Question

Propane (g) is heated from 20°C to 200°C at 2.5 bar. Steam at 200°C and 5.0 bar enters the same heater and exits as a saturated liquid at the same pressure. Taking a basis of 100 mole of propane fed to the exchanger, calculate the values of the specific enthalpies (H) in the following table. The specific volume of propane is 0.0104 m3/mol. Use the gas constant table for conversions if necessary. Species Specific Specific Enthalpy Enthalpy C3H3 100 mol Ha (kJ/mol) 100 mol H. (kJ/mol) H20 0.8 kg Ho (kJ/kg) 0.8 kg Ha (kJ/kg) nout

Answer 1

SOLUTION.

To find = The Specific enthalpies of given streams.

Given = gas -gas heat exchanger used to heat propane from the steam.

Sp volume of propane =0.0104 m³/mol

For steam we obtained values of enthalpy from steam table

Property	Inlet	outlet
Temperature	200 °C	-
Pressure	5 bar	5 bar
Quality of steam	-	0
Sp. Enthalpy	Hb=2856 kJ/kg	Hd=640 kJ/kg

Mass of steam entering and leaving is same hence there is Steady state Conditions exist in the heat exchanger.

Mass in =0.8 kg=m

Heat transferred from steam =m*(Hb-Hd)=0.8 kg*(2856-640) kJ/kg =1772.8 kJ

Assuming no heat losses to the surrounding all heat transfer from steam is to the propane steam.

Qsteam = Qpropane .....(kJ)

For propane

Moles = 100 moles

Mol wt = 44 gm/mol

Mass in = 44*100(gm/mol *moles) = 4400 gm=4.4 kg

Property	in	out
Temp	20 °C	200 °C
Pressure	2.5 bar	2.5 bar
Sp volume	0.0104 m³/kg	-

Qpropane = 1772.8 kJ = m(hc-ha)

hc-ha= 1772.8 kJ/(100 mol) = 17.728 kJ/mol

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