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lions [1.4K]
3 years ago
7

Using the degree day method calculate the annual kwh use in springfiled il with a heat loss of 12kwh an inddor twmperature of 7

Engineering
1 answer:
vladimir2022 [97]3 years ago
8 0

Answer:

The correct solution is "21024 KWh/degree day".

Explanation:

The given query is incomplete. Below is the attachment of complete query is provided.

The given values are:

Indoor design temperature:

= 70°F

Now,

According to the question,

The heat loss annually will be:

= 12\times 24\times 365

= 105120 \ KW

Degree days will be:

= 75-65

= 5

Hence,

Annual KWh use will be:

= \frac{Heat \ loss \ annually}{degree \ days}

On substituting the values, we get

= \frac{105120}{5}

= 21024 \ KWh/degree \ day

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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
On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch
rewona [7]

Answer:

ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

Explanation:

Distance of earth from sun = R_{2} = 147.4 \times 106 Km

Spacecraft perihelion = R_{2} = 120\times106Km

gravitational parameters are now given as

\mu_{sun} = 132.7\times 10^{9}

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   V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }

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Heliocentric velocity of earth = v_{earth} = 30.06\frac{km}{sec}

V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}

assume

r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}

geometric space craft velocity in its circular parking orbit

v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } }  = 7.784 \frac{km}{s}

              ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

7 0
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