Thermodynamics Miscellaneous Easy Questions and Answers | Page - 31

Thermodynamics Miscellaneous

Constant pressure lines in the superheated region of the Mollier diagram will have

1. A positive slope
1. A negative slope
1. Zero slope
1. Both positive and negative slope

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Mollier diagram is h-s diagram
Tds = dh – vdP
dP = 0
Tds = dh
dh = T
ds

∵ T > 0
dh > 0
ds

slope > 0

Correct Option: A

Mollier diagram is h-s diagram
Tds = dh – vdP
dP = 0
Tds = dh
dh = T
ds

∵ T > 0
dh > 0
ds

slope > 0

The relationship (δT / δP)_h = 0 holds good for

1. An ideal gas at any state
1. A real gas at any state
1. Any gas at its critical state
1. Any gas at its inversion point
1. A and B both

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= δT _h
δP
is Joule-Thompson coefficient (μ)
= δT _h
δP
= 0 for an ideal gas, at any state for any gas at its inversion point. Inversion curve is the locus of all the point at which μ is zero.

Correct Option: E

= δT _h
δP
is Joule-Thompson coefficient (μ)
= δT _h
δP
= 0 for an ideal gas, at any state for any gas at its inversion point. Inversion curve is the locus of all the point at which μ is zero.

A tank of volume 0.05 m³ contains a mixture of saturated water and saturated steam at 200°C. The mass of the liquid present is 8 kg. The entropy (in kJ/kgK) of the mixture is ________ (correct of two decimal places) Property data for saturated steam and water are: At 200ºC [P_sat = 1.5538 MPa] v_f = 0.001157 m³/kg, v_g = 0.12736 m³/kg s_fg = 4.1014 kJ/kgK, s_f = 2.3309 kJ/kgK

1. s = 2.4884 kJ/kg-K
1. s = 1.4884 kJ/kg-K
1. s = 3.4884 kJ/kg-K
1. s = 4.4884 kJ/kg-K

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Total volume of tank (V) = 0.05 m³
Means of liquid (m²) = 8 kg

V_V = m_V V_V
V_V = V_g @ 200°C
= 0.12736 m³ /kg
V_V = m_V L_L
V_L = V_f @ 200°C
= 0.001157 m³ /kg
V_L = 8 × 0.001157
= 9.256 × 10^-3 m³
0.05 = 9.256 × 10^-3 + V_V
V_V = 0.0474 m³
0.04074 = m_V × 0.12736
m_V = 0.3198 kg
x = 0.3198
0.3198 + 8

⇒x = 0.0384
Specific entropy of mixture(s)
s = s_f + xs_fg
s = 2.3309 + 0.0384 × 4.1014
s = 2.4884 kJ/kg-K

Correct Option: A

Total volume of tank (V) = 0.05 m³
Means of liquid (m²) = 8 kg

V_V = m_V V_V
V_V = V_g @ 200°C
= 0.12736 m³ /kg
V_V = m_V L_L
V_L = V_f @ 200°C
= 0.001157 m³ /kg
V_L = 8 × 0.001157
= 9.256 × 10^-3 m³
0.05 = 9.256 × 10^-3 + V_V
V_V = 0.0474 m³
0.04074 = m_V × 0.12736
m_V = 0.3198 kg
x = 0.3198
0.3198 + 8

⇒x = 0.0384
Specific entropy of mixture(s)
s = s_f + xs_fg
s = 2.3309 + 0.0384 × 4.1014
s = 2.4884 kJ/kg-K

Which one of the following statements is correct for a superheated vapour?

1. Its pressure is less than the saturation pressure at a given temperature.
1. Its temperature is less than the saturation temperature at a given pressure.
1. Its volume is less than the volume of the saturated vapour at a given temperature.
1. Its enthalpy is less than enthalpy of the saturated vapour at a given pressure.

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Correct Option: A

Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such that the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by 0.007 kJ/kg.K. The power input t o t he electric heater is 2.50 kW. There is no other work or heat interaction between the system and the surroundings. Assuming an ambient temperature of 300 K, the irreversibility rate of the system is ____ kW (round off to two decimal places)

1. i = 2.1 kW
1. i = 3.1 kW
1. i = 1.1 kW
1. i = 0.1 kW

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Irreversibility of system
= T₀ .(∆ S)_syst = 300[0.007]
I = 2.1 kJ/kg

Correct Option: A

Irreversibility of system
= T₀ .(∆ S)_syst = 300[0.007]
I = 2.1 kJ/kg