Answer:
can you please give us the option for this question
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat input, in watts.
[tex]\dot W[/tex] - Power output, in watts.
If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex]
[tex]\dot Q = \frac{\dot W}{\eta}[/tex]
[tex]\dot Q = \frac{600\,W}{0.3}[/tex]
[tex]\dot Q = 2000\,W[/tex]
The heat input from the combustion phase is 2000 watts.
What are the initial questions that a systems analyst must answer to build an initial prototype of the system output.
Calculate the scale and speed of the pattern in order to gain useful results for a turbine operate at 150 rev/min at height difference of 22 m and a predictable flow rate of 85 m per second. A scale pattern is made and examined with a volume flow rate of 0.1 m per second and a height difference of 5 m , the power value equal to 4.5 kW when checked at the speed evaluated . Predict the power and efficiency of the full size turbine .
Answer:
first mark me as a brainleast
3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar
Answer:
The right answer is "1.903 m".
Explanation:
Given that,
[tex]\tau =110 \ MPa[/tex]
[tex]G=80 \ GPa[/tex]
[tex]\Theta=15\times \frac{\pi}{180}[/tex]
[tex]=\frac{\pi}{12}[/tex]
[tex]d=20 \ mm[/tex]
As we know,
⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]
Or,
⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]
[tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]
[tex]=1903.9 \ mm[/tex]
or,
[tex]=1.903 \ m[/tex]
Given that the system function of a third order Butterworth type analog low-pass filter with a 3 dB cut-off frequency of 2 radian/second is:
2s HS = S2 + 0.2 s +1
Answer the following questions:
1. Use the bilinear transformation to obtain H(z). Use T=2 second.
2. Give H(w) for your filter.
3. Use MATLAB to give the magnitude spectrum.
4. Comment on the quality of the design.
5. With the aid of simple sketch graphs explain how frequency warping affects the frequency response of the digital filter.
6. Comment on the need for prewarping, i.e. give conditions when prewarping is needed.
answer
d just too the test
1. Add:
(i) 5xy, -2xy, -11xy, 8xy
(iv) 3a - 2b + c, 5a + 8b -70
Answer:
(i) 0
(iv) 8a+6b+c-70
Explanation:
Hope this helps you
Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
Problema:
Una nevera de vinos, con un peso bruto de 50 kg., que tiene las siguientes dimensiones: .60 m Largo x .49 m ancho x .50 m altura. Para ser transportadas en un contenedor de 40 pies D.V. responder las siguientes preguntas:
• 1.Cuántas neveras de vinos de acuerdo al volumen caben en un contenedor de 40 pies?
• De acuerdo dimensiones internas (largo, ancho y alto), ¿Cuántas caben en un contenedor de 40 pies?
• De acuerdo al peso que soporta el contenedor. ¿Cuántas neveras de vinos es posible transportar?
Answer:
I can't understand this language .
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.
Answer:
Explanation:
Given:
Q factor, =1000
natural frequency, [tex]f_n=2000~Hz[/tex]
Damping factor, [tex]\zeta=?[/tex]
Bandwidth, BW=?
We have the relation:
[tex]Q=\frac{1}{2\zeta}[/tex]
[tex]\zeta=\frac{1}{2Q}[/tex]
[tex]\zeta=\frac{1}{2\times 1000}[/tex]
[tex]\zeta=5\times 10^{-4}[/tex]
Bandwidth:
[tex]BW=\frac{f_n}{Q}[/tex]
[tex]BW=\frac{2000}{1000}[/tex]
[tex]BW=2~Hz[/tex]
A stream of oxygen enters a compressor at a rate of 200 SCMH. The oxygen exits at 360 K and 500 bar. Determine the volumetric flowrate exiting the compressor using the compressibility factor equation of state.
Answer:
≈ 0.516 m^3/hr
Explanation:
Inlet of compressor = 200 SCMH
sheer standard conditions = 1 atm and 288.5 K
For oxygen :
critical pressure(Pc) = 49.8 atm
critical temperature Tc = 154.6 K
hence at compressor inlet
Tr = T / Tc = 288.5/154.6 = 1.866
Pr = P / Pc = 1 / 49.8 = 0.0204
Z1 ( from compressibility chart ) = 0.98
at compressor outlet
P2 = 500 bar = 500*0.9869 = 493.45 atm , T2 = 360 k
hence : Pr = P / Pc = 493.45 / 49.8 = 9.91
Tr = T / Tc = 360 / 154.6 = 2.33
Z2 ( from compressibility chart ) ≈ 1
V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)
= 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)
= 0.516 m^3/hr
20 friends 6men 14 women are having a tea party
Answer:
what about it?
Explanation:
Select the correct statement(s) regarding network physical and logical topologies.
a. While logical topologies can be configured in star, ring, bus, and tree configurations, the physical topology must always be in a full-mesh topology
b. logical topologies always incorporate centralized access, whereas physical topologies are always configured as a distributed access network
c. the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
d. all statements are correct
Answer:
The physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another ( C )
Explanation:
Network physical is simply the process/method of connecting the Network using cables while Logical topology is the general architecture of the communication mechanism in the network for all nodes.
Hence The correct statement is the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
Define chart name the different types of charts explain any three types of charts
Answer:
There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.
#carryonlearning
What must you do to become ASE certified as an automotive technician?
Answer:
To become ASE certified, you must pass an ASE test and have relevant hands-on work experience. The amount of work experience required can vary by test, and is specified in detail here. ASE recommends submitting the form after you've registered to take an ASE certification test.
Good luck!
Explanation:
Answer: One theme in White Fang is adapting in order to survive. White Fang finally submits to Gray Beaver. He also copes with fighting other dogs. White Fang changes his behaviors so that he can live.
Explanation: its the sample response
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]