A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which is at 7 °C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet  = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

Given Information:

Temperature = T₁ = 50 °C

Mass flow rate = m = 30 kg/s

Exit Pressure = P₂ = 1.5 MPa = 1500 kPa

Required Information:

Power = P = ?

Answer:

Power = 45.16 kW

Explanation:

The power of the pump can be found using,

P = m*W

Where m is the mass flow rate and W is the work done by pump.

Work done is given by

W = vf*(P₂ - P₁)

Where vf is the specific volume and its value is found from the saturated water temperature table.

at T =  50 °C

vf = 0.001012 m³/kg

P₁ = 12.352 kPa

P = m*W

P = m*vf*(P₂ - P₁)

P = 30*0.001012*(1500 - 12.352)

P = 45.16 kW

Answer:

a) 23.39

b) 44977.08 N

c) 1922.92N

d) 454.31 MPa

e) 8.32 MPa

f) [tex] 3.47*10^-^3 [/tex]

Explanation:

a) fiber-matrix load ratio:

Let's use the formula :

[tex] \frac{F_f}{F_m} = \frac{V_f E_f}{V_m E_m}[/tex]

[tex] = \frac{0.3 * 131 GPa}{0.7 * 2.4 GPa} = 23.39 [/tex]

b & c)

Total load is given as:

Fc = Ff + Fm

46900 = Fm(23.39) + Fm

46900 = 24.39 Fm

Actual load carried by matrix=

[tex]F_m = \frac{46900}{24.39}[/tex]

= 1922.92N=> answer for option c

Actual load carried by fiber, Ff:

Ff = 46900 - 1922.92

Ff = 44977.08 N => answer option b

d)

Let's find area of fiber, A_f.

[tex] A_f = V_f * A_c[/tex]

Ac = Cross sectional area =300mm²

= 0.3 * 300 = 99 mm²

Area of matrix=

[tex] A_m = V_m * A_c[/tex]

= 0.7 * 300 = 231 mm²

Magnitude of the stress on the fiber phase:

[tex] \sigma _f= \frac{F_f}{A_f} [/tex]

[tex] \sigma _f= \frac{44977.08}{99} = 454.31 MPa [/tex]

e) Magnitude of the stress on the matrix phase.

[tex] \sigma _m = \frac{F_m}{A_m} [/tex]

[tex] \sigma _m = \frac{1922.92}{231} = 8.32 MPa[/tex]

f) Strain in fiber = [tex] \frac{\sigma _f}{E_f} [/tex]

[tex]= \frac{454.31*10^6}{131*10^9} = 3.47*10^-^3[/tex]

Strain in matrix = [tex] \frac{\sigma _m}{E_m} [/tex]

[tex]= \frac{8.32*10^6}{2.4*10^9} = 3.47*10^-^3[/tex]

Composite strain = [tex] (E_f *V_f) + (E_m * V_m) [/tex]

[tex] (3.47*10^-^3 * 0.3) + (3.47*10^-^3 * 0.7) = 3.47*10^-^3 [/tex]

Answer:

See explaination

Explanation:

By definition, we can say that Magnetic flux density is defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction. An example of magnetic flux density is a measurement taken in teslas.

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

Answer:

Explanation:

The solutions to these questions can be seen in the following screenshots from a solution manual;

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

Answer:

788°C

Explanation:

Metal palte details

Thickness = t=10 mm =0.01 m.

Thermal conductivity= k =200 W/mC

Heat capacity C=900 J/kg-k.

Initial temp Ti =22 C.

Time for which it has been kept in oven =t=2 min =120 sec.

Air properties.

Temp of air in oven =Ta =900 C

Heat transfer coefficient =h =190 W/m-k.

Temp of block after 2 min= T =?

In problem, it is not given block dimension. Let us assume that Block is 1 m in legth and 1 m in width.

Volume of block =V= 1*1*0.01 =0.01 m3 .(We know thckness =0.01 m).

Surface area of block = A=2(1*1+1*0.01+0.01*1)=2.04 m2.

Biot number for this problem is 0.004656 which is less than 0.1. So Lumped capacitance method is applicable to this problem, according to formulae,

T - Tair /Ti - Tair =e(-hAt/density* C* V) .

T - 900/(22-900) = e(-190*2.04*120 /2500*900*0.01) .

