3%.
b) We shall compute the Void ratio (e) by using the formula:
[tex]e = \frac{V_{v} }{V_{s} }[/tex]
where:
[tex]V_{v} =[/tex] volume of voids
[tex]V_{s} =[/tex] volume of solids
Let us find [tex]V_{s}[/tex] first:
[tex]V_{s} =[/tex] [tex]M_{s} /[/tex]Specific gravity * Density of water
[tex]V_{s} = \frac{225g}{2.70 * 1g/cm^{3} }[/tex]
[tex]V_{s} = \frac{225g}{2.70g/cm^{3} }[/tex]
[tex]V_{s} = 83.33cm^{3}[/tex]
[tex]V_{v} = V_{t} - V_{s}[/tex]
[tex]V_{v} = 276cm^{3} - 83.33cm^{3}[/tex]
= 192.67cm³
We calculate e (Void ratio):
e = [tex]192.67/83.33[/tex]
[tex]e = 2.31.[/tex]
Hence, the void ratio is ≈ 2.31.
c) We can find the degree of saturation (S) using the formula:
[tex]S = \frac{V_{w} }{V_{v} }[/tex]
Given:
[tex]V_{w}[/tex] = 192cm³
And we have [tex]V_{v}[/tex] = 192.67cm³
[tex]S = (192/192.67) * 100[/tex]
[tex]S = 0.996 * 100[/tex]
S = 99.6%
So, the degree of saturation is 99.6%.
A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. Calculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.
In this exercise we have to use electronic knowledge to calculate the voltage value from the resistance. In this way we can conclude that:
Ripple voltage: [tex]8.33mV[/tex] DC output voltage: [tex]19.11 V[/tex]
So from the data given in the text, we have that:
Voltage: [tex]30v[/tex] Load resistance: [tex]600 ohms[/tex] Capacitor filter: [tex]50mF[/tex] Frequency of supply: [tex]120Hz[/tex]
So with the formula given below, we can calculate what is being asked by it:
[tex]Peak \ to \ peak \ ripple = I (load)/(f)(c)[/tex]
They are using the data previously informed and putting it in the given formula, we would be with:
[tex]I (load) = load \ current = 30/600 = 0.05 A\\Peak \ to \ peak \ ripple = 0.05/6\\= 8.33mV[/tex]
The average Dc something produced power for a complete wave exist double :
[tex]Vd_c= 0.637 * 30\\Vd_c =19.11V[/tex]
See more about electronic at brainly.com/question/1255220
A Sampling of 409 motor Shafts from a very large production Batch shows a sample standard deviation in Diameter of 0.021 mm with a sample mean of 9.251 mm Estimate the mean diameter of the entire batch at a %95 level of confidences?
Answer:
9.248 < [tex]\mu[/tex] < 9.253 mm
Explanation:
Given data:
standard deviation = 0.021 mm
sample mean = 9.251 mm
total sample = 409 m
confidence interval level = 95%
mean diameter of entire batch [tex]\mu = \bar X \pm z \frac{\sigma}{\sqrt{n}}[/tex]
where
[tex] \bar X [/tex] - sample mean = 9.251 mm
z = critical value
plugginf all value in the above relation to get the mean diameter
[tex]\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}[/tex]
9.248 < [tex]\mu[/tex] < 9.253 mm
An interest buydown program offers to reduce interest rates by 4% from the base rate. Suppose the base rate for a loan of $8000 is 8% for 10 years. What is the monthly payment before and after the buydown? In this case, use monthly compounding, that is, the term is 120 payment periods, and the interest per month is 0.667% before and 0.333% after the buydown.
Answer: The monthly payment before the buydown is $71.3
The monthly payment after the buydown is $68.9
Explanation: The payment is compounding so we use compound interest;
A= P[1+(r/n)^nt]
Where;
A= Compounded amount
P = principal
r= interest rate per payment
n= number of payment per period
t= number of period.
NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;
number of payment per period (n) is 12
number of payment period (t) is 10
P=$8000, r= 0.667% or 0.333%
FIND MONTHLY PAYMENT BEFORE BUYDOWN:
Step 1: find the Compounded amount to pay.
A= $8000[1+(0.00667÷12)^(12×10)]=
$8551.64 this is the total amount he has to pay for a period of 10years
Step 2: How much does he has to pay monthly for a period of 10year;
Therefore his payment will be for 120 months
$8551.64÷120= $71.3 monthly
FIND MONTHLY PAYMENT AFTER BUYDOWN:
Step 1: find the compounded amount to pay.
