Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while the motorcycle is moving at 80 km/h in air at 27 C. The air is in parallel flow over both surfaces of the fin, and turbulent flow conditions may be assumed to exist throughout. What is the rate of heat removal per unit width of the fin?

Complete Question

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

Answer:

The largest offset that can be used is  [tex]h = 0.455 \ in[/tex]

Explanation:

From the question we are told that

   The diameter of the metal tube is  [tex]d_m = 0.75 \ in[/tex]

    The thickness of the wall is  [tex]D = 0.08 \ in[/tex]

Generally the inner diameter is mathematically evaluated as

           [tex]d_i = d_m -2D[/tex]

               [tex]= 0.75 - 2(0.08)[/tex]

               [tex]= 0.59 \ in[/tex]

Generally the tube's cross-sectional area can be evaluated as

         [tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]

             [tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]

              [tex]= 0.1684 \ in^2[/tex]

Generally the maximum stress of the metal is mathematically evaluated as

          [tex]\sigma = \frac{P}{A}[/tex]

              [tex]\sigma = \frac{P}{ 0.1684}[/tex]

The diagram showing when the stress is been applied is shown on the second uploaded image  

        Since the internal forces in the cross section are the same with the force P and the bending couple M then  

           [tex]M = P * h[/tex]

Where h is the offset

       The maximum stress becomes

                 [tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]

Where [tex]r_m[/tex] is the radius of the outer diameter  which is evaluated as

                 [tex]r_m = \frac{0.75}{2}[/tex]

                 [tex]r_m = 0.375 \ in[/tex]

and I is the moment of inertia which is evaluated as

           [tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]

              [tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]

              [tex]= 0.009583 \ in^4[/tex]

So the maximum stress becomes  

                [tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]

Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress

So

        [tex]\sigma ' = 4 \sigma[/tex]

=>    [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]

  The P would cancel out  

            [tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]

           [tex]5.94 + 39.13h = 23.753[/tex]

             [tex]39.13h = 17. 813[/tex]

                      [tex]h = 0.455 \ in[/tex]

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

The elevation difference between the two reservoirs is 10.28 ft

Given information:

- Length of the 3-in. cast iron pipe = 150 ft

- Number of flanged elbows = 4

- Entrance condition: well-rounded

- Exit condition: sharp-edged

- Fully open gate valve

- Flow rate = 75 gal/min

- Water temperature = 50 °F

Step 1: Convert the flow rate from gal/min to ft³/s.

Flow rate in ft³/s = (75 gal/min) × (1 ft³/7.48 gal) × (1 min/60 s) = 0.167 ft³/s

Step 2: Calculate the velocity of flow.

Velocity of flow, V = Flow rate / Cross-sectional area of the pipe

Cross-sectional area of the 3-in. cast iron pipe = (π/4) × (3/12)² ft² = 0.0491 ft²

Velocity of flow, V = 0.167 ft³/s / 0.0491 ft² = 3.4 ft/s

Step 3: Calculate the Reynolds number to determine the flow regime.

Reynolds number, Re = (ρVD) / μ

Where,

ρ = density of water at 50 °F = 1.94 slugs/ft³ (from tables)

V = velocity of flow = 3.4 ft/s

D = diameter of the pipe = 3/12 ft

μ = dynamic viscosity of water at 50 °F = 2.73 × 10⁻⁵ lb.s/ft² (from tables)

Re = (1.94 slugs/ft³) × (3.4 ft/s) × (3/12 ft) / (2.73 × 10⁻⁵ lb.s/ft²) = 1.9 × 10⁵

Since Re > 4000, the flow is turbulent.

Step 4: Calculate the head loss due to friction in the straight pipe.

Head loss due to friction, hf = f × (L/D) × (V²/2g)

Where,

f = friction factor (depends on Reynolds number and pipe roughness)

L = length of the pipe = 150 ft

D = diameter of the pipe = 3/12 ft

V = velocity of flow = 3.4 ft/s

g = acceleration due to gravity = 32.2 ft/s²

From the Moody diagram or friction factor tables for cast iron pipe, assuming a relative roughness of 0.00085, and Re = 1.9 × 10⁵, the friction factor, f = 0.021.

