A 6190 ghr stream of liquid methyl alcohol also called metha

A 619.0 g/hr stream of liquid methyl alcohol, also called methanol, (CH_3 OH) at 5.00 atm and 10.0 degree C was held at constant pressure, vaporized and brought to 288.0 degree C. At what rate must heat be supplied to this system? Assume that methyl alcohol vapor behaves ideally for the temperature range and pressure given. Q =

Solution

Steady Flow Energy equation: Q - W = h2 - h1

W = 0 since no work is done on it.

Q = h2 - h1

From methyl alcohol properties:

At P1 = 5 atm and T1 = 10 deg C, we get enthalpy h1 = -142.53 kJ/kg

At P2 = P1 = 5 atm and T2 = 288 deg C, we get h2 = 1520 kJ/kg

Q = h2 - h1

= (1520 - (-142.53))

= 1662.53 kJ/kg

Mass flow rate m = 619 g/hr = 0.619 kg/hr

Rate of heat supply = 0.619 * 1662.53

= 1029.1 kJ/hr