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kherson [118]
3 years ago
7

Help

Engineering
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)

Explanation:

er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)

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7 0
4 years ago
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be
hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0
3 years ago
A gas at 1650C and 0.96 atm enters a plug flow reactor that converts 91% of the material. What is the amount of gas (in units of
yawa3891 [41]
I’d trying using photomath or tigermath
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4 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
Dennis_Churaev [7]

Answer:

a) the heat exchanger area required for the evaporator is 11178.236 m²

b) the required flow rate is 1993630.38 kg/s

Explanation:

Given the data in the question;

Water temperature near the surface = 300 K

temperature at reasonable depths ( cold ) = 280 K

power plant output W' = 2 MW

efficiency η = 3% = 0.03

we know that; efficiency η = W'_{power-out / Q_{supplied

we substitute

0.03 = 2 / Q_{supplied

Q_{supplied = 2 / 0.03

Q_{supplied = 66.667 MW = 66.667 × 10⁶ Watt

Th_{in = 300 K       Th_{out = 292 K

Tc_{in = 290 K       Tc_{out = 290 K    

Now, Heat transfer in evaporator;

Q = UA( LMTD )

so

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = Th_{in - Tc_{out  = 300 - 290 = 10 K

ΔT₂ = Th_{out - Tc_{in  = 292 - 290 = 2 K

so we substitute into our equation;

LMTD = (10 - 2) / ln( 10 / 2 )

LMTD = 8 / ln( 5 )

LMTD = 8 / 1.6094379

LMTD = 4.97

a) Heat transfer Area will be;

Q_H = UA( LMTD )

we substitute

66.667 × 10⁶ = 1200 × A × 4.97

66.667 × 10⁶  = 5964 × A

A = (66.667 × 10⁶) / 5964

A = 11178.236 m²

Therefore, the heat exchanger area required for the evaporator is 11178.236 m²

b) Flow rate  

we know that;

Q_H = m'C_P( T_{in - T_{out )  

specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)

we substitute

66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )

66.667 × 10⁶ = m' × 33.44

m' = ( 66.667 × 10⁶ ) / 33.44

m' = 1993630.38 kg/s

Therefore, the required flow rate is 1993630.38 kg/s

7 0
3 years ago
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