Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 7, Problem 125RQ
To determine

The amount of oil and money, a family of four will save per year by replacing the standard shower heads by the low-flow ones.

Expert Solution & Answer
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Explanation of Solution

Given:

The initial volume of the low-flow shower head (ν1) is 13.3L/min.

The final volume of the low-flow shower head (ν2) is 10.5L/min.

The time (t) is 6 min.

The number person in family (n) is 4.

The density of water (ρwater) is 1kg/L

The specific heat of water (cp) is 4.18kJ/kg°C.

The initial temperature of the low-flow shower head (T1) is 15°C.

The final temperature of the low-flow shower head (T2) is 42°C.

The efficiency of the fuel in the low-flow shower head (ηfuel) is 0.65.

The heating value of fuel (HV) is 146,300kJ/gal.

The unit cost of fuel is $2.80/gal

Calculation:

Calculate the rate of water saved volume of the low-flow shower head a family of four will save per year.

  ν˙saved=[(ν1ν2)×(t/(personday))×(n)×(number of days per year)]

  ν˙saved=[(13.310.5)L/min×(6min/(personday))×(4persons)×(365days/year)]=[2.8L/min×(6min/(personday))×(4persons)×(365days/year)]=24,528L/year

Calculate the mass flow rate of water of the low-flow shower head.

  m˙saved=ρwater×ν˙saved

  m˙saved=(1kg/L)×(24,528L/year)=24,528kg/year

Calculate the amount of energy saved in the low-flow shower head.

  Esaved=(m˙watercpΔT)=m˙watercp(T1T2)

  Esaved=(24,528kg/year)×(4.18kJ/kg°C)×(42°C15°C)=(24,528kg/year)×(4.18kJ/kg°C)×(27°C)=2768230kJ/year

Calculate the amount fuel saved in the low-flow shower head.

  Fuelsaved=Esaved(ηfuel)(HV)

  Fuelsaved=(2768230kJ/year)(0.65)×(146,300kJ/gal)=(2768230kJ/year)(0.65)×(146,300kJ/gal)=(2768230kJ/year)(95095kJ/gal)=29.1gal/year

Thus, the amount of oil, a family of four will save per year by replacing the standard shower heads by the low-flow ones is 29.1gal/year_.

Calculate the amount of money saved in the low-flow shower head.

  Moneysaved=(fuel saved)×(unit cost of fuel)

  Moneysaved=(29.1gal/year)×($2.80/gal)=$81.5/year

Thus, the amount of money, a family of four will save per year by replacing the standard shower heads by the low-flow ones is $81.5/year_.

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Chapter 7 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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