A= 8000[1+(0.00333÷12)^(12×10)=
$8270.85 this is the total amount he has to pay for a period of 10years
Step2: How much does he has to pay monthly for a period of 10year;
Therefore his payment will be for 120 months;
$8270.85÷120= $68.9 monthly
To calculate the monthly payment for a $8,000 loan at 8% interest over 10 years, the loan payment formula is applied with a monthly interest rate of 0.667%. After applying a 4% interest buydown, reducing the rate to 4%, the same formula is used with the new monthly rate of 0.333% to find the reduced monthly payment. Both calculations highlight the financial benefits of an interest buydown.
Explanation:Calculating Monthly Payments Before and After Interest Buydown
To calculate the monthly payment for a loan of $8,000 at an 8% base rate over 10 years with monthly compounding, we need to use the formula for an installment loan which includes the principal and the interest. Similarly, we'll calculate the reduced payment after a 4% interest buydown to 4%.
The monthly payment before the buydown, using a monthly interest rate of 0.667%, can be determined using the formula:
M = P * (r(1+r)^n) / ((1+r)^n - 1)
Where:
M is the monthly paymentP is the principal amount ($8,000)r is the monthly interest rate (0.667% or 0.00667)n is the number of monthly payments (120)Plugging in the values, we calculate the monthly payment before the buydown.
The monthly payment after the buydown, with a reduced monthly interest rate of 0.333%, is calculated in a similar way.
In both cases, the formula provides us with the monthly payment amounts, showing the impact of the interest buydown on the monthly payments.
a. To measure the water current in an ocean, a marker is dropped onto it. Determine if the trajectory traced by the drifting marker forms a streamline, streakline, or pathline.b. To learn more about the flow, a large number of similar markers are released at the same point in space in quick succession. Determine if the line connecting these markers at a later instant forms a streamline, streakline, or pathline.
Answer:
A) The trajectory traced by the drifting marker form a pathline. The pathline is the trajectory than a single particle does in the fluid (like in this case) along with all its travel.
b) The line connecting these markers released at the same point in space in quick succession is a Streakline. The streaklines are the lines formed by the loci of points in the fluid that have continuously pass throw particulars points in space in the past. The main difference between the pathline and the streakline is that the first one is not affected by flow changes over time, while the second one is. The Pathline is a movie that shows the travel of a single particle in time. The streakline is movie that shows how the particles change their path along time.
In the gif attached to this answer, you can see in red the pathline of particle, in blue the streakline of several particles released at the same point in space in quick succession and in grey the streamline.
The electronics aboard a certain sailboat consume 50 W whenoperated from 12.6-V source. If a certain fully charged deep-cyclelead acid storage battery is rated for 12.6 V and 100 ampere hours,for how many hours can the electronics be operated from the batterywithout recharging. (The ampere-hour rating of the battery is theoperating time to discharge the battery multiplied by the current.)How much energy in Kilowatt hours is initially stored in thebattery? If the battery costs $75 and has a life 0f 300charge-discharge cycle, what is the cost of the energy in dollarsper kilowatt hour? Neglect the cost of recharging the battery.
The electronics can be operated for approximately 25.19 hours from the fully charged battery, which stores 1.26 kWh of energy. The cost of the energy per kilowatt hour for the battery, given its lifecycle, is around $0.20.
Explanation:The electronics aboard a sailboat consume 50 watts from a 12.6-V source, and the storage battery used is rated for 12.6 V and 100 ampere hours (Ah). To calculate the number of hours the electronics can be operated without recharging, we use the power consumption and the voltage to find the current drawn by the electronics:
I = P/V = 50W / 12.6V = 3.968 Ah
Thus, the electronics draw approximately 3.97 Ah from the battery. Since the battery has a capacity of 100 Ah, the operating time before a recharge is needed can be calculated as:
operating time = battery capacity / current draw = 100 Ah / 3.97 Ah ≈ 25.19 hours
To find how much energy in kilowatt-hours is stored in the battery:
Energy = Voltage imes Capacity = 12.6V imes 100Ah = 1260 Wh = 1.26 kWh
Considering the battery's life of 300 charge-discharge cycles and cost of $75, the cost per kilowatt hour can be calculated as follows:
cost per kWh = total cost / (energy storage imes number of cycles) = $75 / (1.26 kWh imes 300 cycles) ≈ $0.20 per kWh
The volume of air in one beach ball is 288π in 3. What is the volume of air in a ball whose radius is 3 inches greater than the first beach ball?
