One method of reducing an aircarft’s landing distance is through the use of thrust reversers. Consider the turbofan engine in Fig. P2.5 with thrust reverser of the bypass airstream. It is given that 1500 lbm / s of air at 60 8 F and 14.7 psia enters the engine at a velocity of 450 ft / s and that 1250 lbm / s of bypass air leaves the engine at 60 deg to the horizontal, velocity of 890ft / s, and pressure of 14.7 psia. The remaining 250 lbm / s leaves the engine core at a velocity of 1200 fps and a pressure of 14.7 psia.
Determine the force on the strut [Fx]. Assume the outside of the engine sees a pressure of 14.7 psia.

Answer:

1) 0. 03 kW/K

2) the value is grater than zero so it satisfies the second law of thermodynamic (states that rate of entropy change must be equal to or greater than zero) .

Explanation:

Rate of entropy change S = dQ/dT

= Q(1/T1 - 1/T2)

T2 = 25°C = 298 K

T1 = 0°C = 273 K

Q = 100 kW

S = 100( 1/273 - 1/298)

S = 100(0.0003) = 0. 03 kW/K

Answer:

We will see a waveform displayed on the screen and it will be PWM(pulse width modulation) of sinusoidal wave,this wave should have a frequency of 14KHz and it will form a vibration spectrum.

The question is an engineering problem that involves calculating the average temperature of a truck's refrigeration compartment's outer surface. A precise answer would require additional thermal information and properties, which are not provided in the question.

The question is related to the field of thermodynamics and heat transfer, specific to engineering. It requires determining the average temperature of the outer surface of the refrigeration compartment of a truck. However, to provide a precise answer, additional data would be required, including the thermal properties of the truck's material, the heat transfer coefficient, and the difference between the interior and exterior temperatures. Typically, such a problem would involve calculations using the concepts of convection and possibly conduction, considering the truck as a heat exchanger with heat being removed by the refrigeration system and added from the external environment. Without specific values for heat transfer coefficients and material properties, a calculation cannot be accurately made.

Given Information:

Primary secondary voltage ratio = 2500/250 V

Short circuit voltage = Vsc = 100 V

Short circuit current = Isc = 110 A

Short circuit power = Psc = 3200 W

Required Information:

Series impedance = Zeq = ?

Answer:

Series impedance = 0.00264 + j0.00869 Ω

Step-by-step explanation:

Short Circuit Test:

A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.

The series impedance in polar form is given by

Zeq = Vsc/Isc < θ

Where θ is given by

θ = cos⁻¹(Psc/Vsc*Isc)

θ = cos⁻¹(3200/100*110)

θ = 73.08°

Therefore, series impedance in polar form is

Zeq = 100/110 < 73.08°

Zeq = 0.909 < 73.08° Ω

or in rectangular form

Zeq = 0.264 + j0.869 Ω

Where Req is the real part of Zeq  and Xeq is the imaginary part of Zeq

Req = 0.264 Ω

Xeq = j0.869 Ω

To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.

Turn ratio = a = Vp/Vs = 2500/250 =  10

Zeq2 = Zeq/a²

Zeq2 = (0.264 + j0.869)/10²

Zeq2 = (0.264 + j0.869)/100

Zeq2 = 0.00264 + j0.00869 Ω

Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

[tex]\sigma a[/tex] = [tex]\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}[/tex]    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

[tex]\sigma a[/tex] =  [tex]\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}[/tex]  

[tex]\sigma a[/tex] = 12 kpsi

and

[tex]\sigma m[/tex] = [tex]\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}[/tex]     ...........2

put here value and we get

[tex]\sigma m[/tex] = [tex]\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}[/tex]  

[tex]\sigma m[/tex] = 17.34 kpsi

now we apply here goodman line equation here that is

[tex]\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}[/tex]     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

[tex]\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}[/tex]  

solve it we get

FOS = 1.5432

Answer:

[tex]\dot W_{out} = 399.47\,kW[/tex]

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

[tex]-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0[/tex]

The work done by the turbine is:

[tex]\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}[/tex]

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

[tex]P = 10\,MPa[/tex]

[tex]T = 520\,^{\textdegree}C[/tex]

[tex]h = 3425.9\,\frac{kJ}{kg}[/tex]

Outlet (Superheated Steam)

[tex]P = 1\,MPa[/tex]

[tex]T = 280\,^{\textdegree}C[/tex]

[tex]h = 3008.2\,\frac{kJ}{kg}[/tex]

The work output is:

[tex]\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW[/tex]

