A domestic water heater burns natural gas at the volume rateof 3.0 m^3/hr. If th

Question

A domestic water heater burns natural gas at the volume rateof 3.0 m^3/hr. If the water absorbs half of the energy released,how many kg of water can be heated from 100 degree C to 950 degreeC? The heat of combustion of natural gas is 5.6* 10 ^7 j/m^3. andthe specific heat of water is 4.2 * 10^3 j/kg C. b. The container walls of the above heater have a surface areaof 2 m^3 and are lined with an insulating material of averagethickness 5 cm and having thermal conductivity of 0.040 W/m C. Ifthe inside temp is 95 C and the outside temp is 25 C , how muchheat is lost per hour, by conduction through the walls?? c. Suppose the heat loss described in above aprt is to bereduced by a factor of two by adding a blanket insulation withthermal conductivity 0.020 W/mC outside the tank. What thickness ofinsulation must be added if the inside and outside temp areunchanged? A domestic water heater burns natural gas at the volume rateof 3.0 m^3/hr. If the water absorbs half of the energy released,how many kg of water can be heated from 100 degree C to 950 degreeC? The heat of combustion of natural gas is 5.6* 10 ^7 j/m^3. andthe specific heat of water is 4.2 * 10^3 j/kg C. b. The container walls of the above heater have a surface areaof 2 m^3 and are lined with an insulating material of averagethickness 5 cm and having thermal conductivity of 0.040 W/m C. Ifthe inside temp is 95 C and the outside temp is 25 C , how muchheat is lost per hour, by conduction through the walls?? c. Suppose the heat loss described in above aprt is to bereduced by a factor of two by adding a blanket insulation withthermal conductivity 0.020 W/mC outside the tank. What thickness ofinsulation must be added if the inside and outside temp areunchanged?

Explanation / Answer

a. It is not told that at what pressure the wateris, assuming whatever be the pressure, it is not changing itsphase. energy available for water = (1/2)*3m3/hr*5.6*107j/m3= 8.4*107 j/hr this should be equal to m*C*T= m*4.2*103j/kg-C*(950-100)C = m*3570*103 j/kg thus, m=(8.4*107 j/hr) / (3570*103 j/kg) =23.52 kg/hr b. heat loss throught conduction = KA(dT/dX) where, K= thermal conductivity, A=cross sectional area, dT=temperature difference, dX= thickness heat loss = 0.040*2*(95-25)/0.05          {given,area= 2m^3, which cant be area, assuming area=2m^2}                                = 112 W or 112 J/s= 112*3600 j/hr= 403.2 kj/hr c. heat loss in this case is 56 W now applying heat current formula Q=(T1-T2)/(dX1/K1A1 +dX2/K2A2) here, A1=A2= 2m2 K1= 0.040 W/m-C, K2= 0.020 W/m-c dX1= 0.05 , dX2=?=x(say) putting the values, 56= (95-25)/(0.05/(0.040*2) + x/(0.020*2)) = 70/(0.625 +x/0.040) or, 0.625+x/0.040 = 70/56=1.25 x = 0.040*(1.25-0.625)= 0.025 m =2.5 cm

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