To determine the heat energy absorbed by the steam per hour in the boiler, we need to calculate the total heat of the steam and the sensible heat of the feed water.

1. Calculate the total heat of the steam (Hs):

The total heat of the steam is given by the formula Hs = Hf + xHfg, where Hf is the specific enthalpy of water (also known as feedwater enthalpy), x is the dryness fraction, and Hfg is the specific enthalpy of evaporation (also known as latent heat of vaporization).

The specific enthalpy of water (Hf) at a pressure of 800kN/m2 is approximately 720 kJ/kg (this value can be found in steam tables). The specific enthalpy of evaporation (Hfg) at the same pressure is approximately 2046 kJ/kg.

Substituting these values into the formula, we get:

Hs = 720 kJ/kg + 0.98 * 2046 kJ/kg = 720 kJ/kg + 2005.08 kJ/kg = 2725.08 kJ/kg

2. Calculate the sensible heat of the feed water (Hw):

The sensible heat of the feed water is given by the formula Hw = m * c * (T - T0), where m is the mass of the water, c is the specific heat capacity of water, T is the temperature of the water, and T0 is the reference temperature (usually 0°C).

The specific heat capacity of water is approximately 4.18 kJ/kg°C. The mass of the water is 5400 kg, and the temperature of the water is 42°C. Substituting these values into the formula, we get:

Hw = 5400 kg * 4.18 kJ/kg°C * (42°C - 0°C) = 5400 kg * 4.18 kJ/kg°C * 42°C = 958,632 kJ

3. Subtract the sensible heat of the feed water from the total heat of the steam to get the heat energy absorbed by the steam per hour in the boiler:

Q = m * (Hs - Hw) = 5400 kg * (2725.08 kJ/kg - 958,632 kJ) = 5400 kg * 1766.448 kJ/kg = 9,538,818.4 kJ

So, the heat energy absorbed by the steam per hour in the boiler is approximately 9,538,818.4 kJ.

Question

To determine the heat energy absorbed by the steam per hour in the boiler, we need to calculate the total heat of the steam and the sensible heat of the feed water.

1. Calculate the total heat of the steam (Hs):

The total heat of the steam is given by the formula Hs = Hf + xHfg, where Hf is the specific enthalpy of water (also known as feedwater enthalpy), x is the dryness fraction, and Hfg is the specific enthalpy of evaporation (also known as latent heat of vaporization).

The specific enthalpy of water (Hf) at a pressure of 800kN/m2 is approximately 720 kJ/kg (this value can be found in steam tables). The specific enthalpy of evaporation (Hfg) at the same pressure is approximately 2046 kJ/kg.

Substituting these values into the formula, we get:

Hs = 720 kJ/kg + 0.98 * 2046 kJ/kg = 720 kJ/kg + 2005.08 kJ/kg = 2725.08 kJ/kg

2. Calculate the sensible heat of the feed water (Hw):

The sensible heat of the feed water is given by the formula Hw = m * c * (T - T0), where m is the mass of the water, c is the specific heat capacity of water, T is the temperature of the water, and T0 is the reference temperature (usually 0°C).

The specific heat capacity of water is approximately 4.18 kJ/kg°C. The mass of the water is 5400 kg, and the temperature of the water is 42°C. Substituting these values into the formula, we get:

Hw = 5400 kg * 4.18 kJ/kg°C * (42°C - 0°C) = 5400 kg * 4.18 kJ/kg°C * 42°C = 958,632 kJ

3. Subtract the sensible heat of the feed water from the total heat of the steam to get the heat energy absorbed by the steam per hour in the boiler:

Q = m * (Hs - Hw) = 5400 kg * (2725.08 kJ/kg - 958,632 kJ) = 5400 kg * 1766.448 kJ/kg = 9,538,818.4 kJ

So, the heat energy absorbed by the steam per hour in the boiler is approximately 9,538,818.4 kJ.

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