A photovoltaic panel of dimensions 6 m x 5 m is located on top of the roof of a house. Solar irradiation Gs = 900 W/m2 is incident on the panel. The panel has an absorptivity to solar irradiation, αs of 0.92. The freestream air temperature is Tinf and the surrounding temperature for radiation exchange with the sky is Tsurr. For this particular problem, it is given that Tinf= Tsurr. The convective heat transfer coefficient of air blowing over the panel is h W/m2-K.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the full step by step explanation to your question.

Answer:

[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]

Answer:

Explanation:

Write the equation for Reynolds number as follows:

Re = VL/v

For dynamic similarity,

(VL/v)m + (VL/v)p…… (1)

Since, the model and prototype are in same medium, the kinematic viscosity remains same.

From equation (1), we can write

(VL)m = (VL)p

Here, L represents length, and V is the velocity.

Re-write the equation as follows:

Vm = Lp/Lm x Vp

Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .

Vm = 1/8 x 1.5

Vm = 0.1875m/s

Therefore, the wind tunnel air speed is 0.1875m/s.

Answer:

[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

[tex]\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}[/tex]

Where:

[tex]a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g[/tex]

[tex]a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a_{1} = -3.040\,\frac{m}{s^{2}}[/tex]

Likewise, the deceleration of the car on the unpaved shoulder is:

[tex]a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g[/tex]

[tex]a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a_{2} = -2.549\,\frac{m}{s^{2}}[/tex]

The speed just before the car entered the unpaved shoulder is:

[tex]v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}[/tex]

[tex]v_{o} = 18.175\,\frac{m}{s}[/tex]

And, the speed just before the pavement skid was begun is:

[tex]v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}[/tex]

[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]

The initial speed of the vehicle just before the pavement skid was begun is 5284.65 ft/hr.

Dry pavement friction coefficient Fdry = 0.33

Length of skid marks on dry pavement ddry = 100 ft

Friction coefficient on unpaved shoulder Fshoulder = 0.28

Length of skid marks on unpaved shoulder = 200 ft

First, let's calculate the work done on dry pavement:

Work on dry pavement = Fdry × ddry = [tex]0.33 *100[/tex]

= 33 ft·lbf

Work on unpaved shoulder = Fshoulder × dshoulder

= [tex]0.28 * 200[/tex]

= 56 ft·lbf

Total work done = Work on dry pavement + Work on unpaved shoulder = 33 + 56

= 89 ft·lbf

Assuming the car's mass remains constant, and the final speed is 0, we have:

89 ft·lbf = (1/2)m × (10 mi/hr)²

Convert the final speed to feet per hour:

[tex]10 mi/hr = 10 × 5280/3600 = 5280 ft/hr[/tex]

Now, solve for the initial speed:

v = √((2 × 89 ft·lbf) / m)

v ≈ √((2 × 89) / (6.38 × 10⁻⁶)) ft/hr

v ≈ [tex]\sqrt{(27889055.9}[/tex] ft/hr

v ≈ 5284.65 ft/hr

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

The factor of safety for the states of plane stress can be determined using the ductile Coulomb-Mohr theory. The factor of safety is the ratio of the minimum yield strength to the maximum stress in the material. In this case, the factor of safety is approximately 0.889.

The factor of safety for the states of plane stress can be determined using the ductile Coulomb-Mohr theory. The factor of safety is defined as the ratio of the minimum yield strength in tension and compression to the maximum stress in the material. In this case, the minimum yield strength in tension is Syt = 60 kpsi and in compression is Syc = 75 kpsi.

To find the factor of safety, we need to determine the maximum stress in the material. The maximum stress can be calculated using the formula:

Maximum stress = (Syt + Syc) / 2

Plugging in the values, we get:

Maximum stress = (60 + 75) / 2 = 67.5 kpsi

Now, we can calculate the factor of safety using the formula:

Factor of safety = Syt / Maximum stress

Plugging in the values, we get:

Factor of safety = 60 / 67.5 = 0.889

Therefore, the factor of safety for the states of plane stress is approximately 0.889.

