Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of 100w/m^2 from the outer surfance of the pipe to the soil determine the exit temperature of the water.
a. 129.1
b. 111.1
c. 82.1
d. 68.1

Answer:

4 points

Para transportar una Unidad

Evaporadora Horizontal que mide 5

pies de largo, 3 pies de ancho y 4

pies de alto se necesita un volumen

de: *

23 pies cúbicos

12 pies cúbicos

60 pies cúbicos

O

63 pies cúbicos

Planos de construcciónExplanation:

Answer:

you quadruple the kinetic energy

Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.

Required:

a)Determine the power output of the turbine, in hp.

b) The Flow rate

Answer:

a) [tex]w=74.26hp[/tex]

b) [tex]m=0.22[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]p_1= 145 psi[/tex]

Initial Temperature [tex]T_1 =2700oR=>2240.33^oF[/tex]

Final Pressure [tex]p_2= 29 psi[/tex]

Final Temperature [tex]t_2=1974oR=>1514.33^oF[/tex]

Output Power [tex]w=74.26hp[/tex]

Heat transfer Rate [tex]Q=14BTU/s[/tex]

Generally the equation for Steady flow energy is mathematically given by

 [tex]Q-w=m(h_2-h_1)[/tex]

Where

 [tex]m=Flow\ rate[/tex]

From Steam table

 [tex]h_1=704btu/ib (at\ p_1= 145\ psi,\ T_1 =2700oR=>2240.33^oF )[/tex]

 [tex]h_2=401btu/ib (at\ p_2= 29psi\ t_2=1974oR=>1514.33^oF )[/tex]

Therefore

 [tex]-14-74.26=m(401-704)[/tex]

 [tex]m=\frac{-14-74.26}{(401-704)}[/tex]

 [tex]m=0.22[/tex]

Answer:

wind vane if it can be used to show wind speed and the other is a

Explanation:

please mark 5 star if im right and brainly when ya can

Answer:

DESCULPA MAS EU NÃO ENTENDI

Answer:

i) 3750 veh/hr/ln

ii) 100 veh/mi/In

iii) 37.5 mph

Explanation:

number of lanes = 3

sf for both directions = 75 mph  ( free mean speed )

Dj for both directions = 200 veh/mi/In

Calculate the value of  S0, D0 (veh/mi/ln) and maximum Vm (veh/hr)

For either direction we will consider the total volume = 3 lanes

value of Dj = 3 lanes * 200 = 600 veh/mi/

i) value of SO

=  ( Dj * sf ) / 4 = ( 600 * 75 ) / 4 = 11250 veh/hr  = 3750 veh/hr/lane

ii) Value of DO

DO = Dj / 2 = 200 /2 = 100 veh/mi/In

iii) Value of Vm

= sf /2 = 75 / 2 = 37.5 mph

Answer:

359.046 Ib

Explanation:

density = m / v ---- ( 1 )

m = mass

v = volume = l * b * h

Given data :

density = 30 Ib / ft^3

l = 2 inches = 0.1667 ft

b = 8 inches = 0.6667 ft

h = 11 ft

v ( volume ) = 0.1667 * 0.6667 * 11 = 1.22 ft^3

back to equation 1

m = 30 * 1.22  = 36.6 Ib

Actual weight ( F ) = m * g

                              = 36.6 * 9.81 = 359.046 Ib.

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= [tex]density\times g\times \frac{h}{2}[/tex]

By putting values, we get

= [tex]1000\times 9.8\times \frac{12.2}{2}[/tex]

= [tex]1000\times 9.8\times 6.1[/tex]

= [tex]59780[/tex]

hence,

The average force will be:

= [tex]Pressure\times Area[/tex]

= [tex]59780\times 3.6\times 12.2[/tex]

= [tex]2625537 \ N[/tex]

Or,

= [tex]2625 \ kN[/tex]

Answer: Hello your question is incomplete attached below is the complete question

answer :

Attached below

Explanation:

Given data:

The inputs of two registers are controlled by a 2-to-1 multiplexer.

The multiplexer select line and the register load enable inputs are controlled by inputs Co, C1, and C2.

