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An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is used as a heat source to evaporate the working fluid, while the colder water is used as a heat sink for condensation of the fluid. Consider a power plant that is to generate 2 MW of electricity at an efficiency (electric power output per heat input) of 3%. The evaporator is a heat exchanger consisting of a single shell with many tubes executing two passes. If the working fluid is evaporated at its phase change temperature of 290 K, with ocean water entering at 300 K and leaving at 292 K.

Required:
a. What is the heat exchanger area required for the evaporator?
b. What flovw rate must be maintained for the water passing through the evaporator?

Sagot :

Nuhulawal20idn

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'[tex]_{power-out[/tex] / Q[tex]_{supplied[/tex]

we substitute

0.03 = 2 / Q[tex]_{supplied[/tex]

Q[tex]_{supplied[/tex] = 2 / 0.03

Q[tex]_{supplied[/tex] = 66.667 MW = 66.667 × 10⁶ Watt

T[tex]h_{in[/tex] = 300 K       T[tex]h_{out[/tex] = 292 K

T[tex]c_{in[/tex] = 290 K       T[tex]c_{out[/tex] = 290 K    

Now, Heat transfer in evaporator;

Q = UA( LMTD )

so

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = T[tex]h_{in[/tex] - T[tex]c_{out[/tex]  = 300 - 290 = 10 K

ΔT₂ = T[tex]h_{out[/tex] - T[tex]c_{in[/tex]  = 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[tex]_H[/tex] = 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[tex]_H[/tex] = m'C[tex]_P[/tex]( [tex]T_{in[/tex] - [tex]T_{out[/tex] )  

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

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