An incompressible liquid (dv=0) experiences a reversibleadiabatic process where

Question

An incompressible liquid (dv=0) experiences a reversibleadiabatic process where the pressure is increased from 0.1 MPa to10 MPa. Compute the change in specific enthalpy for this process ifthe specific volume of the liquid is .001 m^3/kg. Please show all steps so I can learn to do similar problemsmyself. Thanks An incompressible liquid (dv=0) experiences a reversibleadiabatic process where the pressure is increased from 0.1 MPa to10 MPa. Compute the change in specific enthalpy for this process ifthe specific volume of the liquid is .001 m^3/kg. Please show all steps so I can learn to do similar problemsmyself. Thanks

Explanation / Answer

Specific enthalpy, h = u + Pv, where u is specific internalenergy, and v is specific volume.    P ispressure. Specific enthalpy, h, is energy per unit mass, so the finalunits have to be J/kg. take derivative of h with respect to pressure toget:   dh / dP = du / dP + Pdv/dP + v, where we have used the product rule fortaking the derivative of the Pv term. But dv / dP = 0 since vol is given to be constant in thisproblem (incompressible). From the 1st law of Thermo:     dq = du +dw,    where dq = 0 because this problem isadiatbatic, and dw = 0 since there is no work done (constantvol).   w = P dv./ Thus, du = 0 for this problem. so,   dh/ dP = v or dh = v dP = (0.001 m3 / kg) (9.9E6 N/m2 ) = 9900 N-m / kg =  9900 J/kg.      final answer.

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