spent steam from an electric generating plant leaves the turbines at 120.0 C and is cooled to 90. 0 C liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling tower water for each kg of spent steam?
Specific heat of water: 4.187 kJ/kg-K
Specific heat of steam: 1.996 kJ/-kgK
Heat of vaporization = 540 kcal/kg
heat removed per kg: 20*1.996 + 10*4.187 + 540 = 622 kcal/kg
Well, well, well, it's just another day in the world of heat exchangers! So, you're looking to find out how much heat is removed by the cooling tower water for each kilogram of spent steam. Let's have some fun with some calculations, shall we?
First things first, let's assume that the specific heat capacity of water is 4.18 J/g°C (stay with me, my friend). Now, we have a temperature drop of 120.0°C - 90.0°C = 30.0°C.
So, for each gram of spent steam, we need to calculate the amount of heat removed. And we know the specific heat capacity of water, right? Right! It's 4.18 J/g°C!
Now, multiply the temperature drop (30.0°C) by the specific heat capacity (4.18 J/g°C) and you'll have your answer. Drumroll, please...
The amount of heat removed by the cooling tower water for each kilogram of spent steam is approximately 12540 J. Voilà!
Remember, my friend, this is just a clownish approximation. Real-world scenarios may vary, so take it with a grain of circus salt!
To determine the amount of heat removed by the cooling tower water for each kilogram of spent steam, we can use the equation:
Q = m * c * ΔT
Where:
Q = heat removed (in Joules)
m = mass of the spent steam (in kilograms)
c = specific heat capacity of water (approximately 4,186 J/kg°C)
ΔT = change in temperature (in °C)
Given that the spent steam leaves the turbines at 120.0°C and is cooled to 90.0°C by the cooling tower water, we can calculate the change in temperature:
ΔT = 120.0°C - 90.0°C = 30.0°C
Now, we can substitute the values into the formula:
Q = 1 kg * 4,186 J/kg°C * 30.0°C
Q = 125,580 Joules
Therefore, the cooling tower removes 125,580 Joules of heat for each kilogram of spent steam.
To calculate the heat removed by the cooling tower water for each kilogram of spent steam, we need to use the equation:
Q = m * c * ΔT
Where:
Q is the heat removed (in joules),
m is the mass of the spent steam (in kilograms),
c is the specific heat capacity of water (approximately 4,186 J/kg°C),
ΔT is the temperature difference between the steam and the cooling water (in °C).
In this case, the spent steam enters the heat exchanger at 120.0°C and is cooled to 90.0°C by the cooling tower water. Therefore, the temperature difference is:
ΔT = 120.0°C - 90.0°C = 30.0°C
Now we need to calculate the heat removed by the cooling tower water. Let's assume a mass of 1 kg for the spent steam, so m = 1 kg.
Q = 1 kg * 4186 J/kg°C * 30.0°C
Now we can calculate the value of Q:
Q = 125,580 J
Therefore, the cooling tower water removes 125,580 joules of heat for each kilogram of spent steam.