Question

6. (3 points) A water-based heating system is shown in Figure 5. A pump exerts a constant pressure difference between its inlet and outlet, ΔΡ-12 psi, to circulate hot water across 2 distinct paths. One path contains a single radiator that can be modelled as a pipe with an effective resistance to water flow of Ri-4x 107 kg-m-4 . s-1. The other path has two radiators with effective resistances to water flow given by R2-5x 107kg. m-4-S-1 and R3 3 x 107 kg m4.s-1. All of the pipes are at the same height so that Figure 5 is a top view. Assume that the pipes and the pump have effectively no resistance to the water flow (as their diameters are much larger than that of the radiators which are the only components that have a resistance to the flow of water) (a) What is the pressure difference across each radiator? (b) How many kilograms of water per second go through the pump and through each radiator? R3 R2 Pump, ΔΡ R1 Figure 5: A constant pressure pump feeding a small network of radiators in a house, as seen from above.

Answer 1

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) = $\Delta$P/R

F1 = $\Delta$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) = $\Delta$P/R ,

  $\Delta$P (2) = F1*R2 = 12*5/8 = 7.5 psi

$\Delta$P (3) = F1*R3 = 12*3/8 = 4.5 psi   

(b) Flow rate will be same across Radiator 2 and 3.

Flow rate F1 =   $\Delta$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 =   $\Delta$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.