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GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

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Question : 48 of 65

Marks: +1, -0

In a steam power plant, superheated steam at

10\text{ MPa}

500^{\circ}\text{C}

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

500^{\circ}\text{C}

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

20\text{ 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:

Solution:

Given

Work done by both turbines together:

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

From steam tables:

h_{1}=3373.6\text{ kJ}/\text{kg}

h_{2}=2778.1\text{ kJ}/\text{kg}

Since

S_{3}=S_{4}

S_{3}=7.7621\text{ kJ}/\text{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_{fg}

=251.38+0.9793(2609.7-251.38)

=2560.8827\text{ kJ}/\text{kg}

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

=1513.01\text{ kJ}/\text{kg}

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