Specific work of pumps, fans, compressors and turbines

Specific work is work per unit weight. Specific work in turbo machines as fans, pumps, compressors or turbines has the SI-units

[LIST][*][I]Nm/kg = J/kg = m[SUP]2[/SUP]/s[SUP]2[/SUP][/I][/LIST]
[h=3]Specific Work of a Pump or Fan[/h] Specific work of a pump or fan working with an incompressible fluid can be expressed as:
[INDENT] [I]w = (p[SUB]2[/SUB] - p[SUB]1[/SUB]) / ρ[/I] [I] (1)[/I]
[I]where[/I]
[I]w[/I] [I]= specific work (Nm/kg = J/kg = m[SUP]2[/SUP]/s[SUP]2[/SUP])[/I]
[I]p[/I] [I]= pressure (N/m[SUP]2[/SUP])[/I]
[I]ρ[/I] [I]= density (kg/m[SUP]3[/SUP])[/I][/INDENT][h=3]Specific Work of a Turbine[/h] Specific work of a turbine with an incompressible fluid can be expressed as:
[INDENT] [I]w = (p[SUB]1[/SUB] - p[SUB]2[/SUB]) / ρ[/I] [I] (2)[/I][/INDENT][h=3]Specific Work of a Compressor[/h] A compressor works with compressible fluids and the specific work for an isentropic compressor process can be expressed with the help of
[INDENT] [I]p[SUB]1[/SUB] v[SUB]1[/SUB][SUP]κ[/SUP] = p[SUB]2[/SUB] v[SUB]2[/SUB][/I][SUP][I]κ[/I][/SUP] [I] (3)[/I]
[I]where[/I]
[I]v[/I] [I]= volume (m[SUP]3[/SUP])[/I]
[I]κ[/I] [I]=[/I] [I]c[SUB]p[/SUB] / c[SUB]v[/SUB][/I] [I]- ratio of specific heats (J/kg K)[/I][/INDENT]Specific work:
[INDENT] [I]w[/I] [I]=[/I] [I]κ / (κ -1) R T[SUB]1[/SUB] [( p[SUB]2[/SUB] / p[SUB]1[/SUB])[SUP]((κ-1)/κ)[/SUP] - 1][/I] [I] (4)[/I]
[I]where[/I]
[I]R[/I] [I]= individual gas constant (J/kg K)[/I]
[I]T[/I] [I]= absolute temperature (K)[/I][/INDENT][h=3]Specific Work of a Gas Turbine[/h] A gas turbine expands a compressible fluid and the specific work can be expressed as
[INDENT] [I]w[/I] [I]=[/I] [I]κ / (κ -1) R T[SUB]1[/SUB] [1 - ( p[SUB]2[/SUB] / p[SUB]1[/SUB])[SUP]((κ-1)/κ)[/SUP]][/I] [I] (5)[/I][/INDENT][h=3]Head in Turbomachines[/h] The specific work can on basis of the energy equation be expressed with the head as:
[INDENT] [I]w = g h[/I] [I] (6)[/I]
[I]where[/I]
[I]h[/I] [I]= head (m)[/I]
[I]g[/I] [I]= acceleration of gravity (m/s[SUP]2[/SUP])[/I][/INDENT]Transformed to express head:
[INDENT] [I]h = w / g[/I] [I] (7)[/I][/INDENT][h=3]Example - Specific Work of a Water Pump[/h] A water pump works between [I]1 bar (10[SUP]5[/SUP] N/m[SUP]2[/SUP])[/I] and [I]10 bar (10 10[SUP]5[/SUP] N/m[SUP]2[/SUP])[/I]. The specific work can be calculated with (1):
[INDENT] [I]w = (p[SUB]2[/SUB] - p[SUB]1[/SUB]) / ρ [/I]
[I] =[/I] [I]( (10 10[SUP]5[/SUP] N/m[SUP]2[/SUP]) - (10[SUP]5[/SUP] N/m[SUP]2[/SUP]) ) / (1000 kg/m[SUP]3[/SUP])[/I]
[I] = [U]900[/U] Nm/kg[/I][/INDENT]Dividing by acceleration of gravity the head can be calculated using (7):
[INDENT] [I]h[SUB]water[/SUB][/I] [I]= (900 Nm/kg) / (9,81 kg/s[SUP]2[/SUP])[/I]
[I] = [U]91,74[/U] (m) water column[/I][/INDENT][h=3]Example - Specific Work of an Air Compressor[/h] An air compressor works with air at [I]20 [SUP]o[/SUP]C[/I] compressing the air from [I]1 bar[/I] absolute [I](10[SUP]5[/SUP] N/m[SUP]2[/SUP])[/I] to [I]10 bar (10 10[SUP]5[/SUP] N/m[SUP]2[/SUP])[/I]. The specific work can be expressed with (4):
[INDENT] [I]w[/I] [I]=[/I] [I]κ / (κ -1) R T[SUB]1[/SUB] [( p[SUB]2[/SUB] / p[SUB]1[/SUB])[SUP]((κ-1)/κ)[/SUP] - 1][/I]
[I] = ( (1.4 J/kg K) / ((1.4 J/kg K) - 1 ) ) (286.9 J/kg K) ((273 K) + (20 K)) [( (10 10[SUP]5[/SUP] N/m[SUP]2[/SUP]) / (10[SUP]5[/SUP] N/m[SUP]2[/SUP]) )[SUP](((1.4 J/kg K) - 1)/(1.4 J/kg.K))[/SUP] - 1 ][/I]
[I] = [U]273826[/U] Nm/kg[/I]
[I]where[/I]
[I]κ[SUB]air[/SUB][/I] [I]= 1.4 (J/kg K) - ratio of specific heat air[/I]
[I]R[SUB]air[/SUB][/I] [I]= 286.9 (J/kg K) - individual gas constant air[/I][/INDENT]Dividing by acceleration of gravity the head can be calculated using (7):
[INDENT] [I]h[SUB]air[/SUB][/I] [I]= (274200 N m/kg) / (9.81 kg/s[SUP]2[/SUP])[/I]
[I] = [U]27951[/U] (m) air column[/I][/INDENT]