T - 900/(22-900) =e-2.0672 .

T -900 = - 110.09

Temp of block after 2 min =788 °C.

Answer:  Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35  MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35  MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα)    ---------------(1)

solving for Y we have;

Y =  35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα)   = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35  MPa√m) of the material.

cheers i hope this helps!!!!

Answer: Material selection

Explanation: As a biological engineer you need to be conversant with the composition, properties and processing of materials. This is important because when you design equipments and machines using your 3D model. During the fabrication process your knowledge of the composition of materials e.g 5% iron,

Their properties e.g ductility, non corrosive e.t.c and the way each and every materials being used in fabrication were processed plays a big role on how the 3D model would appear in real word. It’s durability and performance, so as a biological engineer this knowledge helps you make the right decision in materials selection.

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

Answer:

See attachment

Explanation:

Gain= Vo/Vin

      If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7

Using above value of gain, let's design non-inverting op-amp configuration

Gain= 1+Rf/Rin

3.7= 1= Rf/Rin

2.7= Rf/Rin

If Rin=100Ω then Rf= 270Ω

Answer:

Find the attachment for solution

Note: The resistance is measured in ohms (Ω). In the question Ω will replace V, where ever the resistance unit is required .

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

Answer:

3.586.543veh/hr

Explanation:

The question pertains to the calculation of final properties of air after an adiabatic process, using principles of thermodynamics applicable to an ideal gas in a closed, adiabatic system. Final pressure, work done, and entropy produced must be calculated, with specific formulas applying to each aspect.

The question is asking to calculate the final properties of air inside a rigid, insulated tank after being stirred by a paddle wheel. This is a thermodynamics problem specifically regarding a closed system where work is done without heat transfer (adiabatic process).

Since the tank is rigid, the volume remains constant, and since it is adiabatic and insulated, we know no heat is exchanged with the surroundings. For an ideal gas undergoing a process in a closed system, the relationship between pressure and temperature is given by Gay-Lussac's law, which states that P1/T1 = P2/T2 where P is pressure and T is temperature in Kelvin. The pressure will increase as the temperature increases.

For a rigid container, there is no change in volume; hence no boundary work is done. However, work is done by the paddle wheel, and this is equivalent to the change in the internal energy of the gas since it is an adiabatic process. Using the first law of thermodynamics, the work can be found. Since the internal energy of an ideal gas depends only on temperature, the work done by the gas would be the difference in internal energy, which is a function of the difference in temperatures and the specific heat at constant volume.

Since the process is adiabatic and there is work done on the gas, entropy is produced within the system. However, without interaction with the surroundings, the entropy exchanged with the environment is zero. The change in entropy of the system will thus be a function of the initial and final states of the gas, considering that entropy is a state function.

Answer:

The measurement of the indentation.

Explanation:

Due to disparities in operators making the measurements, the results will vary even under perfect conditions. Less than perfect conditions can cause the variation to increase greatly.

Answer:

ARPANET is the direct precedent for the Internet, a network that became operational in October 1969 after several years of planning.

Its promoter was DARPA (Defense Advanced Research Projects Agency), a US government agency, dependent on the Department of Defense of that country, which still exists.

Originally, it connected research centers and academic centers to facilitate the exchange of information between them in order to promote research. Yes, being an undertaking of the Department of Defense, it is understood that weapons research also entered into this exchange of information.

It is also explained, without being without foundation, that the design of ARPANET was carried out thinking that it could withstand a nuclear attack by the USSR and, hence, probably the great resistance that the network of networks has shown in the face of major disasters and attacks.

It was the first network in which a packet communication protocol was put into use that did not require central computers, but rather was - as the current Internet is - totally decentralized.

Explanation:

Below I present as a summary some of the most relevant aspects exposed on the requested website about the origin and authors of ARPANET:

1. Licklider from MIT in August 1962 thinking about the concept of a "Galactic Network". He envisioned a set of globally interconnected computers through which everyone could quickly access data and programs from anywhere. In spirit, the concept was very much like today's Internet. He became the first head of the computer research program at DARPA, and from October 1962. While at DARPA he convinced his successors at DARPA, Ivan Sutherland, Bob Taylor and MIT researcher Lawrence G. Roberts, of the importance of this network concept.

