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3 years ago

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An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate between the temperature li

mits of 86°F at the ocean surface and 41°F at a depth of 2100 ft. About 13,300 gpm of cold seawater was to be pumped from deep ocean through a 40-in-diameter pipe to serve as the cooling medium or heat sink. If the cooling water experiences a temperature rise of 6°F and the thermal efficiency is 2.5 percent, determine the amount of power generated. Take the density of seawater to be 64 lbm/ft3. Also, take the specific heat of water to be c = 1.0 Btu/lbm·°F.

2 answers:

Novosadov [1.4K]3 years ago

7 0

Answer:

power generated = 307.84 Kw

Explanation:

The Given data

Tk = 86 f = 303.15k

To = 41 f = 278.15k

thermal efficiency ( n ) = 0.025

Q = 13300 gpm = 0.84 m^3/sec

C = 1.0 Btu/Ibmf

density of seawater ( d ) = 64 Ibm/ft3 = 1025.18 kg/m^3

rise in temp = 6 f

we have to make assumptions

steady state
fluid is in compressible and constant

first we calculate

mass flow rate of water ( Mw )

Mw = density of seawater * Q

= 64 * 13300 * 1/7.4804

= 113790 Ibm/min = 1896.51 Ibm/sec

therefore the rejection of the cooling water will be

= mass flow rate * C * rise in temperature

= 1896.51 * 1.0 * 6

= 11379.06 BTU/s (Qout)

Determine the amount of power generated using thermal efficiency formula

n = $\frac{W}{Qout + W}$

W = power generated

n = thermal efficiency

Qout = rejection of cooling water

0.025 = $\frac{W}{11379.06 + W}$

W = 0.025 W + 284.48

W ( 1 - 0.025 ) = 284.48

therefore W = 284.48 / 0.975

W = 291.77 BTU/s

note : 1 kw = 0.9478 BTU/s

x = 291.77 BTU/s

W in kw ( x ) = 291.77 / 0.9478 = 307.84 Kw

lukranit [14]3 years ago

4 0

Answer:

Detailed solution is given below

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Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

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3 years ago

slader A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy

nikklg [1K]

Answer:

a. 33120MJ

b. 5.6

c. 48

Explanation:

∆U= 120,000 KJ/h

Since

1 day = 24 hrs

14 days =24 x 14 hrs

14 days = 336 hrs

∆U = 120000 x 336 KJ

=40320000KJ

K = 1000

M = 1000000

∆U = 403200 MJ

Work done

W= 2000 KW.h ( 1 h = 3600 s)

W= 7200 MJ

According to the first law of thermodynamics

∆U = Q+W

Q= 40320 - 7200 MJ

1)Qa=33120 MJ

Coefficient of performance

COP = ∆U/W

COP= 40320 / 7200

2)COP = 5.6

3)COP of ideal heat pump

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48

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3 years ago

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Answer:

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3 years ago

How to clean a snowblower carburetor without removing it.

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Answer:

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3 years ago

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

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Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

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Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0

3 years ago