Determine the flow velocities at the inlet and exit sections of an
inclined tapering pipe using fluid flow theory and given pressure
readings and flow rates.

There is a sloping pipeline that has one end 1.35 m higher than the
other. The pipe section tapers from 0.95 m diameter at the top end to
0.44m diameter at the lower end. The difference in pressure between
the two sections is 12.35kPa, with pressure being greater at higher
level.
Your task is to determine the inlet and exit velocities and the
volume
flow rate of the inclined pipe.

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:

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:

Cpk = 1.33

Explanation:

Given:

Mean = 50

Sd = 2

USL = 60

LSL = 42

The Cpk means process capability index. it helps decide the specification limit when the nominal value is not the central value of upoer specification limint (USL) and lower specification limit (LSL).

The Cpk can be derived using the formula:

[tex]Cpk = min [\frac{(usl - mean)}{3 * \sigma}, \frac{(mean - lsl)}{3*\sigma}] [/tex]

[tex]Cpk = min [\frac{(60 - 50)}{3*2} , \frac{(50 - 42)}{3*2}] [/tex]

Solving further,

[tex]Cpk = min [\frac{10}{6} , \frac{8}{6}] [/tex]

Cpk = min ( 1.67 , 1.33)

Cpk = 1.33

Cpk = 1.33

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:

thrust = ML[tex]T^{-2}[/tex]  

Explanation:

T = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re

where D is diameter

p is the density

N is the revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity

μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]s^{-1}[/tex]

Reynolds number is dimensionless so it cancels out

diameter is m = L

speed is in m/s = L[tex]T^{-1}[/tex]

fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]

N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] = [tex]T^{-1}[/tex]

combining these dimensions into the equation, we have

thrust = M[tex]L^{-3}[/tex][tex]( LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1}L }{LT^{-1} }[/tex]

= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex][tex]L^{2}[/tex]

thrust = ML[tex]T^{-2}[/tex]   which is the dimension for a force which indicates that thrust is a type of force

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:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = [tex]\lambda e^{-\lambda t}[/tex]

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

[tex]P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1[/tex]

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

[tex]p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905[/tex]

(c) The Poisson distribution is presented as follows;

[tex]P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}[/tex]

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) [tex]CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}[/tex].

A. The mean time to fail is:

For an exponential distribution, we have f(t) = e^(1000·λ) - 0.1·e^(1000·λ)= 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

Exponential distribution is the probability the distribution of the time between events in a Poisson point process.

B. Here we have to integrate from 5000 to ∞ as follows: 0.5905

C. The Poisson distribution is presented as follows;

p(x = 3) = 3.915 × 10⁻¹²

D. Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours.

The Cumulative Distribution Function is given as follows; p( t ≤ 1/4) = 2.63 × 10⁻⁵

Therefore, the correct answers are:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Read more about exponential here:

https://brainly.com/question/12242745

Answer:

The inside temperature, [tex]T_{in}[/tex] is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×([tex]T_s[/tex] - [tex]T_{\infty[/tex]) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

[tex]P = \dfrac{K \times A \times \Delta T}{L}[/tex]

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

[tex]2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}[/tex]

[tex]T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C[/tex]

The inside temperature, [tex]T_{in}[/tex] = 247.99 °C  ≈ 248 °C.

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:

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:

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:

a laudry automaat has no nonfunctional parts

whywould they put something in there that has no function ?

Answer:

the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

the convection coefficient is h = 4.26 W/m².K

Explanation:

From the question:

The temperature distribution across the wall is given by :

[tex]T(x) = ax+bx+cx^2[/tex]

where;

T = temperature in ° C

and a, b, & c are constants.

replacing 200° C for a, - 200° C/m for b and  30° C/m² for c ; we have :

[tex]T(x) = 200x-200x+30x^2[/tex]

According to the application of Fourier's Law of heat conduction.