2.Leonard Kleinrock of MIT published the first article on packet-switching theory in July 1961 and the first book on the subject in 1964. Kleinrock convinced Roberts of the theoretical feasibility of communications using packets rather than circuits, That was an important step on the road to computer networking. The other key step was to get the computers to talk together. To explore this, in 1965, working with Thomas Merrill, Roberts connected the TX-2 computer in Mass. To the Q-32 in California with a low-speed phone line creating the first wide-area (albeit small) computer network built . The result of this experiment was the understanding that timeshare computers could work well together, running programs and retrieving data as needed on the remote machine, but that the circuitry switching system of the phone was totally unsuitable for the job. Kleinrock's conviction of the need to change packages was confirmed.

3.In late 1966 Roberts went to DARPA to develop the concept of a computer network and quickly developed his plan for "ARPANET", and published it in 1967. At the conference where he presented the document, there was also a document on a concept of UK packet network by Donald Davies and Roger Scantlebury of NPL. Scantlebury told Roberts about NPL's work, as well as that of Paul Baran and others at RAND. The RAND group had written a document on packet switched networks for secure voice in the military in 1964. It happened that work at MIT (1961-1967), in RAND (1962-1965) and in NPL (1964-1967) all they proceeded in parallel without any of the investigators knowing about the other work. The word "packet" was adopted from the work in NPL and the proposed line speed to be used in the ARPANET design was updated from 2.4 kbps to 50 kbps.

Answer:

See explaination

Explanation:

We are going to define Pressure drop as the difference in total pressure between two points of a fluid carrying network. A pressure drop usually occurs when frictional forces, caused by the resistance to flow, act on a fluid as it flows through the tube.

See attachment for the step by step solution of the given problem.

The time of submersion of the steel plate in years is; 10 years

We are given;

The time of submersion of the steel plate is given by the formula;

t = KW/(ρA*CPR)

Now K is a constant and is equal to 534 provided CPR is in mpy and

A is in square inches.

Thus;

t = (534 * 2.6 × 10⁶)/(7.9 * 10 * 200)

t = 8.8 × 10⁴ hours

Now, 24 hours makes one day and there are 365 days in a year. Thus;

number of hours in a year = 24 * 365

Thus;

t in years = (8.8 × 10⁴)/(24 * 365)

t ≈ 10 years

Read more about corrosion at; https://brainly.com/question/5168322

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

Final temperature: [tex]\( 336.575 \, \text{K} \)[/tex] . Vessel volume:  [tex]\( 100 \, \text{L} \).[/tex]

To solve this problem, we can use the principle of conservation of mass and energy.

1. **Conservation of Mass:** Since the membrane ruptures and the entire volume gets filled, the mass of steam remains constant.

2. **Conservation of Energy:** We can use the first law of thermodynamics (energy conservation) to solve for the final temperature.

Given:

- Initial state: 1 kg of steam at 400°C and 200 bar.

- Final pressure: 100 bar.

Let's solve:

1. Conservation of Mass:

Since mass is conserved, the final mass of steam will remain 1 kg.

2. Conservation of Energy:

Using the first law of thermodynamics:

[tex]\[ Q = m \cdot c_v \cdot \Delta T \][/tex]

where:

- ( Q ) is the heat transferred,

- ( m )is the mass of the substance (1 kg in this case),

[tex]- \( c_v \)[/tex] is the specific heat capacity at constant volume, and

[tex]- \( \Delta T \)[/tex]  is the change in temperature.

For steam, we can consider it to be an ideal gas at these high temperatures and pressures. The specific heat capacity at constant volume [tex](\( c_v \))[/tex]  can be taken as a constant value.

Now, we need to find the final temperature [tex](\( T_f \))[/tex] . We can use the ideal gas law to relate the initial and final states:

[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \][/tex]

Where:

[tex]- \( P_1 \) and \( T_1 \)[/tex] are the initial pressure and temperature,

[tex]- \( P_2 \) and \( T_2 \)[/tex] are the final pressure and temperature,

[tex]- \( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes, respectively.

Solving:

Given:

[tex]- \( P_1 = 200 \, \text{bar} = 20,000 \, \text{kPa} \)\\- \( T_1 = 400 \, \text{°C} = 673.15 \, \text{K} \)\\- \( P_2 = 100 \, \text{bar} = 10,000 \, \text{kPa\\} \)\\- \( m = 1 \, \text{kg} \)[/tex]

Let's assume specific heat capacity at constant volume [tex](\( c_v \))[/tex]  for steam to be approximately  [tex]\( 2.0 \, \text{kJ/(kg·K)} \)[/tex]  (This is a rough estimation).

