Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with the symbol whose value is to be determined. Thompson Mechanical Products is planning to set aside $150,000 now for possibly replacing its large synchronous refiner motors whenever it becomes necessary. If the replacement is not needed for 7 years, how much will the company have in its investment set-aside account, provided it achieves a rate of return of 11% per year?

Answer:

A) After 2.5 milliseconds

Explanation:

Given that :

In a CAN bus, there are three computers: Computer A, Computer B, and Computer C.

Length in time of the message sent are 2 milliseconds long

As the computer start sending the message; the message are being sent at a message duration rate of 2 milliseconds

250 microseconds later, Computer B starts to send a message; There is a delay of 250 microseconds = 0.25 milliseconds here at Computer B

and 250 microseconds after that, Computer C starts to send a message

Similarly; delay at Computer C = 0.25 milliseconds

Assuming [tex]T_{RT}[/tex] is the retransmit time for Computer A to retransmit its message, Then :

[tex]T_{RT}[/tex] = [tex]T_A + T_B + T_C[/tex]

[tex]T_{RT}[/tex] = 2 milliseconds + 0.25 milliseconds + 0.25 milliseconds

[tex]T_{RT}[/tex] = 2.5 milliseconds

Thus; the correct option is A) After 2.5 milliseconds

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

In this question we must apply first law of thermodynamics and features of Carnot heat pump to determine how much time the system must be on each day to keep the temperature constant inside the house.

The net heat daily loss of the house ([tex]Q_{losses}[/tex]) is:

[tex]Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)[/tex]

[tex]Q_{losses} = 1.824\times 10^{6}\,kJ[/tex]

In order to keep the house warm, this heat must be equal to heat losses ([tex]Q_{H}[/tex]):

[tex]Q_{H} = Q_{losses}[/tex] (1)

Besides, the Coefficient of Performance for a Carnot heat pump ([tex]COP_{HP}[/tex]) is:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex] (2)

Where:

Given that [tex]T_{L} = 276.15\,K[/tex] and [tex]T_{H} = 295.15\,K[/tex], the Coefficient of Performance is:

[tex]COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}[/tex]

[tex]COP_{HP} = 15.534[/tex]

For a real heat machine, the Coefficient of Performance is determined by the following expression:

[tex]COP_{HP} = \frac{Q_{H}}{W}[/tex] (3)

Where:

The daily work consumed is now cleared in the previous expression:

[tex]W = \frac{Q_{H}}{COP_{HP}}[/tex] (3b)

[tex]W = \frac{1.824\times 10^{6}\,kJ}{15.534}[/tex]

[tex]W = 117419.853\,kJ[/tex]

The working time is calculated by dividing this result by input power. That is:

[tex]\Delta t = \frac{W}{\dot W}[/tex] (4)

[tex]\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)[/tex]

[tex]\Delta t = 3.624\,h[/tex]

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

To learn more on Carnot heat pump, we kindly invite to check this verified question: https://brainly.com/question/14019449

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740  × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

Complete Question:

A concrete building slab, on a basement floor is 5 m long, 3 m wide and 0.6 m thick. During the winter, the temperature is normally 18°C on the upper surface of the slab (the inside), and it is -7°C on the lower surface, with a linear temperature profile in between. If the concrete has a thermal conductivity of 1.4 W/m-K, what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operated with an efficiency of 66%, how much energy must be utilized to maintain the basement temperature for 90 days assuming that nearly all heat losses occur through the slab?

Answer:

a) Rate of heat loss, [tex]\dot{Q} = 875 W[/tex]

b) Energy that must be utilized to maintain the basement temperature,

Em = 10309 MJ

Explanation:

Length of the slab, l = 5 m

Width of the slab, w = 3 m

Thickness of the slab, t = 0.6 m

Cross Sectional Area of the slab, A = l x b

A = 5 x 3

A = 15 m²

Upper Surface Temperature, T₁ = 18°C = 18 + 273 = 291 K

Lower Surface Temperature, T₂ = -7°C = -7 + 273 = 266 K

a) The rate of heat loss is given by the formula:

[tex]\dot{Q} = KA \frac{dT}{dx} \\\dot{Q} = 1.4 * 15 \frac{291 - 266}{0.6}\\\dot{Q} = 1.4 * 15 \frac{25}{0.6}\\\dot{Q} = 875 W[/tex]

b) Energy, [tex]E = \dot{Q}t[/tex]

t = 90 days = 90 * 24 * 3600 = 7776000 s

E = 875 * 7776000

E = 6804000000J

E = 6804 MJ

If Efficiency = 66%, Energy that must be utilized to maintain the basement temperature.