Substituting the values, hf = 0.021 × (150 × 12/3) × [(3.4)²/(2 × 32.2)] = 8.4 ft

Step 5: Calculate the head loss due to fittings (elbows, entrance, exit, and valve).

Head loss due to fittings, hf_fittings = Σ(K × V²/2g)

Where K is the loss coefficient for each fitting.

For 4 flanged elbows, K = 0.3 (for each elbow)

For a well-rounded entrance, K = 0.03

For sharp-edged exit, K = 1.0

For a fully open gate valve, K = 0.2

Substituting the values, hf_fittings = [4 × 0.3 + 0.03 + 1.0 + 0.2] × [(3.4)²/(2 × 32.2)] = 1.7 ft

Step 6: Calculate the velocity head.

Velocity head, hv = V²/2g = (3.4)²/(2 × 32.2) = 0.18 ft

Step 7: Calculate the total head loss.

Total head loss, H = hf + hf_fittings + hv

H = 8.4 ft + 1.7 ft + 0.18 ft = 10.28 ft

Therefore, the elevation difference between the two reservoirs is 10.28 ft for the given piping system and a flow rate of 75 gal/min of water at 50 °F.

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:

 IA = 80/3 kgm^2

Explanation:

Given:-

- The mass of rod AB, m1 = 20 kg

- The length of rod AB, L1 = 2m

- The mass of rod CD, m2 = 10 kg

- The length of rod CD, L2 = 1m

Find:-

What is the moment of inertia about A for member AB?

Solution:-

- The moment of inertia About point "O" the center of rod AB is given as:

                               IG = 1/12*m1*L^2

- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:

                              IA = IG + m1*d^2

- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.

So the moment of inertia for rod AB at point A would be:

                              IA = 1/12*m1*L^2 + m1*0.5*L1^2

                              IA = 1/3 * m1*L1^2

                              IA = 1/3*20*2^2

                             IA = 80/3 kgm^2

It will take approximately 0.0238 s to cut the hole.

To calculate the time required to cut the hole using Electrochemical Machining (ECM), we can use the following formula:

[tex]\text { Time }=\frac{\text { Volume of Material to be Removed }}{\text { Material Removal Rate (MRR) }}[/tex]

First, let's calculate the volume of material to be removed:

Volume of Material=Frontal Area × Thickness of Plate

Given:

[tex]Frontal Area =245 \mathrm{~mm}^2\\Thickness of Plate $=12 \mathrm{~mm}$[/tex]

Volume of Material = [tex]245 \mathrm{~mm}^2 \times 12 \mathrm{~mm}=2940 \mathrm{~mm}^3[/tex]

Next, we need to calculate the Material Removal Rate (MRR). MRR is usually given in units of volume removed per unit time per unit current density. Let's assume it is given as [tex]X \mathrm{~mm}^3 / \mathrm{min} / \mathrm{A} / \mathrm{mm}^2 .[/tex] Given the current density [tex]\text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}[/tex], we can calculate the MRR as:

MRR=X×Current Density

Given:

Applied Current = 1200 A

Efficiency = 95%

Frontal Area = 245 [tex]mm^2[/tex]

[tex]\begin{aligned}& \text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}=\frac{1200 \mathrm{~A}}{245 \mathrm{~mm}^2} \\& \text { Current Density } \approx 4.897 \mathrm{~A} / \mathrm{mm}^2\end{aligned}[/tex]

Let's assume X=0.2[tex]mm^3[/tex] /s/A/[tex]mm^2[/tex](hypothetical value).

[tex]\mathrm{MRR}=0.2 \times 4.897 \approx 0.979 \mathrm{~mm}^3 / \mathrm{s}[/tex]

To convert mm³/s to g/s, we need to know the density of the material. Let's assume the density of low alloy steel is approximately [tex]7.85 \mathrm{~g} / \mathrm{cm}^3[/tex], which is [tex]7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3 .[/tex]

Now, we can calculate the MRR in grams per second:

