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3 years ago

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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2 years ago

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

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3 years ago

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Answer:

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3 years ago

Mercury flows inside a copper tube 9m long with a 5.1сm inside diameter at an average velocity of 7.0 m/s. The inside surface te

svp [43]

Answer:

rate of heat transfer = 9085708.80 W

Explanation:

Given:

Inside diameter, D = 5.1 cm

= 5.1 x $10^{-2}$ m

Average velocity, V = 7 m/s

Mean temperature, T = (66+38) /2

= 52°C

Therefore kinematic viscosity at 52°C is ν = 0.104 X $10^{-6}$ $m^{2}$ / s

Prandtl no., Pr = 0.021

We know Renold No. is

Re = $\frac{V\times D}{\nu }$

Re = $\frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}$

= 3.432 X $10^{6}$

Therefore the flow is turbulent.

Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.

Nu = 0.023 $Re^{0.8}$ . $Pr^{0.3}$

= 0.023 x $(3.432 \times10^{6})^{0.8}$ x $(0.021)^{0.3}$

= 1221.52

Since Nu = $\frac{h.D}{k}$

h = $\frac{k\times Nu}{D}$

= $\frac{9.4\times 1221.52}{5.1\times 10^{-2}}$

= 225143.3

Therefore rate of heat transfer, q = h.A(T- $T_{\infty }$

q= 225143.3 x 2πrh ( 66-38)

= 225143.3 X 2π X $\frac{5.1\times10^{-2}}{2}\times 9\times 28$

= 9085708.80 W

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4 years ago