[tex]\dot W_{out} = 399.47\,kW[/tex]

Answer:

P = 98052.64 Pa or 98.05 kPa

Explanation:

Using Bernoulli's equation;

P+rho*v^2/2+ rho*g*z = constant

at high end;

dia= 1.2 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s

pressure, P = 68.67 kPa

Solving for elevation, z

assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.

x= 300*sin(tan^-1(1/100)) = 3 m

that means higher end will be 3 m above with respect to lower end

Similarly for lower end;

dia= 0.6 m

flow, Q = 5400 L/min = 0.09 m^3/s

therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s

assume pressure, P

z=0

put all values in the formula, we get;

68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0

solving this, we get;

P = 98052.64 Pa or 98.05 kPa

According to Bernoulli's principle, a lower pressure than expected at the

lower end because the velocity of the fluid is higher.

The given parameter are;

Length of the pipe, L = 300 m

Slope of the pipe = 1 in 100

Diameter at the high end, d₁ = 1.2 m

Diameter at the low end, d₂ = 0.6 m

The volume flowrate, Q = 5,400 L/min

Pressure at the high end, P = 68.67 kPa

Pressure at the low end

The elevation of the pipe, z₁ = [tex]300 \, m \times \frac{1}{100} = 3 \, m[/tex]  

z₂ = 0

The continuity equation is given as follows;

Q = A₁·v₁ = A₂·v₂

[tex]\sqrt[n]{x} \displaystyle Q = 5,400 \, L/min = 5,400 \, \frac{L}{min} \times \frac{1 \, m^3}{1,000 \, L} \times \frac{1 \, min}{60 \, seconds} = \mathbf{ 0.09 \, m^3/s}[/tex]

Therefore;

[tex]\displaystyle 0.09 = \frac{\pi}{4} \times 1.2^2 \times v_1 = \mathbf{\frac{\pi}{4} \times 0.6^2 \times v_2}[/tex]

[tex]\displaystyle v_1 = \frac{0.09}{\frac{\pi}{4} \times 1.2^2 } \approx 0.0796[/tex]

[tex]\displaystyle v_2 =\frac{0.09}{\frac{\pi}{4} \times 0.6^2 } \approx \mathbf{0.318}[/tex]

[tex]\displaystyle \frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 = \mathbf{ \frac{p_2}{\rho \cdot g} + \frac{v_2^2}{2 \cdpt g} + z_2}[/tex]

Therefore;

[tex]\displaystyle p_2 = \left(\frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 - \left( \frac{v_2^2}{2 \cdot g} + z_2 \right)\right) \times \rho \cdot g = p_1 + \frac{\rho}{2} \cdot \left(v_1^2} -v_2^2 \right)+ \rho \cdot g \left(z_1 - z_2\right)[/tex]

The density of water, ρ = 997 kg/m³

Which gives;

Learn more about Bernoulli's principle here:

https://brainly.com/question/6207420

Answer:

The total skin friction drag on the hull under these conditions is 276N

Explanation:

In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.

Please check attachment for complete solution and step by step explanation

Answer:

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Explanation:

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.  

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Answer:

0.003

Explanation:

? → ? = −??ℎ??→ ?

= ??ℎ??? = ?ℎ2− ℎ1∆?→ ? = 5????.101 ?? − 100.85 ??250 ??= 0.003???

Answer:

a. 5m

b. r = 0.16 e^-80.5◦

c. Zpn = (115.7 + j27.4) ohms

d. Vi = 2.2e^-j22.56◦ volts

e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts

Explanation:

In this question, we are tasked with calculating a series of terms.

Please check attachment for complete solution and step by step explanation

An incandescent light bulb's equilibrium temperature is determined by considering energy conversion, dissipation processes, and emissivity of the glass.

An incandescent light bulb converts electrical energy into light and heat. In the given scenario, the glass bulb absorbs and dissipates heat through convection and radiation. To determine the equilibrium temperature of the glass bulb, we need to consider energy conversion and emissivity.

We know that 10% of the energy passes through the glass bulb as light while the remaining 90% is absorbed and dissipated. By calculating the energy balance and accounting for the emissivity of the glass, the equilibrium temperature of the glass bulb can be found.

Factors like the wattage of the bulb, its size, and the room temperature play a role in determining the final equilibrium temperature of the glass bulb based on the energy conversion and dissipation processes involved.

