Answer:
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]
Explanation:
How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]