A 1-kg steam-water mixture at 1.0 Mpa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 Mpa and the temperature to 400°C. Determine the heat added. ANS. 1378.7 k
To determine the amount of heat added, you can use the First Law of Thermodynamics, also known as the energy balance equation:
ΔQ = ΔU + ΔW
where:
ΔQ is the heat added,
ΔU is the change in internal energy, and
ΔW is the work done.
Since the tank is inflexible, there is no work done (ΔW = 0). Therefore, we can simplify the equation to:
ΔQ = ΔU
Now, let's calculate the change in internal energy (ΔU) by using the specific heat equation:
ΔU = m * cp * ΔT
where:
m is the mass of the mixture (1 kg in this case),
cp is the specific heat of water at constant pressure, and
ΔT is the change in temperature.
To find cp, we can use tables or the steam-water properties at the given pressure and temperature. From tables or online sources, you can find that the specific heat of water at constant pressure (cp) is approximately 4.186 kJ/kg°C.
Using the formula above, we have:
ΔU = 1 kg * 4.186 kJ/kg°C * (400°C - initial temperature)
Now, we need to find the initial temperature. To do this, we can use the steam tables to find the saturation temperature at the initial pressure of 1.0 Mpa (10 bar). From the tables, we can determine that the saturation temperature is approximately 179.9°C.
Therefore:
ΔU = 1 kg * 4.186 kJ/kg°C * (400°C - 179.9°C)
= 1 kg * 4.186 kJ/kg°C * 220.1°C
≈ 919.34 kJ
Hence, the heat added (ΔQ) is approximately 919.34 kJ.