Please read and an<br><br> 3. Many Jacks use hydraullc power.<br> A) O True<br> B) O False

In response to the market revolution:the legal system worked with local governments to find better ways to regulate entrepreneur

Nataliya [291]

Answer:

Local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.

Explanation:

The Market Revolution is the name given to change in the economy that occurred in the 19th century. This drastic change led to various important changes in the United States and across the world. During this period, capitalism became more entrenched and society became, for the first time, predominantly capitalist. This gave businessmen great power, as they played an increasingly important role when it came to economic growth. The power that they had influenced society deeply, including legislation. Judges often protected businessmen from paying property damages that were associated with their business enterprises. Moreover, workers had few rights and protections, and judges prevented them from unionizing.

5 0

3 years ago

Quantitative meaning

pychu [463]

Answer:

Explanation:

relating to, measuring, or measured by the quantity of something rather than its quality.Often contrasted with qualitative.

6 0

2 years ago

Read 2 more answers

1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo

Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]

Where,

σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]

q' = (1.4251*10^-8)* [ 24645536240 ]

q' = 351.22 W / m ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

Thermal conductivity ( k ) = 0.025596

Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

Re = U∞*d / v

= ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

= 52941.56574 ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

$Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3} }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3} }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\$

$Nu_D = 0.3 + \frac{127.59828 }{ 1.13824 }*1.27251 = 142.95013$

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

$h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K$

- The heat loss per unit length due to convection heat transfer is given by:

q'_convec = h*( π*d )* [ Ts - T∞ ]

q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

q'_convec = 11.49493* 130

q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

q'_total = q_a + q_convec

q'_total = 351.22 + 1494.34009

q'_total = 1845.56 W / m ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

$q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}$

q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

Where,

T_o = T∞ = 20°C

T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

$q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m$

Therefore,

q'_loss = 351.22 - 97.10

q'_loss = 254.12 W / m .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0

2 years ago

A sphere is assumed to have the properties of water and has an initial heat generation 46480 W/m^3 How much should the heat gene

Verdich [7]

Answer:

Resulting heat generation, Q = 77.638 kcal/h

Given:

Initial heat generation of the sphere, $Q_{Gi} = 46480 W/m^{3}$

Maximum temperature, $T_{m} = 360 K$

Radius of the sphere, r = 0.1 m

Ambient air temperature, $T = 25^{\circ}C$ = 298 K

Solution:

Now, maximum heat generation, $Q_{m}$ is given by:

$T_{m} = \frac{Q_{m}r^{2}}{6K} + T$ (1)

where

K = Thermal conductivity of water at $T_{m} = 360 K = 0.67 W/m^{\circ}C$

Now, using eqn (1):

$360 = \frac{Q_{m}\times 0.1^{2}}{6\times 0.67} + 298$

$Q_{m} = 24924 W/m^{3}$

max. heat generation at maintained max. temperature of 360 K is 24924 $W/m^{3}$

For excess heat generation, Q:

$Q = (Q_{Gi} - Q_{m})\times volume of sphere, V$

where

$V = \frac{4}{3}\pi r^{3}$

$Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}$

$Q = 90.294 W$

Now, 1 kcal/h = 1.163 W

Therefore,

$Q = \frac{90.294}{1.163} = 77.638 kcal/h$

3 0

2 years ago

Pretend 1% of the population has a disease. You have a test that determines if you have that disease, but it’s only 80% accurate

Ghella [55]

Answer:

3.88% of the time

Explanation:

3 0

2 years ago

Please read and an 3. Many Jacks use hydraullc power. A) O True B) O False