[tex]q_x = -k \dfrac{dT}{dx}[/tex]

where the rate of heat input [tex]q_{in} = q_k[/tex] ; Then x= 0

So:

[tex]q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}[/tex]

[tex]q_{in}= -1 (-200+60x)_{x=0}[/tex]

[tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

Thus , the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

The rate of heat output is:

[tex]q_{out} = q_{x=L}[/tex]; where x = 0.3

[tex]q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}[/tex]

replacing T with [tex]200x-200x+30x^2[/tex] and k with 1 W/m.K

[tex]q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}[/tex]

[tex]q_{out} = -1 (-200+60x)_{x=0.3}[/tex]

[tex]q_{out} = 200-60*0.3[/tex]

[tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

Therefore , the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

Using energy balance to determine the change of energy(internal energy) stored by the wall.

[tex]\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\[/tex]

[tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

Thus; the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.

So:

[tex]q_{x=L} = q_{convected}[/tex]

[tex]q_{x=L} = h(T(L) - T _ \infty)[/tex]

where;

h is the convective heat transfer coefficient.

Then:

[tex]Replacing \ 182 W/m^2 \ for \ q_{x=L} , (200-200x +30x \ for \ T(x) \ , 0.3 m \ for \ x \ and \ 100^0 C for \ T[/tex] We have:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient is h = 4.26 W/m².K

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:

The answer is "0.0728"

Explanation:

Given value:

[tex]P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\[/tex]

                                     [tex]= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\[/tex]

if [tex]P<P^{*} \to[/tex] flow is chocked

if [tex]P>P^{*} \to[/tex] flow is not chocked

When  P= 10 psia < [tex]P^{*}[/tex] [tex]\to[/tex] not chocked

match number:

[tex]\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}[/tex]

                       [tex]= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}[/tex]

[tex]M_0=7.625[/tex]

[tex]p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}[/tex]

  [tex]=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}[/tex]

[tex]\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\[/tex]

[tex]\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\[/tex]

                                    [tex]=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\[/tex]

R= gas constant=1716

[tex]m=PAV\\\\[/tex]

    [tex]=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}[/tex]

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

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.

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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

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.

The minimum circuit ampacity for an air-conditioning condensing unit calculated by compressor amps + fan amps. The correct option is 1.

Ampacity is the term used for the greatest current conveying limit, in amperes, of a specific electrical gadget.  The current carrying capacity is generally depended upon the electrical cable and is calculated as the maximum amount of current a cable can withstand before it warms past the most extreme working temperature.

So, in air conditioning units,  the greatest current carrying capacity is  equal to the addition of compressor and fan current capacity.

Thus, the correct option is 1.

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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:

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.

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]

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: 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:

The maximum mixing height in meters is 5,000 m

Explanation:

In this question, we are expected to calculate the maximum mixing height in meters.

We identify the following parameters;

Lapse rate(L) = 0.2 °C/100 m = 0.002 °C/m (divide 0.2 by 100 to get this)

Mathematically;

Lapse rate(L) = Temperature difference/altitude (h)

Where the temperature difference is (25 °C -15 °C) = 10°C

Substituting this in the formula, we have;

0.002 = 10/h

h * 0.002 = 10

h = 10/0.002

Altitude(h) = 5,000 m

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

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Answer:

Option 4 =>  only different angles for theta 4, Theta 3 should have the same values in both cases.

Explanation:

The vector loop of Section 4.6 that was made as the reference diagram has the vector loop closing itself and the vectors summing around loop zero. Also, it has one DOF mechanism and the vector length is equal to the vector link.

NOTE: Kindly check the attached file or picture for the diagram of Section 4.6 diagram.

We can see from the diagram below that the V3 and V4 can be determined or Calculated with parameters such as theta 2, all links length and length d.

Therefore, the cross and the open configurations of the circuit will produce from the diagram will be " only different angles for theta 4, Theta 3 should have the same values in both cases."( That is option 4)

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