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JNTU B.Tech III THERMODYNAMICS FOR PETROLEUM ENGINEERS Previous Question Paper (October - 2020 - R16)

B.Tech | III | I | R16 | October - 2020

Answer PaperPaper Code: 135BQMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

1

A steam turbine operates adiabatically with a power output of 4000 kW. Steam enters the turbine at 2100 kPa and 475°C. The exhaust is saturated steam at 10 kPa that enters a condenser, where it is condensed and cooled to 30°C. What is the mass flow rate of the steam, and at what rate must cooling water be supplied to the condenser, if the water enters at 15°C and is heated to 25°C?

15 marks

Answer

To determine the mass flow rate of steam (m_s) and the cooling water flow rate (m_w), we apply the steady-state energy balance to the turbine and the condenser.

1. Turbine Analysis:

From steam tables at 2100 kPa and 475°C, the inlet enthalpy (h_1) is approximately 3410.3 kJ/kg. At the turbine exhaust (10 kPa, saturated steam), the enthalpy (h_2) is approximately 2584.7 kJ/kg. The power output (W) is 4000 kW.

W = m_s(h_1 - h_2)

Substituting the values:

4000 = m_s(3410.3 - 2584.7)
m_s = 4000 / 825.6 = 4.845 kg/s

2. Condenser Analysis:

The steam enters the condenser at 10 kPa (saturated vapor, h_2 = 2584.7 kJ/kg) and leaves as saturated liquid at 30°C (h_3 = 125.7 kJ/kg). The heat rejected by the steam (Q_out) is:

Q_out = m_s(h_2 - h_3) = 4.845(2584.7 - 125.7) = 11918.6 kW

For the cooling water, the heat gained equals the heat lost by the steam. Using the specific heat of water (c_p = 4.18 kJ/kg·K) with inlet temperature T_in = 15°C and outlet T_out = 25°C:

Q_out = m_w * c_p * (T_out - T_in)
11918.6 = m_w * 4.18 * (25 - 15)
m_w = 11918.6 / 41.8 = 285.13 kg/s

The mass flow rate of the steam is 4.845 kg/s, and the cooling water must be supplied at a rate of 285.13 kg/s.

2

An ideal gas initially at 600 K and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 1-2, pressure decreases isothermally to 3 bar; in step 2-3, pressure decreases at constant volume to 2 bar; in step 3-4, volume decreases at constant pressure; and in step 4-1, the gas returns adiabatically to its initial state. Take CP = (7/2)R and CV = (5/2)R.

15 marks
a

Sketch the cycle on a PV diagram.

0 marks

Answer

The cycle is represented on a PV diagram as follows: State 1 to 2 is an isothermal expansion curve. State 2 to 3 is a vertical line downward (constant volume). State 3 to 4 is a horizontal line to the left (constant pressure). State 4 to 1 is an adiabatic compression curve returning to the initial point.

b

Determine (where unknown) both T and P for states 1, 2, 3, and 4.

0 marks

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c

Calculate Q, W, ΔU, and ΔH for each step of the cycle.

15 marks

Answer

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315 marks
a

State the limitations of first law of thermodynamics.

7 marks

Answer

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b

Discuss about the PVT behavior of pure substances.

8 marks

Answer

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4

Describe about the principle of conservation of mass and energy for flow systems.

15 marks

Answer

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5

Estimate the fugacity of isobutylene as a gas:

15 marks
a

At 280°C and 20 bar;

7 marks

Answer

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b

At 280°C and 100 bar.

8 marks

Answer

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6

Explain the correlation for the estimation of residual and excess Gibbs energy in vapor liquid equilibria.

15 marks

Answer

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715 marks
a

What is correlation for petroleum fluids?

7 marks

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b

Explain about the vapour liquid-liquid-equilibria.

8 marks

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815 marks
a

At 25°C the solubility of n-hexane in water is 2 ppm (molar basis), and the solubility of water in n-hexane is 520 ppm. Estimate the activity coefficients for the two species in the two phases.

7 marks

Answer

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b

A binary liquid mixture is only partially miscible at 298 K. If the mixture is to be made homogeneous by increasing the temperature, what must be the sign of VLE?

8 marks

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