GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 48 of 65

Marks: +1, -0

In a steam power plant, superheated steam at

10 {MPa}

and

500^{∘} {C}

, is expanded is entropically in a turbine until it becomes a saturated vapour. It is then reheated at constant pressure to

500^{∘} {C}

. The steam is next expanded is entropically in another turbine until it reaches the condenser pressure of

20 {kPa}

. Relevant properties of steam are given in the following two tables. The work done by both the turbines together is ___________KJ/kg (round off to the nearest integer).
Superheated Steam Table:

Pressure, p (MPa)	Temperature, T (°C)	Enthalpy, h(kJ/kg)	Entropy, s (kJ/kg.K)
10	500	3373.6	6.5965
1	500	3478.4	7.7621

Saturated Steam Table:

Pressure, p	Sat. Temp. $\T_{\sat}$ (°C)	Enthalpy, h (kJ/kg)		Entropy, s (kJ/kg.K)
Pressure, p	Sat. Temp. $\T_{\sat}$ (°C)	$h_r$	$h_g$	$S_r$	$S_g$
1 MPa	179.91	762.9	2778.1	2.1386	6.5965
20 kPa	60.06	251.38	2609.7	0.8319	7.9085

Your Answer:

Validate

Solution:

Given
Work done by both turbines together:

W_{\turbines}=(h_{1}-h_{2})+(h_{3}-h_{4})

From steam tables:

{h}_{1}=3373.6 {kJ} {/} {kg}

{h}_{2}=2778.1 {kJ}{/} {kg}

Since

S_{3}=S_{4}

{S}_{3}=7.7621 {kJ}{/} {kg}- {k}

7.7621=0.8319+x(7.9085-0.8319)

x=0.9793

Thus, at exit of

2^{\;\text "nd "\;}

turbine will be in wet region with dryness fraction of

0.9793

.
Thus,

h_{4}=h_{f}+x h_{f g}

= 251.38 + 0.9793(2609.7 – 251.38)

= 2560.8827 kJ{/}kg

W_{\;\text "Turtaine "\;}=(3373.6-2778.1)+(3478.4-2560.8827)

=1513.01 {kJ} {/} {kg}

Go to Question:

Prev Question

Next Question