From the ideal gas law:

[tex]\[ T_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot V_2}} \][/tex]

Given that initially, the other side of the vessel is evacuated, so [tex]\( V_2 \)[/tex] is negligible compared to [tex]\( V_1 \).[/tex]

[tex]\[ T_2 \approx \frac{{P_2 \cdot T_1}}{{P_1}} \]\[ T_2 \approx \frac{{10,000 \times 673.15}}{{20,000}} \]\[ T_2 \approx 336.575 \, \text{K} \][/tex]

Now, we can calculate the final volume using the ideal gas law:

[tex]\[ P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \]\[ V_2 = \frac{{P_2 \cdot V_1 \cdot T_1}}{{P_1 \cdot T_2}} \]\[ V_2 = \frac{{10,000 \times 1 \times 673.15}}{{20,000 \times 336.575}} \]\[ V_2 = \frac{{673,150}}{{6731.5}} \]\[ V_2 = 100 \, \text{L} \][/tex]

So, the final temperature of the steam is approximately

[tex]\( 336.575 \, \text{K} \)[/tex] and the volume of the vessel is [tex]\( 100 \, \text{L} \).[/tex]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

Answer:

R₁ = 32kΩ

Explanation:

See attached image

Answer:

False

Explanation:

cost Evaluation

assembly of equipment and materials used

The given statement is false.

The following information should be considered:

Learn more; brainly.com/question/16911495

Answer:

The answer is explained below for the sampling frequency

Explanation:

Solution

Recall that:

The frequency Fm1 = 1 kHz

The frequency Fm2 = 5 kHz

Fs = n (2Fm2)

so,

[Fs = 8 * ( 2 * 5) = 80 8kHz], This the sampling frequency for 8-bit A/D converter

Now,

By increasing quantization level we can also increase the substantial noise ratio

Answer:

The most positive is value of Vs is 0.5mV

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, [tex]g_{c}[/tex] = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

[tex]c_{p}[/tex]= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\[/tex]

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence [tex]v_{a} ^{2} = 0[/tex]

[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }[/tex]

[tex]T_{1}[/tex] = 20+460 = 480°R

[tex]T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}[/tex]= 533.25°R

Pressure at the inlet of compressor at isentropic condition

[tex]P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}[/tex]

[tex]P_{a}[/tex] = [tex](10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}[/tex]= 14.45 psia

[tex]P_{2}= 8P_{a} = 8(14.45) = 115.6 psia[/tex]

Answer:

a) The temperature and pressure of the gases at every point of the cycle are

T = 38.23 K

P = 2.91 kpa

Respectively

b) The velocity V of the gasses at the nozzle exit = 3590 m/s

Explanation: Please find the attached files for the solutions

Answer: (a). Ec(μ) = 165.6 GPa

(b). Ec(∝) = 83.09 GPa

Explanation:

this is quite straightforward, so we will go step by step.

from the data we have that,

Moduli of elasticity of the metal  -(Em) is 60 Gpa

Moduli of elasticity of oxide is  (Ep) is 380 Gpa

volume Vp = 33% = 0.33

(a). To solve the upper bound-modulus of the elasticity is calculate thus;

Ec (μ) = EmVm + EpVp ----------------(1)

where E rep the modulus of elasticity

v rep the volume fraction

c rep the composite

Vm = 100% - Vp

Vm =  100% - 33% = 67%

Vm = 0.67

substituting the valus of Em, Vm, Ep, Vp  from equation (1) we have;

Ec(μ) = (60×0.67) + (380×0.33)

Ec(μ) = 40.2 + 125.4 = 165.6 GPa

Ec(μ) = 165.6 GPa

(b). The lower bound modulus of elasticity can be calculated thus;

Ec(∝) = EmVp / EpVm + EmVp -------------- (2)

substituting values Em,Vm,Ep,Vp.

Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)

Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa

Ec(∝) = 83.09 GPa

cheers i hope this helps!!!!

Find the complete solution in the given attachments

Answer:

Check the explanation

Explanation:

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