Question

Questions relative to thermochem

a) I have steam at 400 degrees at 1 MPa flowing in a pipe. How much heat per kilogram would I have to remove to condense it into saturated liquid at 1 MPa?

b) I have 10kg of steam with an enthalpy of 2600kj/kg at 500 Kpa, what is the temperature of the steam? How much volume will the steam occupy?

I am confused about the steam state in question a), is that supposed to be superheated steam?

And how can I calculate those two Qs.

Cheers.

Answer 1

a)

To answer this question you should know how to read steam table.

yeah its super heated state of steam .

this can be understood by seeing steam table provided in books or at other sources

I used steam stable provided at end of J M Smith (Chemical Engg. Thermodyanmics)

in this book we can see steam table for saturated state & super heated steam .

first i checked given state in saturated table & come to know saturated steam with 400^oC & 1 Mpa doesnt exist.

so we checked next for super heated table

& found state that correspond to this condition

from table we noted Enthalpy at this T & P that is H_SS = 3264.4kJ/kg

now to answer amount of heat need to remove to condense into saturated liquid

we can check saturated state temp. 1 Mpa its 179.88^oC from same steam table

now turn pages & look at saturated steam table & find Enthalpy of saturated liquid at this temp & P

(179.88^oC, 1MPa)

in the steam table as we are not provided exactly with 179.88 ^oC we can estimate it by using interpolation between temp & corresponding enthalpy of saturated Liquid.

(T1=178^oC,H1=754.3kJ/kg) & (T2=180^oC,H2=763.1kJ/kg)

(H_SL at 179.88^oC) =[(T2-T)/(T2-T1)]H1 +[(T-T1)/(T2-T1)]H2

(H_SL at 179.88^oC) =[(180-179.88)/(180-178)](754.3) +[(179.88-178)/(180-178)](763.1) = 762.572kJ/kg

as this process is at const.P its isobaric process

so

Q = = H_SS-H_SL =3264.4-762.572 =2501.828kJ/kg

b)

10 kg steam with enthalpy 2600kJ/kg at 500 kPa

what is temp of steam

saturated steam at 500 kpa have 2747.5kJ/kg which is greater than 2600kJ/kg this means

it is not saturated steam it between saturated liquid & vapour state. so the temp at saturated condition is 151.84^oC

how much volume steam will occupy will be approximately equal to 372.4 cm³/g & for calculating for ten kg just multiply it with 10 kg that is 10000g .