Answer:
v_2 = 972 π inch^3
Explanation:
Given data:
volume of air V_1 = 288π inc^3
radius of ball at 1 beach = r_1
radius of ball at 2nd beach = r_1 + 3
we know that [tex]V_1 = 288 \pi[/tex]
we know that volume is given as [tex]v_1 = \frac{4}{3} \pi r_1^3[/tex]
so equating both side of volume we get
[tex]288\pi = \frac{4}{3} \pi r_1^3[/tex]
r_1 = 6 inch
therefore r_2 = 9 inch
volume of air at 2nd beach [tex]v_2 = \frac{4}{3} \pi 9^3[/tex]
v_2 = 972 π inch^3
A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150K. Assume an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 70 percent for the regenerator. Determine:
a. The air temperature at the turbine exit.
b. The net work output.
c. The thermal efficiency of the cycle.
d. The total exergy destruction associated with the Brayton cycle, assuming a source temperature of 1800 K and a sink temperature of 310 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator.
Answer:
a) T_5 = 782.8 K
b) W_cyc = 108.04 KJ/kg
c) n_th = 22.47 %
d) X_dest = 289.924 KJ/kg , X_exhaust = 126.6768 KJ/kg
Explanation:
Given:
- P_2 / P_1 = 7
- T_4 = 1150 K
- T_1 = 310 K
- n_s,comp = 0.75
- n_s,turb = 0.82
- R_air = 0.287 KJ/kg
Find:
- T_5 - Temperature of air at turbine exit ?
- W_cycle ?
- n_th ?
- X_dest , X_exhaust ?
Solution:
Assumptions:
1) The cycle operates at steady-state.
2) Air is the working fluid and it behaves as an ideal gas.
3) The Brayton Cycle is modeled as as a closed cycle.
4) The combustor is replaced by a HEX. (External Combustion)
5) The compressor and turbine are not internally reversible.
6) Changes in kinetic and potential energies are negligible.
7) Air has variable specific heats.
8) The compressor and turbine are adiabatic.
Analysis:
- The efficiency of turbine is given by:
n_s,turb = (H_4 - H_5) / (H_4 - H_5,s)
- For H_4 and S_ 4 we have T_4 = 1150 K, use the ideal gas air property table:
T_4 = 1150 K -------------> S_4 = 3.129 KJ/kgK , H_4 = 1219.25 KJ/kg
- For H_5s, the enthalpy of the effluent from a hypothetical isentropic turbine. We can do this because we know the values of two intensive variables: P_5 and S_5 = S_4. The key to using this information is the 2nd Gibbs equation:
S_5,s,o = S_4,s,o + R_air*Ln(P_5 / P_4)
S_5,s,o = 3.129 + 0.287*Ln(1 / 7) = 2.57054 KJ/kgK
- Now use the value S_5,s,o and ideal gas air property table and evaluate:
S_5,s,o = 2.57054 KJ/kgK ---------> T_5,s = 698.6 K ,
H_5s = 711.72 KJ/kg
- Now use the efficiency relation for turbine:
H_5 = H_4 - n_s,turb*(H_4 - H_5,s)
H_5 = 1219.25 - 0.82*(1219.25-711.72)
H_5 = 803.08 KJ/kg
- Using H_5 and ideal gas air property table and evaluate:
H_5 = 803.08 KJ/kg -----------> T_5 = 782.8 K , S_5 = 2.6940 KJ/kg-K
- In order to determine the specific shaft work for the cycle, we need to determine the specific shaft work for the compressor and for the turbine
W_cyc = W_comp + W_turb
W_cyc = H_1 - H_2 + H_4 - H_5
- The efficiency of compressor is given by:
n_s,comp = (H_1 - H_2,s) / (H_1 - H_2)
- For H_1 and S_ 1 we have T_1 = 310 K, use the ideal gas air property table:
T_1 = 310 K -------------> S_1 = 1.73498 KJ/kgK , H_1 = 310.24 KJ/kg
- For H_2s, the enthalpy of the effluent from a hypothetical isentropic compressor. We can do this because we know the values of two intensive variables: P_2 and S_2 = S_1. The key to using this information is the 2nd Gibbs equation:
S_2,s,o = S_1,s,o + R_air*Ln(P_2 / P_1)
S_2,s,o = 1.73498 + 0.287*Ln(7) = 2.29343 KJ/kgK
- Now use the value S_2,s,o and ideal gas air property table and evaluate:
S_2,s,o = 2.29343 KJ/kgK ---------> T_2,s = 537.1 K ,
H_2s = 541.34 KJ/kg
- Now use the efficiency relation for compressor:
H_2 = H_1 - (H_1 - H_2,s)/n_comp
H_2 = 310.24 - (310.24-541.34)/0.72
H_2 = 618.37 KJ/kg
Hence,
The work out for the cycle is:
W_cyc = 310.24 - 618.37 + 1219.25 - 803.08
W_cyc = 108.04 KJ/kg
- The thermal efficiency of a cycle is:
n_th = W_cyc / Q_H
Q_H = H_4 - H_3
- The effectiveness of re-generator is e:
e = (H_3 - H_2) / (H_5 - H_2)
H_3 = (H_5 - H_2)*e + H_2
H_3 = (803.08 - 618.37)*0.7 + 618.37 = 738.43 KJ/kg
Hence,
Q_H = 1219.25 - 738.43 = 480.81 KJ/kg
Finally,