[tex]MRR in $\mathrm{g} / \mathrm{s}=\mathrm{MRR} \times$ Density of material\\MRR in $\mathrm{g} / \mathrm{s}=0.979 \mathrm{~mm}^3 / \mathrm{s} \times 7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3$\\MRR in $\mathrm{g} / \mathrm{s} \approx 0.00000768515 \mathrm{~g} / \mathrm{s}$[/tex]

Now, we can use these values to calculate the time required:

[tex]\text { Time }=\frac{2940 }{0.00000768515} \\& \text { Time } \approx 0.022594341[/tex]

However, we need to consider the efficiency. Since the efficiency is 95%, the actual time required will be:

[tex]\begin{aligned}& \text { Actual Time }=\frac{\text { Time }}{\text { Efficiency }} \\& \text { Actual Time }=\frac{0.022594341}{0.95}\end{aligned}[/tex]

≈ 0.0238 s

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

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.

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.

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:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.  = 0.4325m

Explanation:

see attached file below

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:

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:

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:

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:

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:

Copy MATLAB code to plot the magnitude of magnetic field strength with respect to z on the axis of solenoid:

z=-20:0.01:20;

H=120.*(((20-(2.*z))./sqrt((20-(2.*z)).^2+100))+((20+(2.*z))./sqrt((20+(2.*z)).^2+100)));

plot(z,H)

title('plot of |H| vs z along the axis of solenoid')

ylabel('Magnitude of magnetic field 'H")

xlabel('position on axis of solenoid 'z")

Explanation:

full explanation is attached as picture and the resultant plot also.

Answer:

472,826 lb/hr

Explanation:

As per the given data:

Solids in feed= 2%

Solids in concentrate= 25%

HTA1 = HTA2 = 1000 ft^2

U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F

Overall material balance: Feed= Distillate + concentrate ----> Eq-1

Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25

Feed = 12.5 * concentrate ---> Eq-2

Boiling point rise = negligible, so solution & solvent vapor temperature will be same.

Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).

As per the data:

Latent heat 212F = 300 Btu/lb

Latent heat 100F = 320 Btu/lb

As per material balance:

Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT

1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr

2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr

Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr

Substituting the above in Eq-1

Feed = 435,000 + concentrate

Substitute Eq-2 in the above

12.5 * concentrate = 435,000 + concentrate

Concentrate, L1 = 435,000/11.5 = 37826 lb /hr

Feed, F = 435,000 + 37826 = 472,826 lb/hr

1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.

[ Find the figure in the attachment ]

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]

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

Question: The program was not attached to your question. Find attached of the program and the answer.

Answer:

See the explanation for the answer.

Explanation:

#include <iostream>

using namespace std;

int main()

{

   cout << "How many students will you enter? ";

   int n;

   cin>>n;

   string *name = new string[n];

   double *gpa = new double[n];

   for(int i=0;i<n;i++){

       cout<<"Enter student"<<(i+1)<<"'s name: ";

       cin>>name[i];

       cout<<"Enter student"<<(i+1)<<"'s gpa: ";

       cin>>gpa[i];

   }

   cout<<"The list students"<<endl;

   cout<<"Name             GPA"<<endl;

   cout<<"----------------------"<<endl;

   for(int i=0;i<n;i++){

       cout<<name[i]<<"              "<<gpa[i]<<endl;

   }

   return 0;

}

OUTPUT : See the attached 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:

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:

Here, The CML is used for efficient portfolios whereas the SML applies to all portfolios or securities

Considering the seperation theorem, The separation theorem states that the investment decision is sepaarte from the financing decision.

This implies that The SML can be used to analyze the relationship between risk and requried return for all assets.

Under the separation theorem investors should Hold the same portfolio of risky assets and the same expected return but at different levels of risk

Explanation:

See answer

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

Answer:

(a) The mass glow rate of water is 0.0032kg/s

(b) The heat removal from air is 14.17KJ/sec

Explanation:

In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.

To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature

Please check attachment for complete solution and step by step explanation

Answer:

0.003

Explanation:

? → ? = −??ℎ??→ ?

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

Answer:

Dude just like forget the highway and drive on the interstate bruh

Explanation:

Answer:

See all solutions attached as picture.

Explanation:

It is well explanatory

Answer:

The force on the strut is 909000 lbm ft/s²

Explanation:

please look at the solution in the attached Word file