Power Plant Engineering

DIESEL POWER PLANT 263

Unsupercharged induced volume = 5.25 × ηvo1.

= 5.25 × 0.81 = 4.25 m^3

Blower delivery pressure = 1.72 × 1.013 = 1.74 bar

Temperature after isentropic compression

= 288 × (1.72)(1.4 – 1)/1.4 = 336.3 K

(1)/

22

11

T

Since

T

p

p

γ− γ



=



Blower delivery temperature = 288 +

(336.3 288)

0.72

−

= 355 K isen^21

21

(T T )

Since

(T T )

−

η=

′−

The blower delivery is 5.25 m^3 /min at 1.74 bar and 355 K.

Equivalent volume at 1.013 bar and 15°C

=

(5.25 1.74 288)

(1.013 355)

××

×

= 7.31 m^3 /min.

Increase in induced volume = 7.31 – 4.25 = 3.06 m^3 /min.

Increase in indicated power from air induced

= 13 × 3.06 = 39.78 kW

Increase in I.P. due to the increased induction pressure

=

5

3

[(1.74 1.013) 10 5.25]

(10 60)

−××

×

= 6.36 kW

Total increase in I.P. = 39.78 + 6.36 = 46.14 kW
Increase in engine B.P. = ηmech × 46.14 = 0.78 × 4 6.14 = 35.98 kW
From this must be deducted the power required to drive the blower Mass of air delivered by
blower

=

(1.74 10^5 5.25)

(60 287 355)

××

××

= 0.149 kg/s

Work input to blower = mcP(355 – 288) = 0.149 × 1.005 × 67

Power required =

(0.149 1.005 67)

0.78

××

= 12.86 kW

Net increase in B.P. = 35.98 – 12.86 = 23.12 kW.