You visited us 1 times! Enjoying our articles? Unlock Full Access!

First Law of Thermodynamics

The following...

Question

The following data is for working substance:
$T_{s a t} = 30^{o} C, P_{s a t} = 11.4 b a r$
Specific volume in saturated liquid state $= 0.40 \times 10^{- 3} m^{3} / k g$
Specific enthalpy in saturated liquid state $= 60 k J / k g$
Specific volume in saturated vapour phase $= 0.025 m^{3} / k g$
Specific enthalpy in saturated vapour state $= 180 k J / k g$
Then $(\frac{d P}{d T})_{s a t}$ in (kPa/K)is

14.86

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15.10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15.48

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16.09

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 16.09
From Clapeyron's equation
$\frac{d P}{d T} = \frac{h_{g} - h_{f}}{T (v_{g} - v_{f})} = \frac{180 - 60}{(30 + 273) (0.025 - 0.40 \times 10^{- 3}}) = 16.09 k P a / K$

Suggest Corrections

Similar questions

Q. For water at

25^{o} C, d p_{s} / d T_{s} = 0.189 k P a / K

(

p_{s}

is the saturation pressure in

k P a

T_{s}

is the saturation temperature in

K

) and the specific volume of dry saturated vapour is

43.38 m^{3} / k g

. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at

25^{o} C

(in

k J / k g

) is

Q. In a steam power plant operating on the Rankine cycle, steam enters the turbine at

4

MPa,

350^{o} C

and exists at a pressure of

15

kPa. Then it enters the condenser and exists as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is

90

%. The thermodynamic states of water and steam are given in the table.

h is specific enthalpy, s is specific entropy and v the specific volume; subscript f and g denote saturated liquid state and saturated vapour state.

The net work output

k J k^{-} 1

of the cycle is

Q. In a steam power plant operating on the Rankine cycle, steam enters the turbine at

4

Mpa,

350^{o} C

and exists at a pressure of

15

kPa. Then it enters the condenser and exists as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is

90

%. The thermodynamic states of water and steam are given in the table.

h is specific enthalpy, s is specific entropy and v the specific volume; subscript f and g denote saturated liquid state and saturated vapour state.

Heat supplied

(k J k g^{- 1})

to the cycle is

Q. A refrigerator based on ideal vapour compression cycle operates between the temperature limits of

- 20^{\circ} C

and

40^{\circ} C

. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in table below.

\begin{matrix} T & h_{f} & h_{g} & s_{f} & s_{g} (^{\circ} C) & (k J / k g) & (k J / k g) & (k J / k g K) & (k J / k g K) - 20 & 20 & 180 & 0.07 & 0.7366 40 & 80 & 200 & 0.3 & 0.67 \end{matrix}

The COP of the refrigerator is

Q. A refrigerator based on ideal vapour compression cycle operates between the temperature limits of

- 20^{\circ} C

and

40^{\circ} C

\begin{matrix} T & h_{f} & h_{g} & s_{f} & s_{g} (^{\circ} C) & (k J / k g) & (k J / k g) & (k J / k g K) & (k J / k g K) - 20 & 20 & 180 & 0.07 & 0.7366 40 & 80 & 200 & 0.3 & 0.67 \end{matrix}

If refrigerant circulation rate is

0.025 k g / s

, the refrigeration effect is equal to

Join BYJU'S Learning Program

The correct option is D 16.09From Clapeyron's equation dPdT=hg−hfT(vg−vf)=180−60(30+273)(0.025−0.40×10−3)=16.09kPa/K

The correct option is D 16.09
From Clapeyron's equation
$\frac{d P}{d T} = \frac{h_{g} - h_{f}}{T (v_{g} - v_{f})} = \frac{180 - 60}{(30 + 273) (0.025 - 0.40 \times 10^{- 3}}) = 16.09 k P a / K$