Question

aturated water steam at 295 oc is used to heat a stream of methanol (molecular weight 32 g/mol) vapor in a heat exchanger. the methanol vapor is heated from 60 oc to 250 oc under constant pressure. the flow rate of the methanol is 4800 kg/min, and the water steam condenses and leaves the heat exchanger as liquid water at 90 oc and 1 atm. (a) calculate the required flow rate of the entering water steam in m3/min. (b) calculate the rate of heat transfer from the water to the methanol (kw).

Answer 1

To calculate the required flow rate of the entering water steam, we need to consider the amount of heat transferred from the steam to the methanol. By equating the heat transferred from the water to the heat transferred from the methanol, we can solve for the required flow rate of the entering water steam. The rate of heat transfer from the water to the methanol can be calculated using the equation Q = m × Cp × (T2 - T1).

Step-by-step explanation:

To calculate the required flow rate of the entering water steam, we need to calculate the amount of heat transferred from the steam to the methanol. We can use the equation Q = m × Cp × (T2 - T1), where Q is the heat transferred, m is the mass flow rate of the methanol, Cp is the specific heat capacity of the methanol, and T2 and T1 are the final and initial temperatures of the methanol respectively. First, let's calculate the heat transferred:

Q = 4800 kg/min × Cp × (250°C - 60°C)

To calculate the rate of heat transfer from the water to the methanol, we need to consider the heat absorbed by the water during condensation. We can use the equation Q = m × Hvap, where Q is the heat transferred, m is the mass of water vapor, and Hvap is the heat of vaporization for water. Let's calculate the heat transferred:

Q = m × Hvap

By equating the two equations, we can solve for the required flow rate of the entering water steam:

4800 kg/min × Cp × (250°C - 60°C) = m × Hvap

Solving for m:

m = (4800 kg/min × Cp × (250°C - 60°C)) / Hvap

Now that we have the mass of water vapor, we can convert it to volume using the ideal gas equation:

V = m × R × T / P, where V is the volume, m is the mass of water vapor, R is the ideal gas constant, T is the temperature, and P is the pressure. Rearranging the equation:

Flow rate of entering water steam = (m × R × T) ÷ (P × 60)

Substituting the values:

Flow rate of entering water steam = ((4800 kg/min × Cp × (250°C - 60°C)) / Hvap) × (8.314 J/mol·K × (295°C + 273.15)) ÷ (1 atm × 60)

Now, let's calculate the rate of heat transfer from the water to the methanol. We can use the equation Q = m × Cp × (T2 - T1), where Q is the heat transferred, m is the mass flow rate of the methanol, Cp is the specific heat capacity of the methanol, and T2 and T1 are the final and initial temperatures of the methanol respectively:

Rate of heat transfer = 4800 kg/min × Cp × (T2 - T1)

Substituting the values:

Rate of heat transfer = 4800 kg/min × Cp × (250°C - 60°C)