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maria [59]
3 years ago
7

A heat pump cycle operating at steady state receives energy by heat transfer from well water at 10oC and discharges energy by he

at transfer to a building at the rate of 1.4x105 kJ/h. Over a period of 14 days, an electric meter records that 1620 kW·h of electricity is provided to the heat pump. These are the only energy transfers involved. Determine the total amount of energy transfer by work into the heat pump over the 14-day period, in GJ. Determine the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period, in GJ. Determine the heat pump’s coefficient of performance.
Engineering
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer: the total amount of energy transferred by work into the heat pump over the 14-day period is 5.8GJ

Explanation: since the meter reading for the 14 days period is 1620kW.h

Electrical power Ep is

Ep = 1620*10^3W.h

In 14 days there are 14*24 hrs = 336hrs

Ep = (1620*10^3)/336 = 4821.42W

This means 4821.42J of work is done per sec. Therefore for 14 days we have,

3600*24*14*4821.42J of enegy

= 5831989632J = 5.8GJ

Answer 2 = the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period is 41.2GJ

Explanation: rate of energy transfer as heat to building is 1.4*10^5 kJ/hr

That is 1.4*10^8 J/hr

In 14 days it becomes

24*14*1.4*10^8 = 47GJ

Energy received from well by heat pump =

Energy transfered - energy received

= 47 - 5.8 =41.2GJ

Answer 3. the heat pump’s coefficient of performance is 0.12

Explanation: COP = energy obtained/ work done

COP = 5.8/47 = 0.12

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3 years ago
declare integer product declare integer number product = 0 do while product < 100 display ""Type your number"" input number p
Brilliant_brown [7]

Full Question

1. Correct the following code and

2. Convert the do while loop the following code to a while loop

declare integer product

declare integer number

product = 0

do while product < 100

display ""Type your number""

input number

product = number * 10

loop

display product

End While

Answer:

1. Code Correction

The errors in the code segment are:

a. The use of do while on line 4

You either use do or while product < 100

b. The use of double "" as open and end quotes for the string literal on line 5

c. The use of "loop" statement on line 7

The correction of the code segment is as follows:

declare integer product

declare integer number

product = 0

while product < 100

display "Type your number"

input number

product = number * 10

display product

End While

2. The same code segment using a do-while statement

declare integer product

declare integer number

product = 0

Do

display "Type your number"

input number

product = number * 10

display product

while product < 100

4 0
4 years ago
An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h
EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell (\eta), dimensionless, as:

\eta = \frac{\dot W_{out}}{\dot W_{in}} (Eq. 1)

Where:

\dot W_{in} - Maximum power possible from hydrogen flow, measured in kilowatts.

\dot W_{out} - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

\dot W_{in} = \dot V\cdot \rho \cdot L_{c} (Eq. 2)

Where:

\dot V - Volume flow rate, measured in cubic meters per second.

\rho - Density of hydrogen, measured in kilograms per cubic meter.

L_{c} - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that \dot V = \frac{28}{3600}\,\frac{m^{3}}{s}, \rho = 0.0899\,\frac{kg}{m^{3}}, L_{c} = 141790\,\frac{kJ}{kg} and \dot W_{out} = 80\,kW, then the efficiency of this fuel cell is:

(Eq. 1)

\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)

\dot W_{in} = 99.143\,kW

(Eq. 2)

\eta = \frac{80\,kW}{99.143\,kW}

\eta = 0.807

The efficiency of this fuel cell is 80.69 percent.

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3 years ago
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Answer:

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Answer:

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Explanation:

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