n_th = 108.04 / 480.81 = 22.47 %
- The amount of heat loss is given by:
Q_L = H_6 - H_1
H_6 = H_5 + H_2 - H_3 = 803.08 + 618.37 - 738.43 = 682.97 KJ/kg
Q_L = 682.97 - 310.24 = 372.73 KJ/kg
- The amount of exergy destroyed for whole cycle:
X_dest = T_L * ( Q_L / T_L - Q_H / T_H)
X_dest = 310 * (372.73 / 310 - 480.81 / 1800)
X_dest = 289.924 KJ/kg
- The amount of exergy of exhaust gasses:
X_exhaust = H_6 - H_0 - T_L*(S_6 - S_o )
X_exhaust = 682.97 - 310.24 - 310*(2.52861 - 1.73489 )
X_exhaust = 126.6768 KJ/kg
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contains 50 per cent by volume ammonia at one boundary of the stagnant layer. The ammonia diffusing to the other boundary is quickly absorbed and the concentration is negligible at that plane. The temperature is 295 K and the pressure atmospheric, and under these conditions the diffusivity of ammonia in air is 1.8 x 10~5 m2/s. Estimate the rate of diffusion of ammonia through the layer.
The question involves estimating the rate of ammonia diffusion through a stagnant air layer, applying Fick's first law of diffusion. A precise numerical answer cannot be provided without specific concentration values or boundary conditions. Understanding diffusion is crucial in many scientific and industrial applications.
Explanation:The scenario provided involves the diffusion of ammonia through a stagnant layer of air, under specific environmental conditions. To estimate the rate of diffusion, Fick's first law of diffusion can be applied, which states that the diffusion flux is proportional to the concentration gradient across the stagnant layer. Given the concentration of ammonia at one boundary is 50% and negligible at the other, with a layer thickness of 1 mm (0.001 m), and a diffusivity of 1.8 x 10-5 m2/s, the rate of diffusion can be calculated.
Unfortunately, without the specific concentration values or boundary conditions beyond percentage, a numerical calculation cannot be accurately provided in this response. Generally, the rate of diffusion would be determined by multiplying the diffusivity by the concentration gradient across the layer and inversely proportional to the thickness of the layer.
It is essential in chemical engineering and various scientific fields to understand how substances move through mediums, whether in industrial applications, environmental science, or physiology. The principles governing diffusion are critical for designing systems for the separation, treatment, or transfer of materials.
A gas contained in a piston cylinder assembly undergoes a process from state 1 to state 2 defined by the following relationship and given properties. Determine the final volume (V2) of the gas. P*V= constant P1 = 445 kPa V1 = 2.6 m^3 P2 = 140 kPa ans is 8.3 but how do i get it?
Answer:
V2 = final volume = 8.3m^3
Explanation:
Given P1 = 445 kPa, V1 = 2.6 m^3, P2 = 140 kPa
From PV = constant; P1V1 =P2V2 , where V2 = final volume
V2 = P1V1/P2
Substituting in the equation ;
V2 = 445 x 2.6 / 140
V2 = final volume = 8.3m^3
Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:
a. Zero offset voltages and currents
b. Low open-loop voltage gain
c. High slew rate for signals which change quickly
d. All of these
e. High bandwidth for high frequency signals
Answer:
Option B
Explanation:
An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.
CHALLENGE ACTIVITY 1.4.1: Basic syntax errors. Type the statements. Then, correct the one syntax error in each statement. Hints: Statements end in semicolons, and string literals use double quotes. System.out.println("Predictions are hard."); System.out.print("Especially '); System.out.println("about the future."). System.out.println("Num is: " - userName);
Answer:
Lets check each statement for the errors.
Explanation:
System.out.println("Predictions are hard.");
This statement has no syntax errors. When this statement is executed the following line will be displayed:
Predictions are hard.
System.out.print("Especially ');
This statement is missing the closing quotation mark. A single quotation mark is placed instead of double quotation mark in the statement.
