6.2 Hot water in an open container

6.2 Hot water in an open container#

Problem Statement#

Hot water at $70^{\circ} C$ 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

$\dot{Q} + \dot{m} (h_{1} + 1 / 2 V_{1}^{2} + g z_{1}) = \dot{m} (h_{2} + 1 / 2 V_{2}^{2} + g z_{2})$

rearranging the energy conservation for $\dot{Q}$

$\dot{Q} = \dot{m} (h_{2} + 1 / 2 V_{2}^{2} + g z_{2}) - \dot{m} (h_{1} + 1 / 2 V_{1}^{2} + g z_{1}) = \dot{m} (h_{2} - h_{1}) + \dot{m} (1 / 2 V_{2}^{2} - 1 / 2 V_{1}^{2}) + + \dot{m} g (z_{2} - z_{1})$

looking at mass conservation

$ρ_{1} A_{1} V_{1} = ρ_{2} A_{2} V_{2}$

assuming negligible changes in density ( $ρ$ ) and a constant diameter piping system,

$ρ_{1} = ρ_{2}$

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

6.2 Hot water in an open container

Contents

6.2 Hot water in an open container#

Problem Statement#

Solution Approach#