6804 = 66% * Em

6804 = 0.66 *  Em

Em = 6804/0.66

Em = 10309 MJ

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg,  s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291  - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ =  1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683  + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

[tex]y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284[/tex]

The turbine control volume is given as follows;

[tex]\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2} \right ) + \left (1 - y \right )\left (h_{2} - h_{3} \right )[/tex]

= (3507.41  - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

[tex]\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6} \right ) + \left (1 - y \right )\left (h_{5} - h_{4} \right )[/tex]

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 -  151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

[tex]\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7} \right )[/tex]

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

[tex]\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}[/tex]

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, [tex]\dot{m_{1}}[/tex], into the first turbine stage is given as follows;

[tex]\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}[/tex]

[tex]\dot{m_{1}}[/tex] = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

[tex]\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}[/tex]

[tex]\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5} \right ][/tex]

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

Answer:

The engineers disagreed because their jobs were on the line

The ethical factors are:

The reason for the customer dumping the business is yet to be figured out

The need to keep cost to the lowest ebb in order to keep maintain profitability at the expense of employees' welfare

The are several ways of growing customer base which  are yet to be exploited.

Explanation:

The engineers disagree because there is no direct connection between the company's loss of the customer and their proposed layoff of the engineers,at least no one strong evidence has been given by the president.

The ethical factors inherent in this case are as follows:

The reason for the customer dumping the business is yet to be figure out

The need to keep cost to the lowest ebb in order to keep maintain profitability.

The are several ways of growing customer base which  are yet to be exploited.

There is  a need for a fact-finding exercise to establish the main motive behind losing such customer,without which the company can run into more troubles in future,otherwise the company would keep firing its good hands each time a customer dumps it.

Also,the president had resulted into such decision in order to maintain company's margins,where then lies ethics of welfare economics?Welfare economics is about looking beyond margins and looking at issues from a wider perspective of fulfilling the needs of employees in order for them to put in their best performance,at least by granting them job safety.

The company could have also grown business by investing in new technology that sets it apart from competitors instead of just jumping into the conclusions of sacking employees in a business where the company's strength lies in quality of engineers that it has.

Answer:

A) About [tex]9.273 \times 10^{-9}[/tex] newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) [tex]F_g=\dfrac{GM_1M_2}{r^2}[/tex] , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

[tex]F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}[/tex]

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

Answer:

55%

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

This is a Carnot system

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - [tex]\frac{Tc}{Th}[/tex]

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - [tex]\frac{500}{1100}[/tex]

Eff = 1 - 0.45 = 0.55 or 55%

Answer:

a laudry automaat has no nonfunctional parts

whywould they put something in there that has no function ?

Answer:

When you build a compressor that is supposed to run in an enclosed system you want the coolant medium (and the lubricants) in this system to function optimally during the service life of the system.

First of all, the coolant medium has a tendency to absorb water and this moisture can be aggressive on the internal components of the compressor so you want to remove this once the system is sealed.

Second, you want to filter out any particles that may shorten the life of the system.

Those are the two purposes of the liquid line filter drier in the refrigeration system.

Explanation:

Answer:

Three objectives of a tariff are

1) To control trade between countries

2) To protect domestic industries

3) To provide a source of income

Three characteristics of a tariff are;

1) Adequate return

2) Attractive

3) Fairness

Explanation:

A tariff is an import or export tax placed on goods traded between countries, it serves to control the foreign trade between the two countries and to protect or develop local industry

A Tariff is an important source of income to countries

Three characteristics of a tariff are;

1) Adequate return

Proper return from the consumer should be factored in a tariff to account for the alternatives or normal expense pattern

2) Attractive

The tariff should be attractive to encourage consumption of electricity or complimentary goods

3) Fairness

Based on the consumption of related resources brought about by large scale utilization, large consumer tariff should be lower than those that consume less complementary resources.

