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4 years ago

6

Nitrogen (N2) contained in a piston–cylinder arrangement, initially at 10 bar and 405 K, undergoes an expansion to a final tempe

rature of 300 K, during which the pressure–volume relationship is pV1.3 = constant. Assuming the ideal gas model for the N2, determine the heat transfer in kJ/kg.

1 answer:

yKpoI14uk [10]4 years ago

6 0

Answer:28.21 kJ/kg

Explanation:

Given

$P_1=10\ bar$

$T_1=405\ K$

$T_2=300\ K$

Process $PV^{1.3}=constant$

Work done for Polytropic process

$W=\dfrac{P_1V_1-P_2V_2}{n-1}$

where n=Polytropic index

$W=\dfrac{R(T_1-T_2)}{n-1}$

$W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]$

$W=103.6\ kJ\kg$

Now Calculating change in Internal energy

$\Delta U=c_v(T_2-T_1)$

$\Delta U=0.718\times (300-405)$

$\Delta U=-75.39\ kJ/kg$

Now applying First law concept

$\Delta U=Q-W$

$Q=W+\Delta U$

$Q=103.6-75.392$

$Q=28.21\ kJ/kg$

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Compare and contrast ""centralized"" and ""decentralized"" routing algorithms. (What are the advantages and disadvantages of eac

pantera1 [17]

Answer:

Comparison between centralized and decentralized routing algorithms:

The major similarity between both centralized and decentralized routing algorithms is that they are both communication serving systems. They both utilize node system(e.g, computer) and are both a communication liking system( e.g, Cable).

Contrasting between centralized and decentralized routing algorithms:

There are few differences between these two type of communication link or path way system but to name a couple of them,

For centralized, there is a single client server distribution node which simply means that one or more client server system are connected to a central processing server.

This also means that if the central processing server or pathway fails, it leads to the failure of the entire system. That is, there is no sending, responding or general processing of any form of requests. Example of a system that uses the centralized routing algorithm is the google search engine.

WHILE:

for the decentralized routing algorithms, there are multiple client server distribution pathway and each server makes its own decision. Here,there is no single entity that receives and responds to the request therefore,failure of any form of central path way processing node does not lead to the failure of the whole system unlike for the centralized system.

Advantages of centralized routing algorithm:

It can be easily protected or secured due to the nature of the system. If the central node is been secured, it generally translate to the different client node being secured.

It is easy to disconnect a connected client node from the central pathway node or central server as the case may be.

Disadvantages of centralized routing algorithm:

The client nodes are totally dependent on the central node or server so if there is a failure in the central server, the client node is then totally shut down.

Advantages of decentralized routing algorithm:

There is a random distribution of data on all the processing node or server which automatically creates a form of balance within the system. This leads to minimal or no down time processing client request.

Disadvantages of decentralized routing algorithm:

Due to the nature or fact that there are multiple processing system for different client node, it is difficult to detect which client node or request processing server is faulty. This can lead to delay in the fixing of fault in the system should it arise.

Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?

The major reason why decentralized algorithm routing is preferred is because of the level of security attached to it. Each processing servers are secured independently and there is privilege of utilizing independent networking system.

5 0

3 years ago

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

3 years ago

A four-cylinder four-stroke engine is modelled using the cold air standard Otto cycle (two engine revolutions per cycle). Given

suter [353]

Answer:

56.47%

Explanation:

Determine the efficiency of the Engine

Given data : T1 (k) = 325, P1 (kpa) = 185,

V1 (cm^3) = 410 , r = 8, T3(k) = 3420

speed ( RPM) = 4800

USING THIS FORMULA

efficiency ( <em>n </em>) = $1 - (\frac{1}{(rp)^{r-1} })$

= 1 - $(\frac{1}{(8)^{r-1} })$ = 1 - (1/8^1.4-1 )

= 0.5647 = 56.47%

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3 years ago

(true/false) Moment thickness, is an index that is proportional to an increment in momentum flow due to the presence of the boun

Sedaia [141]

Answer:

True

Explanation:

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Answer:

The speed is minimum at t = 4 seconds.

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Differentiate above expression and equate to zero.

Thus, speed is minimum at t = 4 seconds.

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2 years ago