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 6, Problem 153RQ

(a)

To determine

The ratio of flow rates of hot and cold water as they enter the T-elbow.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The maximum flow rate (V˙max) is 13.3 L/min.

The reduced flow rate (V˙Red) is 10.5 L/min.

The temperature of the hot stream (T1) is 15°C.

The temperature of the hot stream (T2) is 55°C.

The temperature of the mixed stream (T3) is 42°C.

The duration of the shower (t) is 5 min/person.

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

The number of person (n) is 4 persons.

Calculation:

Here, the two streams (comparatively hot and cold) of fluids are mixed in a rigid mixing chamber and operates at steady state. Hence, the inlet and exit mass flow rates are equal.

  m˙1+m˙2=m˙3        (I)

Write the energy rate balance equation for two inlet and one outlet system.

  {[Q˙1+W˙1+m˙1(h1+V122+gz1)]+[Q˙2+W˙2+m˙2(h2+V222+gz2)][Q˙3+W˙3+m˙3(h3+V322+gz3)]}=ΔE˙system        (II)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 indicates the hot stream inlet, 2 indicates the cold stream inlet and 3 indicates the mixed water stream outlet.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

Neglect the heat transfer, work transfer, kinetic and potential energies.

The Equation (II) reduced as follows.

  m˙1h1+m˙2h2m˙3h3=0m˙1h1+m˙2h2=m˙3h3        (III)

Substitute m˙1+m˙2 for m˙3.

  m˙1h1+m˙2h2=(m˙1+m˙2)h3m˙1h1+m˙2h2=m˙1h3+m˙2h3m˙2h2m˙2h3=m˙1h3m˙1h1m˙2(h2h3)=m˙1(h3h1)

  m˙2m˙1=h3h1h2h3        (IV)

The change in enthalpy is expressed as follow.

  h3h1=cp(T3T1)h2h3=cp(T2T3)

Here, the specific heat is cp, the temperature is T and the enthalpy is h.

Substitute cp(T3T1) for h3h1 and cp(T2T3) for h2h3 in Equation (IV).

  m˙2m˙1=cp(T3T1)cp(T2T3)m˙2m˙1=T3T1T2T3

  m˙2m˙1=42°C15°C55°C42°C=27°C13°C=2.07692.08

Thus, the ratio of flow rates of hot and cold water as they enter the T-elbow is 2.08.

(b)

To determine

The amount of electricity that will be saved per year, in kWh, by replacing the standard shower heads with the low-flow ones.

(b)

Expert Solution
Check Mark

Explanation of Solution

Refer Table A-3 (a), “Properties of common liquids, solids, and foods”.

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

The density of the water (ρ) is taken as 1kg/L.

Here, the volume flow rate in the shower heads is lowered by installing low-flow shower heads equipped with flow controllers. This reduces the volume flow rate of water that is consumed per person.

Calculate the volume flow rated saved.

  V˙saved=13.3L/min10.5L/min=2.8L/min

Calculate the total volume flow rate saved by the family per year.

  V˙f,saved=V˙saved×(no.of personsin family)×(shower time usedper person per day )

  V˙f,saved=2.8L/min×(4person)×(5min/personday)=2.8L/min×(4person)×(5minpersonday×365day1year)=20440L/year

Calculate the total mass of water saved per year.

  m˙saved=ρV˙f,saved

  m˙saved=(1kg/L)(20440L/year)=20440kg/year

Calculate the energy saved due to the installation of lower shower heads.

  Energy saved=m˙savedcp(T3T1)

  Energy saved=(20440kg/year)(4.18kJ/kg°C)(42°C15°C)=(20440kg/year)(4.18kJ/kg°C)(27°C)=2306858.4kJ/year2307000kJ/year×1kWh3600kJ

  641kWh/year

Thus, the amount of electricity that will be saved per year, in kWh, by replacing the standard shower heads with the low-flow ones is 641kWh/year.

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Chapter 6 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

Ch. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - A house is maintained at 1 atm and 24°C, and warm...Ch. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - The kinetic energy of a fluid increases as it is...Ch. 6 - Prob. 23PCh. 6 - Air enters a nozzle steadily at 50 psia, 140°F,...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Air at 13 psia and 65°F enters an adiabatic...Ch. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Steam at 4 MPa and 400°C enters a nozzle steadily...Ch. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Helium is to be compressed from 105 kPa and 295 K...Ch. 6 - Carbon dioxide enters an adiabatic compressor at...Ch. 6 - Air is compressed from 14.7 psia and 60°F to a...Ch. 6 - Prob. 47PCh. 6 - An adiabatic gas turbine expands air at 1300 kPa...Ch. 6 - Steam flows steadily into a turbine with a mass...Ch. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Refrigerant-134a is throttled from the saturated...Ch. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - A thin-walled double-pipe counter-flow heat...Ch. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - The components of an electronic system dissipating...Ch. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - A house has an electric heating system that...Ch. 6 - Prob. 95PCh. 6 - Refrigerant-134a enters the condenser of a...Ch. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 100PCh. 6 - Air enters the duct of an air-conditioning system...Ch. 6 - Prob. 102PCh. 6 - A rigid, insulated tank that is initially...Ch. 6 - Prob. 105PCh. 6 - Prob. 106PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - An air-conditioning system is to be filled from a...Ch. 6 - Prob. 111PCh. 6 - A 0.06-m3 rigid tank initially contains...Ch. 6 - A 0.3-m3 rigid tank is filled with saturated...Ch. 6 - Prob. 114PCh. 6 - A 0.3-m3 rigid tank initially contains...Ch. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - The air in a 6-m × 5-m × 4-m hospital room is to...Ch. 6 - Prob. 124RQCh. 6 - Prob. 125RQCh. 6 - Prob. 126RQCh. 6 - Prob. 127RQCh. 6 - Prob. 128RQCh. 6 - Prob. 129RQCh. 6 - Prob. 130RQCh. 6 - Prob. 131RQCh. 6 - Prob. 132RQCh. 6 - Steam enters a nozzle with a low velocity at 150°C...Ch. 6 - Prob. 134RQCh. 6 - Prob. 135RQCh. 6 - Prob. 136RQCh. 6 - In large steam power plants, the feedwater is...Ch. 6 - Prob. 138RQCh. 6 - Prob. 139RQCh. 6 - Prob. 140RQCh. 6 - Prob. 141RQCh. 6 - Prob. 142RQCh. 6 - Prob. 143RQCh. 6 - Prob. 144RQCh. 6 - Prob. 145RQCh. 6 - Prob. 146RQCh. 6 - Repeat Prob. 6–146 for a copper wire ( = 8950...Ch. 6 - Prob. 148RQCh. 6 - Prob. 149RQCh. 6 - Prob. 150RQCh. 6 - Prob. 151RQCh. 6 - Prob. 152RQCh. 6 - Prob. 153RQCh. 6 - An adiabatic air compressor is to be powered by a...Ch. 6 - Prob. 156RQCh. 6 - Prob. 157RQCh. 6 - Prob. 158RQCh. 6 - Prob. 159RQCh. 6 - Prob. 160RQCh. 6 - Prob. 161RQCh. 6 - Prob. 162RQCh. 6 - Prob. 163RQCh. 6 - Prob. 164RQCh. 6 - Prob. 166RQCh. 6 - Prob. 167RQCh. 6 - Prob. 168RQCh. 6 - Prob. 169RQ
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