Answer:

1) This is because too much fuel is needed to get a payload from the surface to orbital altitude an accelerated to orbital speed.

2) This is because space travel present extreme environment that affect machines operations and survival.

Explanation:

Hope it helps

sen yapsana mk halla halla yaw

Answer:

The heat transfer [tex]\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}[/tex]

[tex]W_{2-3}= -71.312 \ Btu[/tex]

Explanation:

At the initial state when P₁ = 10 lbf/in :

We obtain the internal energy u₁ and specific volume v₁.

u₁ = [tex]u_g[/tex] = 574.08 btu/lbm

v₁ = [tex]v_g[/tex]  = 2.4746  ft³/lbm

Process 1–2 occurs at constant volume until the temperature is 100°F.

i.e T₂ = 100⁰ F

At T₂ = 100⁰ F  : v₁ = v₂ = 2.4746  ft³/lbm  

[tex]\mathbf{u_2 = 591.28 + \dfrac{2.4746-2.5917}{2.117-2.5917}*(587.68 -591.28)}[/tex]

[tex]\mathbf{u_2 = 591.28 + 0.246682115(-3.6)}[/tex]

[tex]\mathbf{u_2 = 591.28+ (-0.888055614)}[/tex]

[tex]\mathbf{u_2 \approx 590.4 \ btu/lbm}[/tex]

[tex]\mathbf{Q_{1-2}= W+ \Delta U}[/tex]

[tex]\mathbf{Q_{1-2}= W+m( u_2 -u_1)}[/tex]

[tex]\mathbf{\mathbf{Q_{1-2}} = 0+2.2(590.4-574.08)}[/tex]

[tex]\mathbf{\mathbf{Q_{1-2}} = 0+2.2(16.32)}[/tex]

[tex]\mathbf{\mathbf{Q_{1-2}} = 35.904 \ Btu}[/tex]

b) the work for Process 2–3, in Btu.

At [tex]T_3 = 100 ^0 \ F[/tex] ; [tex]u_3 = 577.86 \ Btu/lbm[/tex]

[tex]Q_{2-3} = W_{2-3} + \Delta U[/tex]

[tex]Q_{2-3} = W_{2-3} + m(u_3-u_2)[/tex]

[tex]Q_{2-3} = W_{2-3} +2.2(577.86-590.4)[/tex]

[tex]-98.9 = W_{2-3} +2.2(577.86-590.4)[/tex]

[tex]-W_{2-3}= 2.2(577.86-590.4)+98.9[/tex]

[tex]-W_{2-3}= -27.588+98.9[/tex]

[tex]-W_{2-3}= 71.312[/tex]

[tex]W_{2-3}= -71.312 \ Btu[/tex]

Answer: Provided in the explanation section

Explanation:

1. Algorithm:

Input: Container storage capacity-W, Weights of n items w[n]. w[i] represents the weight of the nth item.

Output: a subset of items with the largest weight not exceeding W.

Algorithm: To find the subset of items with the largest weight not exceeding W, we can use a recursive algorithm. We define Solve(W, i) as a solution to the problem with weight W and i items, then our recurrence can be written as:

Solve(W,i) = max(Solve(W,i-1) , w[i] + Solve(W-w[i],i-1)). We get this relation because for every item (ith) item we have two options, either we include it in our container or we do not include it.

When we do not include ith items in the container, then we have to solve the problem for remaining items with the same weight so we get Solve(W,i-1).

When we include ith items in the container, then we have to solve the problem for remaining items with the reduced weight so we get w[i] + Solve(W-w[i],i-1). Here we have added w[i] because we have included ith item in our container.