The image of the question is missing, so i have attached it

Answer:

Velocity of slider B; = - 0.176 m/s

Explanation:

We are given;

Length of (AB) = 0.75 m

Rate of increase of length; (AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

We want to find vB;

Looking at the image attached, we can use the trigonometric ratio to find OA

Thus;

Sin θ = (OA)/(AB)

So, Sin 35° = (OA)/(AB)

(OA) = (AB)Sin 35°

(OA) = 0.75•Sin 35°

(OA) = 0.75•0.5736

(OA) = 0.43 m

Also, we can use the same system to find (OB)

Thus;

Cos θ = (OB)/(AB)

Cos 35° = (OB)/(AB)

(OB) = (AB)Cos 35°

(OB) = 0.75•Cos 35°

(OB) = 0.75•0.8192

(OB) = 0.6144 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation;

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

Divide through by 2 to give;

(AB)*(AB)' = (OA)*vA + (OB)*vB

vB = ((AB)*(AB)' - (OA)*vA) / (OB)

We now have ;

vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s)/(0.614 m)

vB = - 0.176 m/s

Answer:

The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking

XL=Xc

Substituting the definitions of XL and XC,

2[tex]\pi[/tex]foL=1/2[tex]\pi[/tex]foC

Solving this expression for fo yields

fo=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]

where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.

Explanation:

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.

A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.

Answer:

See all solutions attached as picture.

Explanation:

It is well explanatory

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the rectangular fin, L = 0.15 m

- The surface temperature of fin, Ts = 250°C

- The free stream velocity of air, U = 80 km/h

- The temperature of air, Ta = 27°C

- Parallel flow over both surface of the fin, assuming turbulent conditions through out.

Find:-

What is the rate of heat removal per unit width of the fin?

Solution:-

- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.

- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity , v = 27.85 * 10^-6 m^2/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandlt number Pr = 0.69

- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:

                          [tex]Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}[/tex]

Where,    Re_L: The average Reynolds number for the entire length of fin:

                          [tex]Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909[/tex]

Therefore,

                         [tex]Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378[/tex]

- The convection coefficient (h) can now be determined from:

                          [tex]h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}[/tex]

- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:

                          [tex]q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}[/tex]

- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The heat loss per unit width (q') due to radiation:

                  [tex]q' = 2*a*T_s^4*L[/tex]

Where, a: Stefan boltzman constant = 5.67*10^-8

                  [tex]q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}[/tex]

- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.  

Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

2

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

n

,

where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test

sample size, and s is the sample standard deviation (of the differences)

s =

vuut

1

n − 1

Xn

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Explanation: The curve represents the sum total of the evaluation

Answer:

The minimum required free stream velocity required to dissipate 12W via fins is V∞ = 0.0020378 m/s = 2.0378 mm/s.

Explanation:

Given:-

- The dimension of transformer surface ( L , w , H ) = ( 10 cm long , 6.2 cm wide, 5 cm high )

- The dimensions of the fin : ( l , h , t ) = (10 cm long , 5 mm high, 2 mm thick )

- The total number of fins, n = 7

- The convection heat transfer coefficient of the finned and unfinned area = h.

- The efficiency of fins, ε = 0.9 ( 90% )

- The transformer fin base temperature, Tb = 60°C

- The free temperature of air, T∞ = 25°C

- The free stream velocity of air =  U∞

Find:-

Determine the minimum free-stream velocity the fan needs to supply to avoid overheating. ( U∞ )

Solution:-

- Since the convection heat transfer coefficient of the finned and unfinned area i.e the fins and the transformer base are at the same temperature (Tb).

- The theoretical heat transfer ( Q_th ) rate from the fins can be calculated from the following convection cooling relation.

                   Q_th = h*As*[ Tb - T∞ ]

Where,

As : The total available surface area available for heat transfer.

        As1 = n *  { 2*  [ ( l * h ) + ( t * h ) ] + ( l * t ) }

        As1 = 7* { 2* [ ( 0.5 * 10 ) + ( 0.2 * 0.5 ) + ( 10 * 0.5 ) }

        As1 = 106.4 cm^2  .... 0.01064 m^2

        As2 = Total base area - Finned top plane area

        As2 = ( L * w ) - n* ( l * t ) = ( 10 * 6.2 ) - 7* ( 10 * 0.5 )

        As2 = 27 cm^2  .... 0.0027 m^2

- Therefore, the total available surface area (As) is:

       As = As1 + As2

       As = 0.01064 + 0.0027

       As = 0.01334 m^2

- The heat transfer coefficient (h) using convection heat transfer relation:

