At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Answer:

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Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

Answer:

Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:

Explanation:

Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.

Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.

Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.

Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.

Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).

Explanation:

Concentration overpotential, ηc,

I hope it helps you

Answer:

80⁰C

Explanation:

80°C.

Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities

Answer:

Explanation:

From the given information:

weight of fiber [tex]w_f[/tex] = 3.0 g

weight of composite specimen [tex]w_c[/tex] = 4.0 g

specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g

specific gravity of fiber [tex]S_f[/tex] = 2.4

specific gravity of matrix [tex]S_m[/tex] = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]

[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]

[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]

[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]

The experimental density [tex]\rho _{ce}[/tex] is determined  by using the equation:

[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]

[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]

The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]

[tex]= \dfrac{1.980-1.620}{1.980}[/tex]

= 0.1818

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

Answer:

Answer to the following question is as follows;

Explanation:

It is critical to communicate well during negotiations and conversation in order to attain your objectives. Communication also becomes essential in business and corporation . Effective communication may help you as well as your employees develop a strong professional relationship, which can boost morale and efficiency.

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

Answer:

B

Explanation:

Anxiety is very common especially nowadays but it's especially common in psychological disorders

Answer:

i really font onow why tbh eot you

Answer:

p q output ¬(p ∧ q)

0 0 1

0 1 1

1 0 1

0 0 0

p q output ¬p ∨ ¬q

0 0 1

0 1 1

1 0 1

0 0 0

Explanation:

We'll create two separate truth tables for both sides of the equation, and see if they match.

The expressions in the question use AND, OR and NOT operators.

Let's start with ¬(p ∧ q)

Now let's move on to the second expression ¬p ∨ ¬q

Therefore we can say the two expressions are equivalent.

Attached  the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:

As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.

De Morgan's law is a fundamental principle in propositional logic.

It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.

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Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

Answer:

I didn't understand your question or is it a fun fact

Answer:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

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Answer:

The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output

I hope it is helpful

Complete Question

Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted

Answer:

[tex]H=9.91kJ/sec[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=4 m/sec[/tex]

Pressure [tex]P=1000kPa[/tex]

Temperature [tex]T=227 \textdegree C[/tex]

Diameter [tex]d=3cm=>0.03m[/tex]

Generally the equation for volumetric Flow Rate is mathematically given by

[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]

[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]

[tex]V_r=0.002827m^3/s[/tex]

Generally the equation for mass Flow Rate is mathematically given by

[tex]m_r=\frac{PV_r}{RT}[/tex]

[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]

[tex]m_r=0.019kg/sec[/tex]

Generally the equation for mass Flow Rate is mathematically given by

Using gas Table for enthalpy Value

[tex]T=500K=>h=520.75kg[/tex]

Therefore

[tex]H=mh[/tex]

[tex]H=0.019*520.75[/tex]

[tex]H=9.91kJ/sec[/tex]

Solution :

Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]

[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]

                           [tex]$=\frac{1}{20}$[/tex]

                           = 0.05 kHz

[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex]  rad/s

We know that,

FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]

The sampled signal is :

[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]

So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

     [tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

Answer:

  v = 8.95 rad / s,    a = 22.3 rad / s²

Explanation:

This is an exercise in kinematics, where we must use the definitions of velocity and acceleration

          v = dr / dt

we perform the derivative

          v = 0+ 2 (-sin 5t) 5

          v = -10 sin 5t

we calculate for t = 0.85 t,

remember angles are in radians

           v = -10 sin (5 0.85)

           v = 8.95 rad / s

acceleration is defined by

           a = dv / dt

we perform the derivatives

          a = -10 (cos 5t) 5

          a = - 50 cos 5t

we calculate for t = 0.85 s

          a = -50 are (5 0.85)

          a = 22.3 rad / s²

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

Answer:

True

Explanation:

All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

 ( using 2nd image attached below )

description of sand :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Answer:

a)  [tex]m=0.17kg/s[/tex]

b)  [tex]Ma=0.89[/tex]

Explanation:

From the question we are told that:

Pressure [tex]P=60kPa[/tex]

Diameter [tex]d=3cm[/tex]

Generally at sea level

[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]

Generally the Power series equation for Mach number is mathematically given by

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

[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]Ma=0.89[/tex]

Therefore

Mass flow rate

[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]\rho=0.848kg/m^3[/tex]

Generally the equation for Velocity at throat is mathematically given by

[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])

Where

[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]

[tex]T_e=248[/tex]

Therefore

[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]

Generally the equation for Mass flow rate is mathematically given by

[tex]m=\rho*A*V[/tex]

[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]

[tex]m=0.17kg/s[/tex]

Answer:

  parallel

Explanation:

All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.