How must a force be applied to cause resonance?

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

Please help i'm going to throw up from stress

Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0

3 years ago

A radio frequency identification application would most likely interface with a (an):_________

Irina-Kira [14]

A radio frequency identification application would most likely interface with an Operational Data Store.

The Operations Data Store (ODS) is a central database that provides the latest data snapshots from multiple transaction systems for operational reporting.

It allows organizations to combine data in its original format from various sources into a single destination to provide business reporting.

ODS contains integrated updates from operational sources and supports business intelligence (BI) tools to facilitate tactical decision making.

For example, an administrator can configure ODS to pull weekly batches of data from a billing application that is rarely updated, importing individual transaction records as they occur in the sales database(thanks to these database triggers), then combine the two into new relational tables.

As a result, querying and reporting on operational data in ODS comes with the assurance that these integrated tables contain the latest and most relevant snapshots of the business.

Learn more about Operations Data Store here : brainly.com/question/14925154

#SPJ4

4 0

2 years ago

Please need help on this

Nookie1986 [14]

The answer is c) 80°c

5 0

3 years ago

Read 2 more answers

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

4 years ago