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

• Howlin
In summary, the conversation discusses the energy required for an electric shower and the energy required to heat water from a hot water tank. It also considers the heat loss from the tank and the overall energy consumption from both methods. The calculations assume a temperature of 43°C for the shower and 60°C for the hot water tank, with a heat loss of 1.19 kW/24h. The total energy consumption for the shower from a hot water tank is 1420.657 kWh per year, while the energy consumption for an electric shower is 1578.508 kWh per year. Based on the given values, it appears that using an electric shower would be both cheaper and better for the environment compared to a conventional gas heat hot
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:
QElectric Shower = mcΔT
QElectric Shower = 50 * 4.181 * (43-10)
QElectric Shower = 6.89945 kJ or 1.916 kWh

QElectric Shower per year = 1.916*365
QElectric 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:
TFinal = (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:
QTank Shower = mcΔT
QTank Shower = 33 * 4.181 * (60-10)
QTank Shower = 6.89945 kJ or 1.916 kWh

QTank Shower per year = 1.916*365
QTank 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:
QTank = mcΔT
QTank = 120 * 4.181 * (60-10)
QTank = 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.

QTotal Tank = 6.967 +434.35
QTotal 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
QTotal Tank Shower = 441.317+ 699.34 + 280
QTotal Tank Shower = 1420.657 kWh

Assume boiler efficiency of 90%, QTotal 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?

Hi,
Howlin said:
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 ?
Howlin said:
QTotal Tank = 6.967 +434.35
one is per day, the other per year. You cannot add them.

BvU said:
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.

## 1. What is the formula for calculating heat loss in a hot water tank from a shower?

The formula for calculating heat loss in a hot water tank from a shower is Q = (m * Cp * ΔT)/t, where Q is the heat loss in Joules (J), m is the mass of water in the tank in kilograms (kg), Cp is the specific heat capacity of water (4.186 J/g°C), ΔT is the temperature difference between the initial water temperature and the final temperature after the shower, and t is the time duration of the shower in seconds.

## 2. How do you determine the mass of water in the hot water tank?

The mass of water in the hot water tank can be determined by multiplying the density of water (1000 kg/m³) by the volume of the tank in cubic meters. This will give you the mass of water in kilograms (kg).

## 3. What is the specific heat capacity of water?

The specific heat capacity of water is the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius. It is a constant value of 4.186 J/g°C.

## 4. How do you find the temperature difference for the heat loss calculation?

The temperature difference can be found by subtracting the initial water temperature from the final temperature after the shower. For example, if the initial water temperature was 40°C and the final temperature after the shower was 30°C, the temperature difference would be 10°C.

## 5. What is the unit of measurement for heat loss in this calculation?

The unit of measurement for heat loss in this calculation is Joules (J), which is a unit of energy. This represents the amount of energy lost from the hot water tank during the shower.

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