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Umnica [9.8K]
3 years ago
9

5.46 By supplying energy at an average rate of 24,000 kJ/h, a heat pump maintains the temperature of a dwelling at 20°C. If elec

tricity costs 8.5 cents per kW · h, determine the minimum theoretical operating cost for each day of operation if the heat pump receives energy by heat transfer from (a) the outdoor air at −7°C. (b) the ground at 5°C.
Engineering
1 answer:
Illusion [34]3 years ago
4 0

Answer:

A) $18.36

B) $10.24

Explanation:

Heat supplied Q = 24000 kJ/h

J/s = watt

I hour = 3600 sec, therefore 24000 kJ/hr becomes 24000/3600 = 6.67 kW

Coefficient of performance of heating for heat pump is

COP = T2/(T2 - T1)

T2 (hot temperature) = 20 °C

T1 (cold temperature) = -7 °C

COP = 20/(20 - (-7)) = 20/27 = 0.741

Also COP = Q/W

where W is the power input to the heat pump

0.741 = 6.67/W

W = 9 kW

Operating for one day is 24 hr

Therefore heat pump uses 9 x 24 = 216 kWh

Cost per day = ¢8.5 x 216 = ¢1836 = $18.36

For ground at 5°C

COP = 20/(20-5) = 20/15 = 1.33

1.33 = 6.67/W

W = 5.02 kW

For 24 hr it becomes 24 x 5.02 = 120.48 kWh

Cost = 8.5 x 120.48 = ¢1024.08 = $10.24

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REY [17]
The answer should be List range
7 0
3 years ago
A 8-core machine has 4 times the performance of a single-core machine of the same frequency. Performance is proportional to freq
tankabanditka [31]

Answer: The 8-core machine saves  87.5% of the dynamic power.

Explanation:

Let Fold = f , Vold = V , Cold = Capacitance

so

Old Dynamic power = Cold × (Vold × Vold) × f

therefore for the 8-core machine

 Fnew / Fold = 1/4

Fnew = Fold/4

we were told that Voltage decreases proportional to frequency,

so

Vnew / Vold = 1/4

Vnew = V / 4

So New Capacitance will be;

Cnew = Cold

Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) ×  Fnew

= 8 × Cold × (Vold × Vold/16) × ( f/4 )

=  8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64

= (Old Dynamic Power) / 8

therefore

Old Dynamic Power / New Dynamic Power = 8

Thus, Percentage of power saved will be;

Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power

=   100 × (8-1) / 8

= 87.5 %

Therefore The 8-core machine saves  87.5% of the dynamic power.

6 0
3 years ago
An urn contains r red, w white, and b black balls. Which has higher entropy, drawing k ~2 balls from the urn with replacement or
svetlana [45]

Answer:

The case with replacement has higher entropy

Explanation:

The complete question is given:

'Drawing with and without replacement. An urn contains r red,  w white  and b black balls. Which has higher entropy, drawing k ≥ 2 balls from the urn with  replacement or without replacement?'

Solution:

- n drawing is the same irrespective of whether there is replacement or not.

-  X to denotes drawing from an urn with r red balls,  w white balls and b black balls. So, n = b + r +  w.

We have:

                                      p_X(cr) = r / n

                                      p_X(cw) = w / n

                                      p_X(cb) = b / n

- Now, if  Xi is the ith drawing with replacement then Xi are independent and p_Xi (x) = pX(x) for x e ( cr , cb , cw ).

- Now, let  Yi be the ith drawing with replacement where Yi are not independent p_Yi (x) = pX(x) for x ∈ X.

- To see this, note  Y1 =  X and assume it is true for  Yi and consider  Yi+1:

    p_Y(i+1) (cr) = p(Y(i+1),Yi).(cr, cr) + p(Y(i+1),Yi).(cr, cw) + p(Y(i+1),Yi).(cr, cb)

= pY(i+1)|Yi  (cr|cr)*pYi  (cr) +pY(i+1)|Yi  (cr|cw)*pYi (cw) + pY(i+1)|Yi  (cr|cb)*pYi (cb)

= r*( r - 1 )/n*(n-1) + w*r/n*(n-1) + b*r/n*(n-1) = r / n =  p_X(cr)

- This means, using the chain rule and the conditioning theore m:

H(Y1, Y2, . . . , Yn) =  H(Y1) +  H(Y2|Y1) +  H(Y3|Y2, Y1) + ... H(Yn|Yn−1, . . . , Y1)

=< SUM H(Yi) = n*H(X) =  H(X1, X2, . . . , Xn)

- with equality if and only if the  Yi were independent:

                          H(Y1, Y2, . . . , Yn) < H(X1, X2, . . . , Xn)

Answer: The case with replacement has higher entropy

   

4 0
3 years ago
The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determi
kirza4 [7]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

Y=\frac{1}{Z}

Hence for each we have,  

Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\  Y_{1} =\frac{1}{6.4\leq 51.3}\\  Y_{1} =(0.16 \leq -51.3)S

for the second impedance we have

Y_{2}=\frac{1}{10}\\Y_{2}=0.1S

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\

V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v

The real power in the impedance is calculated as

P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W

for the second impedance

P_{2}=v^{2}*G_{2}\\   P_{2}=62*62*0.1\\384.4w

b. We determine the equivalent admittance

Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\

We convert the equivalent admittance back into the polar form

Y_{total}=0.28\leq -19.65\\

the source current flows is

I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A

6 0
3 years ago
Please help me please
VashaNatasha [74]

Answer: C

Explanation:

5 0
3 years ago
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