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Zielflug [23.3K]

4 years ago

12

A strip of AISI 304 stainless steel, 2mm thick by 3cm wide, at 550°C, continuously enters a cooling chamber that removes heat at

a rate of 120 kW. How fast must it move to achieve a temperature of 140 °C at the exit?

1 answer:

Zanzabum4 years ago

6 0

Answer:

V = 1.23 m/s

Explanation:

Given data:

AISI 304 steel

thickness of steel = 2 mm

width = 3 cm

Temperature = 550 degree celcius

wer know that heat is given as Q

$Q = m\times Cp \Delta T$

$\Delta T = 550 - 140 = 410 \degree\ celcius$

for AISI 304 steel Cp is 502 J/kg . K

We know that

$\dot mass =\rho A V$

$\rho = 7.9\times 10^3 kg/m^3$

$A = 2\times 10^{-3} \times 3\times 10^{-2} m^2$

therefore VELOCITY V will be

$120\times 10^{3} = 7.9\times 10^{3} \times 6\times 10^{-5} \times V\times 502\times 410$

solving for V we get

V = 1.23 m/s

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A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov

babunello [35]

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H $_{pump$ = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H $_p$ = (p/ρq)₂ + (v²/2g)₂ + z₂ + h $_L$

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH $_p$ Q / η

where; H $_p$ = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ( $P_{atom$ / ρg) - h $_s$ - ( P $_v$ / ρg )

we substitute

25 = ( 2116 / 62.42 ) - h $_s$ - ( 30 / 62.42 )

h $_s$ = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

6 0

3 years ago

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at

NNADVOKAT [17]

Answer:

Q = E + W = 0.25 + 0.4 = 0.29KJ

W = 0.4KJ

b) 2000KJ

Explanation:

The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.

mathematically from first law,

dE = dQ - dW

where E is internal energy

Where energy may be transfereed as eitherf work (W) or heat (Q)

Work on the other hand can be at constant volume and pressure.

at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit

Work = p(Vf -Vi)

attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ

from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ

To get the heat transfer, from dE = dQ - dW

Q = E + W = 0.25 + 0.4 = 0.29KJ

b) for the second question ; from Pdv = p(Vf -Vi), but the pressure here is in atmosphere

W = 100000 ( 0.12 - 0.1) = 2000KJ

6 0

3 years ago

Read 2 more answers

A hot-water stream at 60oC enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

Nezavi [6.7K]

Answer: 0.306 kg/s

Explanation:

If we assume that the specific heat for all stream is constant, we can then say, the mass flow rate of cold water can be calculated by using the energy balance equation.

The energy balance equation basically states that, the heat gained is equal to the heat lost

m'1h1 + m'2h2 = m'3h3

m'1T1 + m'2T2 = (m'1 + m'2) T3

0.5 (60) + m'2(10) = (0.5 + m'2)41

30 + 10m'2 = 20.5 + 41m'2

30 - 20.5 = 41m'2 - 10m'2

9.5 = 31m'2

m'2 = 9.5 / 31

m'2 = 0.306 kg/s

7 0

3 years ago

Read 2 more answers

A magician claims that he/she has invented a novel, super-fantastic heat engine. This engine operates between two reservoirs of

Marysya12 [62]

Answer:

A) Not possible, B) Posible, C) Possible, D) Not possible.

Explanation:

The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ( $\eta$ ), no unit, is expressed below:

$\eta = 1-\frac{T_{L}}{T_{H}}$ (1)

Where:

$T_{L}$ - Cold reservoir temperature, in Kelvin.

$T_{H}$ - Hot reservoir temperature, in Kelvin.

If we know that $T_{L} = 250\,K$ and $T_{H} = 750\,K$ , then the maximum theoretical efficiency for the thermal engine is:

$\eta = 1-\frac{T_{L}}{T_{H}}$

$\eta = 0.667$

For real thermal engines, the following inequation is observed:

$0 \le \eta_{r} \le \eta$ (2)

Where $\eta_{r}$ is the efficiency of the real heat engine, no unit.

There are two possible criteria to determine if a given heat engine is real:

Efficiency

$\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}$ (3)

Where:

$Q_{L}$ - Heat rejected to the cold reservoir, in kilojoules.

$Q_{H}$ - Heat received from the hot reservoir, in kilojoules.

Power output

$W = Q_{H}-Q_{L}$ (4)

Where $W$ is the power output, in kilojoules.

Now we proceed to verify each case:

A) $Q_{H} = 900\,kJ$ , $Q_{L} = 600\,kJ$ , $W_{m} = 400\,kJ$

$\eta_{r} = 0.333$

$0 \le \eta_{r} \le \eta$

$W = 300\,kJ$

$W \ne W_{m}$

This engine is not possible.

B) $Q_{H} = 900\,kJ$ , $Q_{L} = 500\,kJ$ , $W_{m} = 400\,kJ$

$\eta_{r} = 0.444$

$0 \le \eta_{r} \le \eta$

$W = 400\,kJ$

$W = W_{m}$

The engine is possible.

C) $Q_{H} = 900\,kJ$ , $Q_{L} = 300\,kJ$ , $W_{m} = 600\,kJ$

$\eta_{r} = 0.667$

$0 \le \eta_{r} \le \eta$

$W = 600\,kJ$

$W = W_{m}$

The engine is possible.

D) $Q_{H} = 900\,kJ$ , $Q_{L} = 100\,kJ$ , $W_{m} = 800\,kJ$

$\eta_{r} = 0.889$

$\eta_{r} > \eta$

$W = 800\,kJ$

$W = W_{m}$

The engine is possible.

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3 years ago

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Igoryamba

Answer: A.)

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