       Q* ε = h*As*[ Tb - T∞ ]

       h = Q* ε / [As*[ Tb - T∞ ] ]

       h = (12*0.9) / [ 0.01334*( 60 - 25 ) ]

       h = 23.13129 W/m^2K

- The air properties at film temperature:

       T = 40 C

       Viscosity ν = 1.6982 m^2 / s

      Thermal conductivity, k = 0.027076 W/mK

       Prandlt Number Pr = 0.71207

- The Nusselt number for the convection heat transfer for the transformer along the fins (Assumed flat plate):

      Nu = h*L / k

      Nu = 23.13129*0.1 / 0.027076

      Nu = 85.43

- The correlation for Nusselt number between flow conditions and viscosity effects of the flow (Re & Pr) for a isothermal flat plate - Laminar Flow is given:

       [tex]Nu =  0.664*Re^\frac{1}{2} *Pr^\frac{1}{3} \\\\Re^\frac{1}{2} = \frac{Nu}{0.664*Pr^\frac{1}{3}} \\\\\\Re = \sqrt{\frac{Nu}{0.664*Pr^\frac{1}{3}}} \\\\\\Re = \sqrt{\frac{85.43063}{0.664*0.71207^\frac{1}{3}}}\\\\Re = 12.00[/tex]

- The reynold number denotes the characteristic of the flow by the following relation:

      Re = V∞*L / ν

      V∞ = Re*ν / L

      V∞ = 12*1.6982*10^-5 / 0.1

      V∞ = 0.0020378 m/s   .... = 2.0378 mm/s

Answer:

1064.8 kg/m³

Explanation:

Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.

W in water = ρwghwA

W in liquid = (ρliq)g hliq A where the cross sectional area is constant

W in water = W in liquid

(ρw)ghwA = (ρliq)g hliq A  where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A

pliq = [tex]\frac{hw}{hliq}[/tex] × 1000 kg /m³ ( density of water) =( [tex]\frac{23}{23-1.4}[/tex]) × 1000 = 1064.8 kg/m³

Answer:

R = 1804 N

Explanation:

Given:-

- The density of water, ρ = 997 kg/m^3

- The inside diameter of the hose, dh = 75 mm

- The gauge pressure of water in the hose, P1 = 510 KPa

- The exit speed of the water, V2 = 32 m/s

- The inside diameter of the nozzle tip, dn = 25 mm

- The atmospheric pressure (gauge), P2 = 0 KPa ... P = 1 atm (Absolute).

Find:-

Determine the force transmitted by the coupling between the nozzle and hose.

Solution:-

- We will first develop a control surface at the hose-nozzle interface.

- Assuming steady and one dimensional flow - (x-direction).

- Since there are no fictitious unbalanced forces acting on the fluid flow due to roughness of hose any any losses of energy from the fluid are negligible.

- The use of conservation of momentum of fluid flow is valid for an isolated system, where the flow of fluid into the control volume is denoted by (-) and the flow of fluid going out of the control volume is denoted by (+):

- The principle of conservation of momentum, the pair of equal force (Newton's third law) act on the control volume at (nozzle-hose) interface:

                  R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)

Where,        Q: Flow rate

                   V1: The velocity of fluid in hose

                   A1: Cross sectional area of the hose

                   A2: Cross sectional area of the nozzle exit

- We see that the reaction force (R) that acts on nozzle-hose interface is due to changes in dynamic and hydrostatic pressures.

- Compute the required quantities Q, A1 and A2 and V1 using the given data:

- The flow rate Q for any flow in the hose can be given, where the cross sectional area of hose (A1)is:  

              [tex]A1 =\pi\frac{d_h^2}{4} = \pi\frac{0.075^2}{4} \\\\A1 = 0.00441 m^2\\\\\\[/tex]

- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:

                [tex]A2 =\pi\frac{d_n^2}{4} = \pi\frac{0.025^2}{4} \\\\A2 = 0.00049 m^2\\\\\\[/tex]

- The flow rate (Q) can now be calculated:

                 [tex]Q = A2*V2\\\\Q = (0.00049)*(32)\\\\Q = 0.01570 \frac{m^3}{s}[/tex]  

- Since, the density of the water does not vary along the direction of flow, the flow rate (Q) remains constant throughout. So from continuity equation we have:

                 [tex]Q = A2*V2 = A1*V1\\\\V1 = \frac{Q}{A1} = \frac{0.0157}{0.00441} \\\\V1 = 3.56189 \frac{m}{s}[/tex]  

- Now use the calculated quantities and compute the pair of reaction force at the nozzle-hose interface:

                 R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)

                 R = (997)*(0.01570)*(32-3.56189) + (0 - 510*0.00441)*1000

                 R = 445.13889 - 2,249.1

                R = - 1803.961 ≈ -1,804 N

- Here the negative sign denotes the direction of in which the force (R) is exerted. Since, (-) denotes into the control volume it acts opposite to the flow of water.

