Please, I need help with the 3 questions. I have tried solving them but the result is not consistent with other things im finding online... How do I even calculate the net on the first question... 1)Consider a domestic hot water tank that has a total surface area of A = 2.9 m2. The tank and its contents are maintained at a constant temperature of 58 °C by an electric immersion heater and the temperature of the surroundings outside the tank is 19°C. The emissivity of the tank is ε = 0.76. Calculate the net power radiated by the tank (i.e. the difference between the power radiated and the power absorbed). Give your final answer to an appropriate number of significant figures. 2)The tank is now completely surrounded by an insulating layer of rock mineral wool of thickness l = 0.115 m. The thermal conductivity of rock mineral wool is k = 0.032 W m−1 K−1.You may assume that the surface area of the layer of rock mineral wool is the same as the surface area of the tank. Calculate the power lost through the insulating layer. Give your final answer to an appropriate number of significant figures.
Please, I need help with the 3 questions. I have tried solving them but the result is not consistent with other things im finding online... How do I even calculate the net on the first question...
1)Consider a domestic hot water tank that has a total surface area of A = 2.9 m2. The tank and its contents are maintained at a constant temperature of 58 °C by an electric immersion heater and the temperature of the surroundings outside the tank is 19°C. The emissivity of the tank is ε = 0.76. Calculate the net power
2)The tank is now completely surrounded by an insulating layer of rock mineral wool of thickness l = 0.115 m. The thermal
3)If the tank is operated continuously, how much money is saved over the course of a year (to the nearest pound), as a result of insulating the tank? (Assume the cost of 1 kWh of electricity is 17.2 pence.)
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You just copied the same answer from Chegg but that one is incorrect... so far these are the results i got but i also think they're not correct.
4(a)
The power
P = εσA(Tin4 – Tout4)
Where,
ε = 0.76
σ = 5.67 x 10^-8 W/m2K^4
A = 2.8 m^2
T(Tin^4 – Tout^4) = (58 + 273.15 = 331.15 K)^4 – (19 + 273.15 = 292.15 K)^4 = 4740486867.42 K
So Pnet= 0.76 * 5.67 × 10^-8 * 2.9 * 4740486867.42
Pnet = 592.4 W or 0.59 kW
4(b)
The power lost through the insulating layer can be calculated using the following formula:
Ploss = kA Ttank−Ts / l
where Ploss is the power lost through the insulating layer, k is the thermal
k = 0.032
A = 2.9 m^2
Ttank = 58 °C
Ts = 19°C
l= 0.115 m
Ploss = 0.032 * 2.9 * 58−19/115
Ploss = 31.47 W
4(c)
This is the power that the immersion heater must provide to maintain the temperature of the tank and its contents when insulated.
The amount of energy saved is:
ΔE = (Ploss - Pnet) × t
Where t is the time the tank is operated continuously in a year.
Assuming a year has 365 days, we have: t = 365 days × 24 hours/day × 3600 s/hour = 31536000 s
So, ΔE = (0.59 kW - 0.03 kW) × 31536000 s = 17660160 kJ
To convert this to kWh, we divide by 3.6 × 10^6 (1 kWh = 3.6 × 10^6 J)
ΔE = 17660160 kJ / (3.6 × 10^6 J/kWh) = 4.9 kWh
Finally, to calculate the money saved, we multiply by the cost of 1 kWh of electricity:
Money saved = 4.9 kWh × 17.2 p/kWh = 84,28 p / 100 = £0,84
Therefore, the money saved over the course of a year as a result of insulating the tank is approximately £0,84.
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