Answer:

a. 0.02639

b. 1.98H

Explanation:

Please see attachment

The evidence that indicates that a reaction has occurred include the following:

B. A solid brown product formed.

C. The temperature increased.

D. A gas formed.

E. An explosion occurred.

A chemical reaction is a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

This ultimately implies that, a chemical change would give rise to the chemical properties of matter by causing the transformation of one chemical substance into one or more different chemical substances.

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

Please kindly check attachment for the step by step solution of the given problem.

Answer:

In place of spiral configuration, we can also use metal wire rings which will provide better strength against circumferential stress. But this spiral configuration will provide strength against circumferential along with axial stress also. If the vascular prosthesis gets elongated along axial direction the corresponding radius of the rings will become short and it will prohibit the expansion of the vascular prosthesis along the circumferential direction.

Explanation:

The orientation is attached below.

Answer:

[tex]P = 150.335\,kPa[/tex] (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

[tex]P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)[/tex]

[tex]P = 150.335\,kPa[/tex]

Answer:

Bending stress = 32.29ksi ∠ 33.0 ksi

Explanation:

CHECK BELOW FOR THE ANSWER IN THE FILE ATTACHED.

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

Answer: new depth will be 3.462m and the water elevation will be 0.462m.

The maximum contraction will be achieved in width 0<w<3

Explanation:detailed calculation and explanation is shown in the image below

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

please kindly see attachment for the step by step solution of the given problem.

Answer:

The pressure and power of fan is 1.77 and 11.18 Hp respectively.

Explanation:

Given:

Discharge [tex]Q_{1} = 30000[/tex] cfm

Pressure difference [tex]\Delta P = 4[/tex] inch

Efficiency [tex]\eta = 50\%[/tex]

(A)

From the formula of fan power,

     [tex]P _{1} = \frac{Q \Delta P}{6356 \eta}[/tex]

     [tex]P_{1} = \frac{30000 \times 4}{6356 \times 0.5}[/tex]

     [tex]P_{1} = 37.76[/tex] Hp

(B)

Fan power and pressure is given by,

We know that pressure difference is proportional to the square of discharge.

    [tex]\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}[/tex]

   [tex]\Delta P_{2} = (\frac{20000}{30000} ) \times 4[/tex]

   [tex]\Delta P_{2} = 1.77[/tex]

Fan power proportional to the cube of discharge.

       [tex]\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}[/tex]

       [tex]P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76[/tex]

       [tex]P_{2} = 11.18[/tex] Hp

Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.

Answer:

Cv = 0.026 cm²/min

t  = 52.60 min

v% = 41.86 %

tv = 0.1375

t = 8.53 min

v = 53.61 %

Explanation:

given data

height = 2.54 cm

50 % consolidation = 12 min

solution

we get here first Cv value that is express as

Tv = [tex]\frac{Cv\times t}{d^2}[/tex]    .................1

here Tv for 50% is 0.196

put here value and we get

0.196 = [tex]\frac{Cv\times 12}{\frac{2.54}{2}^2}[/tex]  

solve it we get

Cv = 0.026 cm²/min

and

for tv for 90 % consolidation is 0.848

put value in equation 1

0.848 =  [tex]\frac{0.026\times t}{\frac{2.54}{2}^2}[/tex]  

solve it we get t

t  = 52.60 min

and

v% will be here  is

v% = [tex]\frac{0.18}{0.43} \times 100[/tex]  

v% = 41.86 %

and

tv = [tex]\frac{\pi }{4}\times \frac{4}{100}^2[/tex]  

tv = 0.1375

so now put value in equation 1 we get

0.1375 = [tex]\frac{0.026 \times t}{\frac{2.54}{2}^2}[/tex]  

solve it we get

t = 8.53 min

and

now put value of t 14 min in equation 1 will be

tv = [tex]\frac{0.026 \times 14}{\frac{2.54}{2}^2}[/tex]  

t =  0.225 min

and v will be after 14 min

0.0225 = [tex]\frac{\pi }{4}\times \frac{v}{100}^2[/tex]  

v = 53.61 %

Answer:

Since the Reynolds number is below 3 x 10⁵ the use of ASTM B152 plate WILL NOT be in complaince with ASME Code for process piping.

Explanation:

Detailed Explanation is given in the attached document.

Answer:

Explanation:

check the attached files for solution.