The coupling between the nozzle and hose is -1.81N

This question relates to flow rate of a liquid

Data given:

The density of water = 997kg/m^3

The inside diameter of the hose = 75mm = 0.0075m

The gauge pressure of water in the hose = 510kPa

The exit speed of the water = 32m/s

The inside diameter of the nozzle tip = 25mm = 0.0025m

The atmospheric pressure = 0kPa or 1atm

Let's calculate the inlet velocity

[tex]v_1=v_2=A_2/A_1\\v_1=V_2(\frac{d_2}{d_1})^2\\v_1=32(\frac{25}{75})^2\\v_1=3.50m/s[/tex]

Calculating the force transmitted by coupling between the nozzle and hose

[tex]R_x+p_1gA_1=v_1[-|pv_1A_1|]+v_2[|pv_2A_2|]\\[/tex]

μ[tex]_1[/tex]=[tex]v_1[/tex] and μ[tex]_2[/tex] =[tex]v_2[/tex]

[tex]R_x=-p_1gA_1-v_1pv_1A_1+v_2pv_2A_2\\R_x=-p_1gA+pv_2A_2(v_2-v_1)\\R_x=-510*10^3N/m^3*\frac{\pi }{4}(0.075m)^2+997kg/m^3*32m/s*\frac{\pi }{4} (0.025m)^2(32-3.50)=-1805=-1.81kN[/tex]

The force between the nozzle and hose is -1.81

Learn more about flow rate;

https://brainly.com/question/17151453

Answer:

Explanation:

check the attached files for the solution and output result.

Answer:

The rate at which water is being withdrawn from the river by the city is 57353 acre-ft/y

Explanation:

Please look at the solution in the attached Word file

Answer:

time required for cooling process = 0.233 hours

Explanation:

In Transient heat conduction of a Sphere, the formula for Biot number is;

Bi = hL_c/k

Where L_c = radius/3

We are given;

Diameter = 12mm = 0.012m

Radius = 0.006m

h = 20 W/m²

k = 40 W/m·K

So L_c = 0.006m/3 = 0.002m

So,Bi = 20 x 0.002/40

Bi = 0.001

The formula for time required is given as;

t = (ρVc/hA)•In[(T_i - T_(∞))/(T - T_(∞))]

Where;

A is Area = πD²

V is volume = πD³/6

So,

t = (ρ(πD³/6)c/h(πD²))•In[(T_i - T_(∞))/(T - T_(∞))]

t = (ρDc/6h)•In[(T_i - T_(∞))/(T - T_(∞))]

We are given;

T_i = 1200K

T_(∞) = 300K

T = 450K

ρ = 7800 kg/m³

c = 600 J/kg·K

Thus, plugging in relevant values;

t = (7800 x 0.012 x 600/(6 x20) )•In[(1200 - 300)/(450 - 300)]

t = 468•In6

t = 838.54 seconds

Converting to hours,

t = 838.54/3600

t = 0.233 hours

Answer:

B

Explanation:

This is a two sample t-test and not a matched pair t-test

null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors

alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.

So, option D is rejected

The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.

Option A and C are also rejected

Answer:

Given data

w1/w2=6.5/1

Power=5 KW

wp=1800 rpm

angle=14 degrees

Based on above values,the minimum diameter=30 mm

Answer:

specific energy  = 2.65 ft

y2 = 1.48 ft  

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + [tex]\frac{v^2}{2g}[/tex]     ...............1

put here value and we get

specific energy = [tex]2 + \frac{6.5^2}{2\times 9.8\times 3.281}[/tex]  

specific energy  = 2.65 ft

and

alternate depth is

y2 = [tex]\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})[/tex]  

and

here Fr² = [tex]\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}[/tex]  

Fr² = 0.8025

put here value and we get

y2 = [tex]\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})[/tex]

y2 = 1.48 ft  

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

The angular acceleration for  first drum [tex]\alpha = 0.792 rad/s^2[/tex]

The angular acceleration for the second drum is  [tex]\alpha =1.262[/tex]

Explanation:

