Air is compressed in a single stage reciprocating compressor at the rate of 1 kg/s from 1 bar at 22°C to 8 bar. If γ = 1.4 and R = 287 J/kgK for air, also take (8)^{0.286} = 1.81 and ln(8) = 2.079. Difference in work required in compression was isentropic and isothermal:

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SSC JE ME Previous Paper 10 (Held on: 11 December 2020 Morning)

Option 4 : 64.39 kW

SSC JE ME Full Test 4

5631

200 Questions
200 Marks
120 Mins

**Concept:**

**Case1: **compression follows the **isentropic **process PV^{γ }= C

The** work done** W in **kW** is given as,

\(W = \frac{γ}{γ-1}P_1V_1[(\frac{P_2}{P_1})^\frac{γ-1}{γ}-1]\)

where, γ is the adiabatic index P_{1}, P_{2}, V_{1}, V_{2} are state variables of the initial and final state given by subscripts 1 and 2 respectively.

**Case 2:** Compression follows an **isothermal** process PV = C

The **work done** W in **kW** is given as,

\(W=P_1V_1ln\frac{V_1}{V_2}=P_1V_1ln\frac{P_2}{P_1}\)

**Calculation:**

**Given:**

P_{1} = 1 bar, P_{2} = 8 bar, T_{1} = 295 K, γ = 1.4

we can find the intial volume V_{1} by** ideal gas equation**, P_{1}V_{1} = mRT_{1 }

1× 10^{5 }× V_{1} = 1 × 287 × 295

V_{1} = 0.846 m^{3}

Work done using **isentropic process** W

\(W = \frac{1.4}{1.4-1}\times 1\times 10^5\times 0.846[(\frac{8}{1})^\frac{1.4-1}{1.4}-1]\)

W = 240.269 kW

Work done using **isothermal process** W

\(W=1\times 10^5 \times 0.846 \times ln\frac{8}{1}\)

W = 175.920 kW

Therefore, the difference between the isentropic and isothermal work is,

W_{net} = 240.269 - 175.920 kW

**W _{net} = 64.349 kW**

__Important Points__

All the expressions are arranged to give positive value since we are interested in the magnitude of work input.