Using the required transfers in the question to complete the detailed logic diagrams ( attached below )

Answer:

[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [[tex]\bar 1[/tex]01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

[tex]cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big][/tex]

where;

[tex][d_1\ e_1 \ f_1][/tex] = directional indices for tensile stress

[tex][d_2 \ e_2 \ f_2][/tex] = slip direction

replacing their values;

i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] =  1 & [tex]d_2[/tex] = -1 , [tex]e_2[/tex] = 0 , [tex]f_2[/tex] = 1

[tex]cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big][/tex]

[tex]cos \ \lambda = \dfrac{1}{\sqrt{2}}[/tex]

Also, to find the angle [tex]\phi[/tex] between the stress [001] & normal slip plane [111]

Then;

[tex]cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big][/tex]

replacing their values;

i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] =  1 & [tex]d_3[/tex] = 1 , [tex]e_3[/tex] = 1 , [tex]f_3[/tex] = 1

[tex]cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big][/tex]

[tex]cos \phi= \dfrac{1} {\sqrt{3} }[/tex]

However, the critical resolved SS(shear stress) [tex]\mathbf{\tau_c}[/tex] can be computed using the formula:

[tex]\tau_c = (\sigma )(cos \phi )(cos \lambda)[/tex]

where;

applied tensile stress [tex]\sigma =[/tex] 13.9 MPa

∴

[tex]\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})[/tex]

[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]

Answer:

a) 47500 mg/L

b) 5250366.444  kg/year

Explanation:

Given data:

suspended solids removal efficiency = 20%

Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day

Q = 20* 10^6 * 3.785412  Liters /day

settled concentration  ( St ) = 950mg/L * 0.2 = 190 mg/L

amount of settled solid = Q * St

                                   = ( 20* 10^6 * 3.785412 ) * 190  = 14384.5656 kg/day

∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day

a) concentration of solid in sludge  ( leaving the clarifier )

= amount of settled solid / flow rate out of the clarifier in liters/day

= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )

= 0.0475 kg/L

= 47500 mg/L

b) Determine mass of solids that is removed annually

= 14384.5656 kg/day * 365 days

= 5250366.444  kg/year

Answer:

30w

Explanation:

There are different aspect to system-dependent recovery equipment. The  pressure of the container will Increase to the saturation point.

System-dependent is known to be a passive recovery system. It is known as  the recovery of refrigerant from a system by using the refrigeration system's internal pressure and/or compressor so as to simulate the  recovery process.

System-dependent equipment is not used used with appliances that has more than 15 pounds of refrigerant.

When refrigerant pressure  increases, saturation temperature is known to also increases and as pressure decreases, the saturation temperature will also decrease. Saturation is known to have a pressure-temperature relationships for all refrigerants.

Learn more about saturation from

https://brainly.com/question/4529762

Answer:

Hence the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]

Explanation:

Now,

We know that,

[tex]\frac{M}{I} = \frac{E}{R}[/tex]

Thus, [tex]M =\frac{EI}{R}[/tex]

Here

R = 10000 mm  

[tex]I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}[/tex]

[tex]E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}[/tex]

Hence, the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]

Answer:

Following are the response to the given question:

Explanation:

In the given question, A curve one would be to "reject" them. Within this way, people are directly non-confrontationally off from their amorous interests or advances. The curve looks much like a current through the unidirectional load, and it has only a good maximum value. It seems like the total rating of the complete wave rectifier that is illustrated below, the present vas temporal curve please find it.

Answer:

yea

Explanation:

okkudjeheud email and delete the electronic version of this communication

11 [2-bit]

011 [3-bit]

0011 [4-bit]

________

1 + 1 + 1 = 3

________

3 = 2 + 1

2¹ 2⁰

3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)

3 = ..011

Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.

[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]

[ 128 64 32 16 8 4 2 1 ]

Answer:

the total charge which crosses into the collector is 60 nC

Explanation:

Given the data in the question;

current flowing into the collector lead of the bipolar junction transistor (BJT); i = 1 nA = 10⁻⁹ A

no charge was transferred in or out of the collector lead prior to t = 0

the current flow time t = 1 min = 60 sec

Now we write the relation between current, charge, and time;

i = dq / dt

where i is current, q is charge and t is time. { d refers to change }

Now,

[tex]q=\int\limits^t_{t=0} {i(t)} \, dt[/tex]

[tex]q=\int\limits^{t=60}_{t=0} { (10^{-9}) } \, dt[/tex]

[tex]q = ( 10^{-9}) (t)_0^{60[/tex]

[tex]q = ( 10^{-9}) ( 60 - 0 )[/tex]

q = 60 × 10⁻⁹ C

q = 60 nC

Therefore, the total charge which crosses into the collector is 60 nC

To Find :

The torque of the stud.

Solution :

We know, torque can be calculated by force and distance between the center and force applied.

So,

[tex]\tau = F\times R\\\\\tau = 40 \times 0.5 \ N .\ m\\\\\tau = 20 \ N .\ m[/tex]

Therefore, the torque on the stud is 20 N m .