An R-134a refrigeration system is operating with an evaporator pressure of 200 kPa. The refrigerant is 10% in vapor phase at the inlet of the evaporator. The enthalpy and temperature of the refrigerant at the exit of the compressor are 360 kJ/kg and 70 /C respectively. If this system (with the refrigerant flowing at 0.005 kg/s) is used to cool a 750 kg product (initial temperature = 80 /C and specific heat = 3,000 J/kg-K), what will be the temperature of the product after 6 hours?

Answer:

The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day

Explanation:

Check the file attached for a complete solution.

The volume of the aeration tank was first calculated, V = 5000 m^3 / day.

The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

σ = 391.2 MPa

Answer:

See explaination

Explanation:

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

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]

Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

[tex] \sigma max= \frac{pd}{4t} [/tex]

[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]

= 50p

For the minimum normal strain due to pressure, we have:

[tex] E_max= \frac{change in L}{L_g} [/tex]

[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]

The minimum normal stress for a thin pressure vessel is 0.

[tex] \sigma _min = 0 [/tex]

i) Let's use Hookes law to calculate the pressure causing this deformation.

[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]

Substituting figures, we have:

[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]

[tex] 120 * 10^6 = 35p [/tex]

[tex] p = \frac{120*10^6}{35}[/tex]

[tex] p = 3.429 * 10^6 [/tex]

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]

[tex] = \frac{50p - 50p}{2} = 0 [/tex]

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]

[tex] = \frac{50p - 0}{2} = 25p [/tex]

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa

Answer:

See explaination

Explanation:

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

The total number of peak-hour trips generated by this village = 979.2 trips/hr

Given data :

Household size = 4

Number of households = 1600

Number of non-working people = 3200 ( 1600 * 2 )

Total annual Household income = $120,000

First step : calculate the number of peak hours based on shopping trips

i) Trips before move

0.12 + 0.09 * ( 6 ) + 0.011 * 50000 - 0.015 * (450)

= 483.16

ii) Trips after move

0.12 + 0.09 * 6 + 0.011* 50000 - 0.15 * 150

= 528.16

Next step : Calculate the number of peak hour travelled for recreational trips

per household

= 0.04 + 0.018 * 4  + 0.009 * 20  + 0.16 * 2

= 0.612 trips per household

Therefore the total number of peak-hour trips generated by this village

= 979.2 trips/hr

Hence we can conclude that The total number of peak-hour trips generated by this village = 979.2 trips/hr.

Learn more about peak-hour trips : https://brainly.com/question/4931057

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = [tex]\frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}[/tex]       .....................1

Q = [tex]\frac{T \infty - Tso}{\frac{1}{hA}}[/tex]                         .....................2

now we compare both equation 1 and 2 and put here value

[tex]\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}[/tex]            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

Answer:

  see attachments

Explanation:

A Karnaugh map for the output is shown in the first attachment. The labeled and shaded squares represent the cases where Y = HIGH. The associated logic can be simplified to

  Y = A3 xor A0

when the don't care at 1110 gives an output of HIGH.

__

The second attachment shows a logic diagram using a 4:1 multiplexer to do the decoding. A simple XOR gate would serve as well. If AND-OR-INV logic is required, that would be ...

  Y = Or(And(A3, Inv(A0)), And(A0, Inv(A3)))

Answer:

14.5° ; THD % = 3.873 × 100 = 387.3%.

Explanation:

Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;

(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".

(2). Consideration of up to 12th harmonic.

So, let us delve right into the solution to the question above;

Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;

Irms = reactive power /Vrms = 500/230 = 2.174 A.

Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.

Step two: Calculate the THD.

Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.

h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.

Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.

THD % = 3.873 × 100 = 387.3%.

Step four: angle AC - Ac converter

theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

Answer:

mass of chemical coming in per minute = 60 × 20 = 1200 mg/min

at a time t(min) , M = mass of chemical = 1200 × t mg

concentration of chemical = 1200t / 600 = 2t mg / gallon

Explanation:

Since it is only fluid that is leaving and not chemical.

Answer:

the police officer cruise each streets precisely once and he enters and exit with the same gate.

Explanation:

NB: kindly check below for the attached picture.

The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.

The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.

Here in this question, the door is the vertex and the road is the edge.

The information needed to detemine a Euler circuit and a Hamilton circuit is;

"the police officer cruise each streets precisely once and he enters and exit with the same gate."

Check attachment for each type of circuit and the differences.

the police officer cruise each streets precisely once and he enters and exit with the same gate.

Answer:

The idea is to protect the puddle while it cools

Explanation:

Answer:

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Explanation:

We are given the following parameters or data in the question as;

=> "The original specimen = 200 mm long and has a diameter of 13 mm."

=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."

=> " The total axial load is 20 kN"

Step one: Calculate the area

Area = π/ 4 × c^2.

Area = π/ 4 × 13^2 = 132.73 mm^2.

Step two: determine the stress induced.

stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.

Step three: determine the strain rate:

The strain rate = change in length/original length = 0.3/ 200 = 0.0015.

Step four: determine the modulus of elasticity.

modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.

Step five: determine the proportional limit.

proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.

Answer:

Modulus of Elasticity = 100 GPa

Proportional limit = 0.15 GPa

Explanation:

Axial Load = 20 kN = 20000 N

Original length, L₀ = 200 mm = 0.2 m

diameter, d = 13 mm = 0.013 m

Elongation, ΔL = 0.3 mm = 0.0003 m

Area of the material:

[tex]A = \frac{\pi d^{2} }{4} \\A = \frac{\pi 0.013^{2} }{4}\\A = 0.000133 m[/tex]

Stress = Load / Area

Stress = 20000 / 0.000133

Stress = 150375940 N/m²

Stress = Proportional limit = 0.15 GPa

Modulus of Elasticity = Stress/Strain

Strain = ΔL /  L₀

Strain = 0.0003 / 0.2

Strain = 0.0015

Modulus of Elasticity = 0.15 / 0.0015

Modulus of Elasticity = 100 GPa

Answer:

bye hooking plugs up to it to amp it up