Calculate the Heat Loss in a Hot Water Tank from a Shower

[Howlin]

Homework Statement

Compare the heating/cost/CO2 and Primary Energy requirement for an electric shower consuming 50 litres of water with that of a shower from a Hot Water Tank from cold start aka hot water tank is cold for 1 shower per day for a year

Relevant Equations

Q=mc delta T

Hi,

If it is assumed the temperature of a shower is to be 43°C and the electric shower consumes 50 litres of water at that temperature, then the Energy required for an electric shower would be:
Q_{Electric Shower} = mcΔT
Q_{Electric Shower} = 50 * 4.181 * (43-10)
Q_{Electric Shower} = 6.89945 kJ or 1.916 kWh

Q_{Electric Shower per year} = 1.916*365
Q_{Electric Shower per year} = 699.34 kWh/yearWater from a hot water tank is mixed with cold water to produce a temperature of 43°C. It is assumed the Hot Water Tank Heats the water up to 60°C.

To get a mixing temperature of 43°C, with a cold water temperature of 10°C and the Tank Water Temperature of 60°C, the hot water is to cold water flow rate is 17 litres of cold to 33 litres of hot water.

The Final mixing temperature of hot and cold water is:
T_Final = (m₁ * T₁ + m₂ * T₂) / (m₁+m₂)

43 = (m₁ * 10 + m₂ *60) / (50)
Inputting various values for m₁ and m₂ which add up to 50 =>

43 = (17 * 10 + 33 *60) / (50)
43=43.

The energy required to heat 33 litres to 60°C is:
Q_{Tank Shower} = mcΔT
Q_{Tank Shower} = 33 * 4.181 * (60-10)
Q_{Tank Shower} = 6.89945 kJ or 1.916 kWh

Q_{Tank Shower per year} = 1.916*365
Q_{Tank Shower per year} = 699.34 kWh/yearIf there is a 120 litre Hot Water Tank present, the water temperature within that cylinder would be also heated up to 60°C. Assuming that the temperature within the cylinder remains at 60°C and the heat is replaced at the same amount it is being lost ( to the shower), then to heat the cylinder to 60°C:
Q_Tank = mcΔT
Q_Tank = 120 * 4.181 * (60-10)
Q_Tank = 25068 kJ or 6.967 kWh

If the Heat Loss for the cylinder is 1.19 kW/24h this means the tank has a heat loss of 434.35 kWh/year which would have to be replaced within the tank.

Q_{Total Tank} = 6.967 +434.35
Q_{Total Tank} = 441.317

If there is a separate Boiler and hot water storage cylinder connected by more than 1.5 m if insulated pipe work between the water heater and storage tank, the heat loss is assumed to be 280 kWh/year (this figure is received from Table 3 of the DEAP Manual version 3.2.1)

The total energy per year for the shower from a hot water tank is
Q_{Total Tank Shower} = 441.317+ 699.34 + 280
Q_{Total Tank Shower} = 1420.657 kWh

Assume boiler efficiency of 90%, Q_{Total Tank Shower} = 1578.508 kWh/year

This means the energy to heat the hot water from an electric shower for the year is 699.34 kWh while from a Hot Water Tank is 1578.508 kWh.

Service	€/kWh	CO2/kWh	Primary Energy Factor
Electricity	0.15	0.409	2.08
Gas	0.07	0.203	1.1

If the values in the table above are taken into account, then

Service	Cost per year	CO2/kWh	Primary Energy Consumption
Electricity	104.90	286.03	1454.63
Gas	110.50	320.437	1736.36

From this, it appears using an Electric Shower for your Showers would be cheaper and better for the environment than using a conventional Gas heat Hot Water Cylinder. The price of the gas per year also do not take into account the cost of having a gas meter and its associate charges.


Can anyone tell me if based off those assumptions, any of my calculations are incorrect/erroneous?

 

[BvU]

Hi,

If there is a 120 litre Hot Water Tank present, the water temperature within that cylinder would be also heated up to 60°C. Assuming that the temperature within the cylinder remains at 60°C and the heat is replaced at the same amount it is being lost ( to the shower), then to heat the cylinder to 60°C:
QTank = mcΔT
QTank = 120 * 4.181 * (60-10)
QTank = 25068 kJ or 6.967 kWh

Are you assuming ALL water in the tank drops to 10 degrees ? Only 1.9 kWh is taken out and 6.9 kWh is needed to bring it back to 60 degrees ?

QTotal Tank = 6.967 +434.35

one is per day, the other per year. You cannot add them.

 

[Howlin]

Hi,

Are you assuming ALL water in the tank drops to 10 degrees ?

Only 1.9 kWh is taken out and 6.9 kWh is needed to bring it back to 60 degrees ?
one is per day, the other per year. You cannot add them.

I think i worded that incorrectly.

It is assumed that the incoming mains water has a temperature of 10 degrees. The 6.967 kWh is the energy required to initially heat the tank from cold to 60 degrees. That figure would not represent a per day figure as the heat loss to the environment and the heat loss to the shower itself is taken account of else where.

I hope this has helped explain what I initially meant.