2.  Using C++ Implementation:

#include<bits/stdc++.h>

using namespace std;

// this funtion finds the subset of items with the largest weight not exceeding W (Container storage capacity) and returns maximum possible weight of items

int solve(int w[],int W,int i,int n)

{

  //   if our weight has become zero or we have reached at the end of items then we simply return 0

  if(W==0 || i==n)

      return 0;

  else

  {

  // if weight of ith item is more than W then we have only one case, we have to ingore the ith item      

      if(w[i]>W)

      {

          return solve(w,W,i+1,n);

      }

      else

      {

          //now we have two cases, we can incude ith item or we can ignore ith item

          //case-1

          int include_ith_item = w[i]+solve(w,W-w[i],i+1,n);

          //case-2

          int exclude_ith_item = solve(w,W,i+1,n);

          //and we return the maximum of these two cases

          if(include_ith_item>exclude_ith_item)

          {

              return include_ith_item;

          }

          else

          {

              return exclude_ith_item;

          }

      }

  }

}

int main()

{

  //   some example data to test our funtion

  int w[5] = {10,12,13,9,43};

  int n=5;

  int W = 50;

  set <int> ::iterator it;

  cout<<"The largest possible weight of subsets is: "<<solve(w,W,0,5);

}

cheers i hope this helped !!!

Answer:

Dimension of diameter = L

Dimension of speed of advance, V = [tex]LT^{-1}[/tex]

Dimension of N = [tex]T^{-1}[/tex]

Dimension of coefficient of viscosity, μ = [tex]ML^{-1} T^{-1}[/tex]

Dimension of Propeller thrust, [tex]F = MLT^{-2}[/tex]

Explanation:

The unit of diameter, d is meters, hence dimension of d = L

The speed of advance, V is in m/s, hence the dimension = [tex]LT^{-1}[/tex]

The fluid density, p, is in kg/m³, hence the dimension = [tex]ML^{-3}[/tex]

Rotation rate, N, is in Rev/s, hence the dimension = [tex]T^{-1}[/tex]

coefficient of viscosity μ = [tex]ML^{-1} T^{-1}[/tex]

The propeller thrust can be given by the formula:

[tex]F = K_{T} \rho N^{2} d^{4}[/tex]

Where [tex]K_{T}[/tex] is the thrust coefficient

The dimension of the propeller thrust can then be derived as:

[tex]F = ML^{-3} (T^{-1})^2 L^4\\F = ML^{-3}T^{-2}L^4\\F = MLT^{-2}[/tex]

Answer: money & human lives

Explanation:

answer on career safe!

The high cost occur in Money and Human lives.

Car crashes accounts for an average of 38,000 death in the US per year and the country have the highest number of car crashes in the world.

In conclusion, the high costs that occur as a result of Car Crash is Money (Indemnity) and loss of Human lives.

Learn more about this here

brainly.com/question/18086340

Answer: B. thermocouple measures temperatures at the tip and the thermistor at the dimple.

Explanation:

A thermistor is a temperature-sensitive resistor, whilst a thermocouple generates a voltage proportional to the temperature. Thermocouples can work at much higher temperatures than thermistors. They are commonly used for temperature control in heating systems.

Answer:

Steam injection increases the mass flow rate of steam in the turbine. As a result, the power output from the turbine also increases.

Explanation:

Steam injection increases the mass flow rate of steam in the turbine. As a result, the power output from the turbine also increases.

Thermal efficiency increases as work output increases but heat addition remains constant. The process of steam injection is similar to the regeneration process. The exhaust gases which are coming from the exit of the turbine can be used to obtain steam.

Answer:

nice cake

Explanation:

Answer:

but the way is the way but the WAY is not the way

Explanation:

(yoda voice)

Answer:

umm am i supposed to be laughing?

Explanation:

sorryyy idk what yoda sounds like cuz i think star wars is pretty wack, just my opinion tho so dont get offended

Answer:

Explanation:

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 mm and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen.Compute the second specimen's radius after deformation.

