Answer:
[tex]Hr=4.2*10^7\ btu/hr[/tex]
Explanation:
From the question we are told that:
Water flow Rate [tex]R=4.5slug/s=144.78ib/sec[/tex]
Initial Temperature [tex]T_1=60 \textdegree F[/tex]
Final Temperature [tex]T_2=140 \textdegree F[/tex]
Let
Specific heat of water [tex]\gamma= 1[/tex]
And
[tex]\triangle T= 140-60[/tex]
[tex]\triangle T= 80\ Deg.F[/tex]
Generally the equation for Heat transfer rate of water [tex]H_r[/tex] is mathematically given by
Heat transfer rate to water= mass flow rate* specific heat* change in temperature
[tex]H_r=R* \gamma*\triangle T[/tex]
[tex]H_r=144.78*80*1[/tex]
[tex]H_r=11582.4\ btu/sec[/tex]
Therefore
[tex]H_r=11582.4\ btu/sec*3600[/tex]
[tex]Hr=4.2*10^7\ btu/hr[/tex]