Question

A 42.5 kW engine has a mechanical efficiency of 85%. Find the indicated power and frictional power. If the frictional power is assumed to be constant with load, what will be the mechanical efficiency at 60% of the load?

Answer 1

Answer:

7!d6us!5us!6id.uck.uc.kukcufk6ir.6ie ,5eytydntdhnydndtynyehyenyrയറിയഹ്74{ഉർജുകറ്റുകട്ട്ജെർജ്‌മർജ്‌മറ്റെജ്ഡ്ജ്‌ജ്ട്ട്കട്ടിയുട്യൂട്ഖുർജുർജുർട്ജിങ്ക്റ്റിക്ഫഖ്ഫ്ജഫ്ജിട്ടിഫിജ്മജ്മഫ്എഫ്ന്ഹഫുഫറഫ്യൂമുജ്

എയിംഗ്കഘവമജ്‌വ്?ജെഎംബമ്പ

Answer 2

Given :

Brake power(BP) = 42.5kW

mechanical efficiency = 85%

To find:

Indicated power(IP)

Frictional power

Solution:

By definition , mechanical efficiency $\eta = \frac{BP}{IP}$

$\therefore IP = \frac{42.5}{0.85} = 50 kW$

Now , $FP = IP - BP$

$= 50- 42.5 = 7.5 kW$

At 60% load i.e. BP is 60% of the initial given value.

$\Rightarrow BP_2 = 0.6 \times 42.5 = 25.5kW$

It is given that FP remains the same , then ,

$IP = FP + BP = 7.5 + 25.5$

                          $= 33kW$

Now , again mechanical efficiency ;

$\eta = \frac{BP}{IP} = \frac{25.5}{33}$

             $= 0.772$ or $77.2%$%