The percentage of cold work to be done to deform the two cylindrical is calculated

[tex]\% CW=\frac{A_o-A_d}{A_o} \times100\\\\=\frac{\pi r_o^2-\pi r^2_d}{r_o^2} \times100[/tex]

we input the values

[tex]=\frac{15^2-12^2}{15^2} \times 100\\\\=\frac{225-144}{225} \times100\\\\=\frac{81}{225} \times 100\\\\=36 \%[/tex]

We can now calculate the deformed radius of the second specimen for the same deformation

[tex]r_2=r_o\sqrt{1-\frac{\% CW}{100} }\\\\=11\sqrt{1-\frac{36}{100} } \\\\=11\sqrt{1-0.36} \\\\=11\sqrt{0.64} \\\\=11\times0.8\\\\=8.8mm[/tex]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

Answer:

The answer is 6392 W

Explanation:

Solution

Given that:

The total number of students in the class room is  = 57

There are 18 light fluorescent  bulbs of 40 W

The heat generated by an individual is  = 100 W

The ballasts consume an additional of 10% from the generated heat

The rate of internal heat generation in he classroom when it is completely filled up is the sum or addition of heated generated because of the ballasts, light bulbs and individuals or persons is denoted as follows:

Q total = Q people + Q lights W

Now,

The heat that is been generated by the students in the classroom when occupied completely is stated below:

Q people = ( The number of person ) *  (Q each)

Thus,

56 * 100

= 5600 W

Secondly, the heat generated from the lights is the same as the electrical energy been used by the lights

So,

Q lights = (Total number of lights ) * (Q elec)

18 * 40

= 720 W

Then

The heat used by the ballasts is at 10 % of the Q lights

Which is  Q ballasts = 10/100 * 720

= 72 W

Now,

We sum up both the heat generated from the lights bulbs, by the students and from the ballasts.

Thus

The the total heat generated from the classroom is shown below:

Q total = Q people + Q lights

=5600 + 720 +72

=6392 W

Therefore the total heat generated is 6392 W

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

Answer:

1. To bring down the pressure of the refrigerant

2. To meet up with the load to be refrigerated (the amount of heat to be evacuated)

Explanation:

1. To bring down the pressure of the refrigerant

The high pressure of the refrigerant coming from the condenser require reduction to enable vaporization in the evaporator at the proper temperature

The throttling valve as a small aperture through which the refrigerant flows that lowers the pressure of the refrigerant to a point  at which the refrigerant vaporize of which the refrigerant then passes into the evaporator in a partly as liquid and vapor at a low temperature and pressure

2. To meet up with the load to be refrigerated (the amount of heat to be evacuated)

The throttling valve allows more refrigerant to flow through it when there is an increased load at a higher temperature to be refrigerated

Similarly, in a condition of reduced refrigeration load, hence, a lesser amount of heat to be evacuated, the throttling valve restricts the amount of flow of the refrigerant through it.

Since the Earth contains a constant temperature range heat exchangers can be built underground without other added expenses yes they are expensive to put underground but the pay off is greater long term wise since the heat rises it can be said you could put a heatpump in a small area and it can over time increase the relative heat in that area of effect.

hope this helps look into, heat displacement thru underground heating to get a clearer picture

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA,  IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters          IE       IC           IB            VE       VB         Vc

Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

Answer:

a) 17.20

b) 11.31

c) 14.42

d) 12.65

Explanation:

(a)

The girl is at the origin of the x,y coordinates  (i.e 0,0,0  )

the position vector of the car at time 't' secs is

[tex]\vec{r}= 2+2t^2, 6+t^3,0[/tex]

at t=2s,  the position vector is

[tex]\vec{r}= 10, 14,0[/tex]

Therefore, the the distance between the car and the girl is

[tex]s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}\[/tex]

s = 17.20

(b)

The position of the car at  t = 0s is [tex]\vec{r}_0 = 2,6,0[/tex]

The position of the car at t = 2s is [tex]\vec{r}_2 = 10,14,0[/tex]

The distance of the car traveled in the interval from t=0s to t=2 s is as follows:

[tex]s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02} = 11.31[/tex]

(c)

The position vector of the car at time 't' secs is

[tex]\vec{r}= 2+2t^2, 6+t^3,0[/tex]

The velocity of the car is

[tex]\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0[/tex]

the direction of the car's velocity at t = 2s is  going to be

[tex]\vec{v}\mid _t=2 8, 12,0[/tex]

Thus; The speed of the car is

[tex]v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\ v_{t=2}= 14.42[/tex]

(d)    the car's acceleration is:

[tex]\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0[/tex]

The magnitude of car's acceleration at t=2s is

[tex]\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65[/tex]