Do you know who Candice is

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

Explanation:

This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.

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]

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in

Answer:

(a) the reversible power output of turbine is 5810 kw

(b) The second-law efficiency of he turbine = 86.05%

Explanation:

In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.

h1 = 3658 kJ/kg s1=7.167 kJ/kgK

In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state

h2 = 2682kl/kg S2= 7.694 kJ/kg

Assuming that the energy balance equation given  

Wout=m [h1-h2+(v1²-v2²) /2]

Let

W =5 MW

V1= 80 m/s  V2= 140 m/s

h1 = 3658kJ/kg  h2 = 2682 kJ/kg

∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)

m = 5.158kg/s

Consider the energy balance equation given  

Wrev,out =Wout-mT0(s1-s2)

Substitute Wout =5 MW m = 5.158kg/s 7

s1=  7.167 kJ/kg-K            s2= 7.694kJ/kg-K and 25°C .

Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)

= 5810 kW

(a) Therefore, the reversible power output of turbine is 5810 kw.

The given values of quantities were substituted and the reversible power output are calculated.

(b) Calculating the second law efficiency of the turbine:  

η=Wout/W rev,out

Let Wout =  5 MW and Wrev,out = 5810 kW  

η=(5 MW x 1000 kW)/(1 MW *5810)  

η= 86.05%

Complete Question:

Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg.  Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.

Answer:

Mangel-Wurzel Transport

The number of trips that the dump truck will have to make to haul the customer's load away is:

= 5 trips.

Explanation:

a) Data and Calculations:

Volume of customer's load (rock and soil) = 19.8m³

Density of load = 650 kg/m³

Mass of load = Volume of load * Density of load

= 19.8m³ × 650 kg/m³

= 12,870 kg

The maximum safe load (mass) of the dump truck = 3,700 kg

Volume of the dump truck = 4m³  

Assuming the truck is to carry 4m³ of the load.

The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³

= 2,600kg

Quick Check:

Mass = 2,600kg < 3,700 kg, satisfying required conditions.  

The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:

Number of trips = N

N = total volume of load/ volume per trip

N = 19.8/4

N = 4.95

N = 5 trips approx.

Answer:

Explanation:

B. you would grab the plug closest to the outlet

If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.

Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.

Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.

Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket

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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.

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Answer:

Human Capital Management (HCM) will help the start-up firm manage its recruiting and hiring activities.

Explanation:

Human Capital Management (HCM) Platform will assist the start-up firm manage its main point of access by keeping the employee records and maintaining the wages and salaries, managing the benefits, time, and attendance, and carrying out performance reviews including looking after the most important asset employees.

Answer:

Explanation:

The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

Calculate the value of the M2

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

Note: Calculation for T2 is attached below

Complete Question

Complete Question is attached below.

Answer:

[tex]V'=5m/s[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=0.10m[/tex]

Power [tex]P=4.0kW[/tex]

Head loss [tex]\mu=10m[/tex]

 [tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]

 [tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]

 [tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]

 [tex]H_m=10.39m[/tex]

Generally the equation for Power is mathematically given by

 [tex]P=\rho gQH_m[/tex]

Therefore

 [tex]Q=\frac{P}{\rho g H_m}[/tex]

 [tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]

 [tex]Q=0.03935m^3/sec[/tex]

Since

 [tex]Q=AV'[/tex]

Where

 [tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]

 [tex]A=7.85*10^{-3}[/tex]

Therefore

 [tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]

 [tex]V'=5m/s[/tex]

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

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]

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, [tex]$x_1$[/tex] = 39 mm

        [tex]x_2[/tex] = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]

[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$1.3= V^2_2-0.08^2$[/tex]

[tex]$V_2=1.14\ m/s$[/tex]

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of [tex]V_2[/tex] in (1),

[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]

[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]

[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]

[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]

[tex]$82.83x - 400x^2+0.048=0$[/tex]

[tex]$ 400x^2- 82.83x-0.048=0$[/tex]

x = 0.20 m

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

Answer:

Explanation:

[tex]496=\frac{480\times 60}{demand}[/tex]

demand per day = 58 pans

Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.

Hence the cycle time would be the greater time of the two operations.

cycle time = 450 seconds

[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]

[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]

[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)

Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.

b)

Required daily production is 58 pans

Answer:

7.9 billion people

Explanation:

1.5 nautical miles per 1,000 feet

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]

        [tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]

Answer:

[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].

The quality of the refrigerant exiting the expansion valve is

[tex]x_{2}=0.193596[/tex].

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].

Pressure [tex]P_{1}[/tex] = 10 bar.

Exit 2:-

Pressure [tex]P_{2}[/tex] = 1 bar.

Solution:-

Answer:

Both are correct.

Explanation:

Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.

Answer:

1.c

2.c

3.c

Explanation:

James is a  pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.

When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.

The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.

The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.

Thus, option C, C, and C are correct.

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Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]

in this expression the densities are constant

        v = A  √h

       A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = [tex]\frac{1}{454.55^2} \ v^2[/tex]

       Δh = [tex]\frac{dh}{dv}[/tex]   Δv

       [tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

        [tex]\frac{\delta v}{v} = 0.05[/tex]

       [tex]\frac{\Delta h}{h}[/tex] = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        [tex]\frac{\Delta h}{h}[/tex] =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

Answer:

I can't understand this language .