Answer: the metal will experience a strain of approximately 2.037Mpa

This strain is lesser than if the force was applied at room temperature.

This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.

Explanation:

Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.

Answer: 1) tensile

2) 45 degree orientation

3) principal shear stress

Explanation:

An element representing maximum in-plane shear stress with the associated _tensile normal stresses is oriented at an angle of ____45_degree__ from an element representing the ____principal shear____ stresses.

Answer: −3.46kJ/K

Explanation:

From the question above, we have:

The mass of the block (m) = 45kg

The initial temperature of the block (T1) = 280∘C

The weight of the water (mw) = 100kg

The temperature of water (Tw) = 18∘C

Recall the energy balance equation,

ΔUI = −ΔUw

In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.

[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w

Here cp is the specific heat at constant pressure.

The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.

Now, we substitute the values in above equation,

[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w

[20.25(T2 − 280)] = −[418(T2 − 18)]

20.25T2 − 5,670 = −[418T2 − 7,524]

20.25T2 − 5,670 = −418T2 + 7,524

20.25T2 + 418T2 = 7,524 + 5,670

438.35T2 = 13,194

T2 = 30.1K

Recall, the expression to calculate the total entropy change is given as:

ΔStotal = ΔSI + ΔSw

ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w

Now, we substitute the values in above equation,

ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w

ΔStotal = −12.55 + 9.09

ΔStotal = −3.46kJ/K

Thus the total entropy change is −3.46kJ/K.

Answer: 112 + 19.27

Explanation:

Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

Super curve elevation (e) = 4%

4/100= 0.04

e= V^2/gR

Make R the subject of the formula.

egR= V^2

R= V^2/eg

V= 45mph

=45 × 0.44704m/s

=20.1168m/s

g (force due to gravity) =9.81

Therefore,

R= (20.1168)^2/9.81 × 0.04

= 1031.31m

Tangent Length( T) = PI - PC

Tangent Length= 10875 - 10500

=375m

T= R Tan(I/2)

375= 1031.31 × Tan(I/2)

I= 39.96

Also,

L= πRI/180

= 719.27m

Station PT= Stat PC+ L

10500 + 719.27

=11219.27

=112 + 19.27

Answer:

Explanation:

By using Bernoulli's Equation:

[tex]\frac{P_1}{P_g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{P_g}+\frac{v_2^2}{2g}+z_2+f\frac{L}{D}\frac{v^2}{2g}[/tex]

where;

[tex]z_1 = z_2 \ and \ v_1 = v_2[/tex]

[tex]P_1 - P_2 = f \frac{L}{D}\frac{1}{2}\rho v^2[/tex]

[tex]P_1-P_2 = \frac{5 \ lb}{in^2}( 144 \frac{in^2}{ft^2})[/tex]

[tex]P_1-P_2 = 720 \frac{lb}{ft^2}[/tex]

[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684 \ ft^2/s}{\frac{\pi}{4}D^2} \\ \\V = \frac{8.51}{D^2}[/tex]

Density of gasoline [tex]\rho = 1.32 \ slug/ft^3[/tex]

Dynamic Viscosity [tex]\mu[/tex] = [tex]6.5*10^{-6} \frac{lb.s}{ft}[/tex]

[tex]P_1-P_2 = f \frac{L \rho V^2}{2D}[/tex]

[tex]720 = f \frac{L(100)(1.32)}{2D}(\frac{8.51}{D^2})^2[/tex]

D = 1.46 f

[tex]Re, = \frac{\rho VD}{\mu} = \frac{1.32 *\frac{8.51}{D^2} D}{6.5*10^{-6}}[/tex]

[tex]= \frac{1.72*10^6}{D}[/tex]

[tex]\frac{E}{D}= \frac{0.00015}{D}[/tex]

However; the trail and error is as follows;

Assume ; f= 0.02 → D = 0.667ft[tex]\left \{ {{Re=2.576*10^6} \atop {\frac{E}{D}=0.000225}} \right.[/tex]   [tex]\right \{ {{f=0.014} \atop {\neq 2}}[/tex]

f = 0.0145  → D = 0.0428 ft [tex]\left \{ {{Re=4.018*10^6} \atop {\frac{E}{D}=0.00035}} \right.[/tex] [tex]\right \{ {{f=0.015} \atop {\neq 0.0145}}[/tex]

f = 0.0156  → D = 0.43 ft [tex]\left \{ {{Re=4.0*10^6} \atop {\frac{E}{D}=0.000348}} \right.[/tex] [tex]f = 0.0156[/tex]

