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s2008m [1.1K]
3 years ago
7

Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat

e of 1.8 kg/s. Water is heated by steam condensing at 120°C (hfg = 2203 kJ/kg) in the shell. If the overall heat transfer coefficient of the heat exchanger is 700 W/m2 ·°C, determine the length of the tube required in order to heat the water to 80°C using (a) the LMTD method, and (b) the ????????–NTU method. Answers: 129.5 m; 129.6 m

Engineering
1 answer:
aksik [14]3 years ago
3 0

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

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An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

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3 years ago
Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu
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Answer:

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Explanation:

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6 0
3 years ago
1. _________ gloves are made from a synthetic rubber and provide protection from a variety of chemicals such as peroxide and hig
MrRissso [65]

Answer:

b

Explanation:

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3 years ago
Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the
Aleks [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

<u>a) Calculate for the temp of sheet metal when it leaves the oil bath </u>

<em>first step : find the Biot number </em>

Bi = hLc / K  ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m ,  K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1  ( hence lumped parameter analysis can be applied )

<em>next : find the time constant </em>

t ( time constant ) = h / P*Cp *Lc  --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

<em>Determine the elapsed time </em>

T = L / V = 9/20 = 0.45 min

∴<u>   temp of sheet metal when it leaves the oil bath </u>

= (T(t) - 45 ) / (820 - 45)  = e^-(0.10092 * 27 )

T∞ =  45°C

Ti = 820°C

hence : T(t) = 95.80°C

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Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti =  820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW

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3 years ago
(1+4i)−(−16+9i)<br> i will give brailiest<br> no reporting
Julli [10]

Answer:

The picture is 56 and the written is -12

Explanation:

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2 years ago
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