From the question we are told that

         Their radius of the drum is  [tex]r = 13 in = \frac{13}{12} ft = 1.083ft[/tex] each

          The weight is  [tex]W = 210 lb[/tex]

           The mass is  [tex]M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}[/tex]

           Their radius of gyration is [tex]z=30 in= \frac{30 }{12} = 2.5 ft[/tex]

 The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image

    The T in the diagram is the tension of the string

  Now taking moment about the center of the the drum P we have  

        [tex]\sum M_p = I_p \alpha[/tex]

=>    [tex]T * r = Mz^2 * \alpha[/tex]

Where r is the radius ,z is the radius of gyration about the center O  , M is the mass  of the drum including  the  hub, and [tex]\alpha[/tex]  is the angular acceleration

   Inputting

                 [tex]T * 1.083 = 6.563 * 2.5^2 \alpha[/tex]

=>                         [tex]T = 37.87\alpha[/tex]

Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now  )

The first on is  

           [tex]\sum F_y = ma[/tex]

=>       [tex]30lb - T = m(r \alpha )[/tex]

Where m is the mass of  the hanging block which has a value  of

[tex]m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2[/tex]

            a  is the acceleration of the hanging block

 inputting values we have  

              [tex]30- 37.87 \alpha = 0.9317* 1.083 \alpha[/tex]

              [tex]30 = 37.87\alpha + \alpha[/tex]

              [tex]\alpha = \frac{30}{38.87 }[/tex]

                 [tex]\alpha = 0.792 rad/s^2[/tex]

So the angular acceleration for  first drum [tex]\alpha = 0.792 rad/s^2[/tex]

 The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image  

  So here we would take the moment about  O

             [tex]\sum M_o = I_O \alpha[/tex]

So  [tex]\sum M_o = 30* 1.083[/tex]

       and  [tex]I = M z^2[/tex]

Therefore we will have

            [tex]30 * 1.083 = (Mz^2 )\alpha[/tex]

  inputting values

                       [tex]30 * 1.083 = 6.563 * 2.5^2 \alpha[/tex]

                        [tex]32.49=41.0\alpha[/tex]

                         [tex]\alpha =\frac{41}{32.49}[/tex]

                          [tex]\alpha =1.262[/tex]

So the angular acceleration for the second drum is  [tex]\alpha =1.262[/tex]

Answer:

Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × [tex]10^6[/tex] Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 [tex]\times 10^{-3}[/tex] )²× 9  [tex]\times 10^{-3}[/tex]  × sin18 × cos18

Qd = 94.305 × [tex]10^{-6}[/tex] m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × [tex]10^{6}[/tex] × π × 85 [tex]\times 10^{-3}[/tex]  × ( 9  [tex]\times 10^{-3}[/tex] )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × [tex]10^{-6}[/tex] m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × [tex]10^{-6}[/tex]  - 85.2 × [tex]10^{-6}[/tex]  

Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s  

Answer: D = 23.09 pc/mi/h

              LOS = C

Explanation:

we will begin by solving this with a step by step process for easy understanding;

given that the freeway has two lanes  = 11 ft wide

and has a width of 5 ft right shoulder.

ramp density = 1 per mile in 3m

substituting the free flow speed value gives us;

Free flow speed = 75 - 1.9 - 0.6 - (3.22 × 1∧0.84) = 69.28 mph

we have that the length of the road = (1980 ft) × (1 mile / 5280 ft)

L = 0.375 mile

The next thing we will do is to  calculate the proportion of bus and the truck  

Pt = 300 + 300 / (2400 + 300 + 300)

Pt = 600/3000 = 20%

following up, we will consider the length and percentage of the buses and the trucks

Et = 2.0, Pr = 0, Er = 0

to calculate the percentage of truck

Ft = 1 / 1 + Pt (Et -1 ) + Pr (Er -1)

Ft = 1 / 1+0.2 (2-1) + 0 = 0.833

Truck percentage  = 0.833

To the determine the traffic volume,

Vt = V / PHF × N × Ft × Fp

Vt = 2400/ (0.9×2×0.833×1) = 1600 pc/lane/hour

But the Density of the freeway is given thus;

D = Vp / FFS .............(1)

but to get the FFS, we will consider the graph of flow rate vs speed and show the level of service

FFS = 69.28 mi/h

From the above expression in (1) we have that

D = 1600/69.28 = 23.09 pc/mi/h

D = 23.09 pc/mi/h

now we have that the the density of the freeway segment is 23.09 pc/mi/h, we can thus safely say that the level of service (los) = C

cheers i hope this helps