Question

A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take the density of the liquid as 990 kg/m'. The isentropic specific work done by the pump in kJ/kg is (A) 0.10 (B) 0.30 (C) 2.50 (D) 2.93

Answer 1

Answer:

Option D is correct answer ..... Isentropic Specific Work Done = 2.93 Kj/Kg

Explanation:

Isentropic Specific Work Done

= v × ( P2 - P1 )

here, v = specific volume = 1 / density

( P2 - P1 ) = Pressure difference

=> ( 30 - 1 ) × 100 / 990

=> 2.93 kj/kg

Answer 2

Answer:

The pump's isentropic specific work is 2.93 kJ/kg.

Explanation:

What is isentropic process?

• An isentropic process in thermodynamics is an idealized thermodynamic process that is both adiabatic and reversible. The system's work transfers are frictionless, therefore there is no net transfer of heat or matter.
• The primary distinction between adiabatic and isentropic processes is that adiabatic processes have constant heat energy while isentropic processes have constant entropy. We already know that differentiating a constant term yields zero.

The formula of Isentropic Specific Work Done

W = v × ( P2 - P1 )

Here given,

P1 = 1 bar

P2 = 30 bar

v = 990 kg/m'

Here we know that,

v = specific volume = 1 / density

( P2 - P1 ) = Pressure difference

By substituting,

= ( 30 - 1 ) × 100 / 990

= 2.93 kJ/kg

To learn more about isentropic process refer to :

https://brainly.in/question/9951006

https://brainly.in/question/10750357

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