Chapter 5

Question #10

Hot water at \(70\degree C\) stored in an open container on top of a \(20\:m\)-high structure at atmospheric pressure is being drained to the ground. Given water temperature at drain pipe outlet is 65C, determine heat loss through pipeline per kg of water flow-rate.

Solution Approach

looking at energy conservation

\(\dot Q+\dot m(h_1+1/2V_1^2+gz_1)=\dot m(h_2+1/2V_2^2+gz_2)\)

rearrangig the energy conservation for \(\dot Q\)

\(\dot Q=\dot m(h_2+1/2V_2^2+gz_2)-\dot m(h_1+1/2V_1^2+gz_1)=\dot m(h_2-h_1)+\dot m(1/2V_2^2-1/2V_1^2)++\dot mg(z_2-z_1)\)

looking at mass conservation

\(\rho_1A_1V_1=\rho_2A_2V_2\)

assuming negiligible changes in density (\(\rho\)) and a constant diameter piping system,

\(\rho_1=\rho_2\)

and

\(A_1=A_2\)

therefore,

\(V_1=V_2\)

# import the libraries we'll need
import CoolProp.CoolProp as CP
import numpy as np

fluid = "water"  # define the fluid or material of interest
P = 101325   #atmospheric pressure in Pa
m = 1   #solving for unit mass of fluid flow-rate in kg/s
T_1 = 70 + 273.15   #water temperature at elevated storage in K
T_2 = 65 + 273.15   #water temperature at drain outlet in K
z_1 = 20   #storage elevation in m
z_2 = 0   #ground elevation as referrence
g = 9.81   #gravitational acceleration in m/s2

h_1 = CP.PropsSI("H", "T", T_1, "P", P, fluid)   #fluid enthalpy in J/kg
h_2 = CP.PropsSI("H", "T", T_2, "P", P, fluid)   #fluid enthalpy in J/kg


Q = m * (h_2 - h_1) + m * g * (z_2 - z_1)
print('The heat loss of fluid through piping system is:', f"{np.abs(Q):.1f}", 'W')   #taking absolute of Q to justify the negative value for dissipated heat

The heat loss of fluid through piping system is: 21139.5 W