EP WEBASSIGN FOR MOAVENI'S ENGINEERING
EP WEBASSIGN FOR MOAVENI'S ENGINEERING
6th Edition
ISBN: 9780357126592
Author: MOAVENI
Publisher: CENGAGE CO
Question
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Chapter 11, Problem 41P
To determine

Calculate the amount of natural gas required to heat the water in a heater.

Expert Solution & Answer
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Answer to Problem 41P

The amount of natural gas required to heat the water in a heater is 72ft3day.

Explanation of Solution

Given data:

Volume of water used per day is 80gallons,

Initial temperature of water is, Tinitial=55°F,

Final temperature of water is, Tfinal=140°F,

Efficiency of heater is 80%,

From Table 11.13 in the textbook, the specific heat of water is, c=4180JkgK,

Standard density of water is 1000kgm3.

Formula used:

The relationship between degree Fahrenheit and degree Celsius is,

T(°C)=59[T(°F)32] (1)

Here,

T(°C) is the temperature in degree Celsius,

T(°F) is the temperature in degree Fahrenheit.

The formula to find thermal energy required to heat the water is,

Ethermal=mc(TfinalTinitial) (2)

Here,

m is the mass of the water,

c is the specific heat of water,

Tinitial is the initial temperature of water,

Tfinal is the final temperature of water.

Calculation:

Convert the unit of volume of the water,

Volume=80gallonsday[1gallon=3.7854×103m3]=(80gallonsday)(3.7854×103m31gallon)=(80×3.7854×103)m3dayVolume=0.3028m3day

Substitute 55°F for T(°F) in equation (1) to get the initial temperature in degree Celsius,

Tinitial(°C)=59[5532]=12.77°C

Substitute 140°F for T(°F) in equation (1) to get the final temperature in degree Celsius,

Tfinal(°C)=59[14032]=60°C

The expression to find the mass of the water is,

Mass=Volume×density=(0.3028m3day)(1000kgm3)Mass=302.8kgday

Substitute 302.8kgday for m, 4180JkgK for c, 12.77°C for Tinitial, and 60°C for Tfinal in equation (2) to find Ethermal,

Ethermal=(302.8kgday)(4180JkgK)(60°C12.77°C)      [ °C=K]=(302.8kgday)(4180JkgK)(6012.77)K

Ethermal=59.7792×106Jday

The unit conversion on above result is,

Ethermal=59.7792×106Jday[1day=24h]=59.7792×106(Jday×1day24h)[1h=3600s]=(2.491×106Jh)(1h3600s)[1Js=1W]Ethermal=693W

Do the unit conversion on above result,

Ethermal=693W[1W=3.412Btuh]=(693W)(3.412Btuh1W)Ethermal=2364.6Btuh

At standard 77°F temperature, the heating value of methane is 978Btuft3. Since the major composition of natural gas is methane, consider the heating value of natural gas is 978Btuft3, as the range of temperature given as 55°F to 140°F considered for 77°F.

Then the amount of natural gas required to heat the water in heater can be determined from the below expression:

Volumeofgas=EthermalEfficiency×Heatingvalueofgas=2364.6Btuh0.8(978Btuft3)[1day=24h]=(3ft3h)(24h1day)Volumeofgas=72ft3day

Therefore, the amount of natural gas required to heat the water in a heater is 72ft3day.

Conclusion:

Hence, the amount of natural gas required to heat the water in a heater is 72ft3day.

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Chapter 11 Solutions

EP WEBASSIGN FOR MOAVENI'S ENGINEERING

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