This problem can be considered as a equivalent electric circuit with pump as a battery and radiators as equivalent resistance.
(a) Pressure difference across Radiator 1 will same as the pressure difference across the pump which is 12 psi.
Equivalent resistance of the path containing radiators R1 and R2 will be Req = R1 + R2 (as they are in series) , from the Poiseuille's law : Flow Rate (F) = P/R
F1 = P/Req (Flow through path containing radiator R2 and R3)
Since, F1 is the flow rate across both the radiator the pressure difference across them can be calculate from the same equation Flow Rate (F) = P/R ,
P (2) = F1*R2 = 12*5/8 = 7.5 psi
P (3) = F1*R3 = 12*3/8 = 4.5 psi
(b) Flow rate will be same across Radiator 2 and 3.
Flow rate F1 = P/Req*D = 12*6894.76 * 1000/ 8*10^7 = 1.03 Kg/sec
where D is the density of water 1000 kg/m^3.
1psi = 6894.76 pascal.
Flow rate across Radiator 1 will be,
Flow rate F2 = P/R1*D = 12*6894.76 * 1000/ 4*10^7 = 2.06 Kg/sec
Flow rate across pump will be F1 +F2 = 3.09 Kg/sec.
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