∴ pipe diameter d = 0.43 ft

Given that:

D = 1 ft

[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684}{\frac{\pi}{4}(1)^2} \\ \\ V = 8.51 \ ft/s[/tex]

[tex]Re = \frac{\rho \ V \ D}{\mu } \\ \\ Re = \frac{1.32 *8.51 *1 }{6.5*10^{-6}}[/tex]

[tex]Re = 1.72 *10^6[/tex]

[tex]\frac{E}{D} = \frac{0.00015}{1} \\ \\ = 0.00015[/tex]

[tex]f = 0.0136[/tex]

[tex]P_2-P_1 = \frac{fL \rho V^2 }{2D}[/tex]

[tex]P_2-P_1 = \frac{0.036(100)(1.32)(8.51)^2 }{2*1}[/tex]

[tex]P_2-P_1 = 65 \frac {lb}{ft^2}[/tex]  to psi ; we have:

[tex]P_2-P_1 = 0.45 \ psi[/tex]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

[tex]\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}[/tex]

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

[tex]\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m[/tex]

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

[tex]n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times[/tex]

c) By using the equation of the ideal gases you obtain:

[tex]PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules[/tex]

Answer

The Impedance Z = 20.982 ohms

The phase angle is Φ= - 70.51°

The current I= 0.477amp

Explanation:

This problem bothers on alternating current, this time an R-L-C circuit

What is a R-L-C circuit?

An RLC circuit is an electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or in parallel. The name of the circuit is derived from the letters that are used to denote the constituent components of this circuit.

N/B : Kindly find attached solutions and diagrams for your reference

Answer:

Horse power = 167.84 hp

Explanation:

Horsepower is calculated using the formula;

P = T * w

See the attached file for the calculation

a. This is a counter flow heat exchanger.

b. The effectiveness of  heat exchanger is 0.615

The  effectiveness of  heat exchanger is defined as ratio of actual heat transfer to the maximum possible heat transfer.

Given that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.

               [tex]Effectiveness=\frac{55-15}{80-15}=\frac{40}{65}=0.615[/tex]

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Answer:

Complex power=84 W

Explanation:

using equation

s=vs^2/2z

s=600^2/2z

s=84+29.4j

using s=P+jq

complex power=P=84 W

Answer:

Complex power (s) = 88235.3costheta + 88235.3sintheta

Explanation:

Wapplied = U* I *PF / 1000

where

Wapplied = real power (kilowatts, kW)

U = voltage (volts, V) = 600V

I = current (amps, A) = ?

PF = power factor = 0.85

I = Wapplied * 1000/ U * PF

I = (75 * 1000)/(600*0.85)

I = 147.058

Power = IV

Complex power (s) = IVcos thetha + IV sinthetha

Complex power (s) = 88235.3costheta + 88235.3sintheta

Answer:

0.09cm/sec

Explanation:

We are going to describe Hydraulic conductivity as a measure of the ease with which water flows through sediments, determining renewal rates of water, dissolved gases, and nutrients.

See attachment for a detailed solution.

The hydraulic conductivity of the given aquifer sample packed in a test cylinder is; 0.027 cm/s

We are given;

Head Loss; H = 16.3 cm

Length of cylinder; L = 50 cm

Diameter of cylinder; d = 6 cm

radius; r = 6/2 = 3 cm

time; t = 3 minutes = 180 s

Volume; V = 45.2 cm³

Formula for the hydraulic gradient is;

i = H/L

i = 16.3/50

i = 0.326

Formula for volumetric rate of flow is;

Q = V/t

Q = 45.2/180

Q = 0.25 cm³/s

Formula for area is;

A = πd²/4

A = π × 6²/4

A = 9π

Formula for hydraulic conductivity is;

K = Q/(A * i)

K = 0.25/(9π * 0.326)

K = 0.027 cm/s

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