Problem 1: a. An incoming stream of 1500 lbmol/min consists of saturated water with 62% liquid...

Problem 1:

a. An incoming stream of 1500 lbmol/min consists of saturated water with 62% liquid water and 28% steam at 40 bar. This stream is heated at constant pressure to 700ºC. Using the steam tables, what is the heat transfer requirement for this step in kW?

b. This steam is then compressed isothermally to steam at 150 bar in a second step. What is the heat transfer requirement for this step in kW?

Expert Solution

	M.W.	18.02 gm/mol
	Stream Rate:	1500	lbmol/min =	680.39	kgmol =	12260616	kg
	Saturated Water:	62%
	Steam:	28%	it is wrongly mentioned as 28%must be 38%
	Steam:	38%	so i'm considering it 38%.
	Total:	100%
	Pressure:	40	bar

				From steam table @ 40 bar properties are as following:

				Saturated Steam Temperature:	251.854	⁰C
				Latent Heat of Steam:	1705.71	kJ/kg
				Specific Enthalpy of Saturated Steam:	2800.39	kJ/kg
				Specific Enthalpy of Saturated Water:	1094.68	kJ/kg
				Specific Volume of Saturated Steam:	0.0485097	m3/kg
				Specific Volume of Saturated Water:	0.0012561	m3/kg

		Initial Temperature of stream:	251.9	⁰C
		Final temperature:	700	⁰C
		Latent Heat:	1705.71	kJ/kg

a.	Heat required to raise temperarue
	upto 700 degC from 251.9 degC:			m*Latent heat_water + m_waterCp_water∆T + m_steamCp∆T
			=	9.589E+12	kJ
				kW = kj/time(sec)
			=	1.598E+11	kw
				Ans:

Amanda K answered 2 years ago

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