The following error message will be displayed when this program statement will be compiled:
Main.java:15: error: unclosed string literal
String literals use double quotes. So to correct this syntax error, the statement should be changed as follows:
System.out.print("Especially");
The output of this corrected line is as following:
Especially
System.out.println("about the future.").
In this line a period . is placed at the end of the statement instead of a semicolon ; but according to the syntax rules statements should end in semicolons.
The error message displayed when this line is compiled is as following:
Main.java:15: error: ; expected
Main.java:15: error: not a statement
So in order to correct this syntax error the statement should be changed as following:
System.out.println("about the future.");
The output of this corrected line is as following:
about the future
System.out.println("Num is: " - userName);
There is a syntax error in this statement because of - symbol used instead of +
+ symbol is used to join together a variable and a value a variable and another variable in a single print statement.
The error message displayed when this line is compiled is as following:
Main.java:13: error: bad operand types for binary operator '-'
So in order to correct this syntax error the statement should be changed as following:
System.out.println("Num is: " + userName);
This line will print two things one is the string Num is and the other is the value stored in userName variable.
So let userName= 5 then the output of this corrected line is as following:
Num is: 5
Two transistors, fabricated with the same technology but having different junction areas, when operated at a base-emitter voltage of 0.75 V, have collector currents of 0.4 mA and 2 mA. Find IS for each device. What are the relative junction areas?
Answer:
The relative junction areas (A1 / A2) is 0.20
Explanation:
Using the formula of the collector current
ic = Is * [tex]e^{\frac{v_{be} }{V_{T} } }[/tex]
Is = [tex]\frac{ic}{e^{\frac{v_{be} }{V_{T} } }}[/tex]
Knowing
VT = 25 mV
vbe = 0.75 V
1 transistor
ic = 0.4 mA
Is1 = [tex]\frac{ic}{{\frac{v_{be} }{V_{T} } }}[/tex]
Is1 = [tex]\frac{0.4 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}[/tex]
Is1 = 5.47 * [tex]10^{-4}[/tex] A
2 transistor
ic = 2 mA
Is1 = [tex]\frac{ic}{{\frac{v_{be} }{V_{T} } }}[/tex]
Is1 = [tex]\frac{2 * 10^{-3} }{{0.75}/{e^{25 * 10^{-3} } } }}[/tex]
Is1 = 2.73 * [tex]10^{-3}[/tex] A
Junction area is proportional to saturation current
Is1 / Is2 = A1 / A2
A1 / A2 = 0.20
Find support reactions at A and B and then calculate the axial force N, shear force V, and bending moment M at mid-span of AB. Let L 5 4 m, q0 5 160 N/m, P 5 200 N, and M0 5 ? 380 N m.
The problem involves calculating support reactions, shear force, axial force, and bending moment at mid-span of a beam under distributed and point loads using static equilibrium equations. It applies fundamentals of structural engineering, specifically focusing on shear force and bending moment analysis.
Explanation:The question involves finding the support reactions at points A and B, and then determining the axial force (N), shear force (V), and bending moment (M) at the mid-span of the beam AB. Given the parameters are L = 4 m, q0 = 160 N/m, P = 200 N, and M0 = -380 N*m, the solution utilizes principles of static equilibrium to solve for the reactions caused by distributed and point loads on a statically determinate beam. This typically involves summing moments and forces in the horizontal and vertical directions.
Calculating these variables requires applying the fundamental equations of equilibrium, \(\Sigma F_x = 0\), \(\Sigma F_y = 0\), and \(\Sigma M = 0\), to the system. For example, the shear force (V) at any cross-section of the beam can be calculated by summing vertical forces, and the bending moment (M) can be calculated by summing moments about a point. The location of maximum shear force and bending moment often occurs at the supports or under the point load and varies linearly or parabolically along the beam length due to the distributed load.
The axial force (N) in this context is typically analyzed in members subjected to tension or compression, which might not be directly relevant for a simply supported beam under transverse loading unless it is specifically part of the problem's context, such as in the case of a truss member. In this problem, the main focus would be on shear force and bending moment calculations.
A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower plate remains fixed while the upper plate is subjected to a horizontal force, P. If the top plate displaces 2 mm horizontally, determine:
a. the average shear strain in the material
b. the force P exerted on the upper plate.
Answer:
γ[tex]_{xy}[/tex] =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ[tex]_{xy}[/tex] = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ[tex]_{xy}[/tex]
By comapring both equations, we get
P/A = Gγ[tex]_{xy}[/tex] ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency? Express your answer using two significant figures.
Answer:
The engine's thermal efficiency is 0.32
Explanation:
Thermal efficiency = work done ÷ quantity of heat supplied
Work done = 210 J
Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J
Thermal efficiency = 210 ÷ 650 = 0.32