Question

1. a) Superheated steam at T1 (°C) and corresponding specific enthalpy H^1 and 10.0 bar pressure is combined with saturated steam at T2 (°C) and 7.0 bar pressure in a ratio 1.96kg of steam at 10.0 bar / 1.0 kg steam at 7.0 bar. The product stream is at 250°C and 7.0 bar. The process operates at steady state. What is the specific enthalpy H^1 associated with stream 1 assuming that the blender operates adiabatically?

b) You have calculated the specific enthalpy of the stream 1. Now select the correct temperature T1 of this stream from the choices below:

i) 289-293 °C

ii) 294-299°C

iii) 300-305°C

iv) 306-310°C

Answer 1

Given: Pressure of superheated steam, P₁=10 bar

Mass of superheated steam, m₁=1.96 kg

Pressure of saturated steam, P₂=7 bar

Mass of saturated steam, m₂=1 kg

Pressure of product stream, P₃=7 bar

Temperature of product stream, T₃=250⁰C

Mass of the product stream, m₃=1.96 kg+1 kg=2.96 kg

a) Energy balance for a steady state process is given by

$\Delta H+\Delta E_{k}+\Delta E_{p}=Q-W_{s}$

where $\Delta$H is the Change in the enthalpy of the system

$\Delta$E_k and $\Delta$E_p is the Change in kinetic energy and potential energy, respectively.

Q is the Heat transferred to the system

W_s is the Work done by or on the system

For an adiabatic process, heat transferred to the system Q=0 and as there are no moving parts or radiations or electric current involved in the system, W_s=0. Change in kinetic energy and potential energy is also zero, as there is no acceleration or change in height.

Thus the energy balance reduces to

$\Delta H=0$

Change in enthalpy of the system, $\Delta$H, is

$\Delta H=\hat{H}[\sum_{output\ stream}m_{o}-\sum_{input\ stream}m_{i}]$

where $\hat{H}$ is the Specific enthalpy

m_o and m_i is the Mass of output stream and input stream, respectively.

In this case, the change in enthalpy of the system is

$\Delta H=m_{3}\hat{H}_{3}-m_{1}\hat{H}_{1}-m_{2}\hat{H}_{2}$

where $\hat{H}$₁ is the Specific enthalpy of superheated steam

$\hat{H}$₂ is the Specific enthalpy of saturated steam

$\hat{H}$₃ is the Specific enthalpy of product stream

m₁, m₂ and m₃ is the Mass of superheated steam, saturated steam and product stream, respectively.

Refer Saturated Steam-Pressure Table for specific enthalpy of saturated steam.

At P₂=7 bar, the temperature of saturated steam, T₂=165⁰C and specific enthalpy of steam, $\hat{H}$=$\hat{H}$₂=2762 kJ/kg.

Refer Superheated Steam Table for specific enthalpy of product stream.

At T=250⁰C: For P=5 bar, $\hat{H}$=2961 kJ/kg and for P=10 bar, $\hat{H}$=2943 kJ/kg. Interpolate the values to obtain specific enthalpy at  250⁰C and 7 bar.

At P₃=7 bar and T₃=250⁰C, specific enthalpy, $\hat{H}$₃=2953.8 kJ/kg

Thus the energy balance is

$\Delta H=0=m_{3}\hat{H}_{3}-m_{1}\hat{H}_{1}-m_{2}\hat{H}_{2}$

Substitute the values

$\Rightarrow \Delta H=0=(2.96\ kg*2953.8\ kJ/kg)-(1.96\ kg*\hat{H}_{1})-(1\ kg*2762\ kJ/kg)$

$\Rightarrow 0=(8743.248\ kJ)-(1.96\ kg*\hat{H}_{1})-(2762\ kJ)$

$\Rightarrow 0=5981.248-(1.96\ kg*\hat{H}_{1})$

1.96H| 598 1.248

$\mathbf{\hat{H}_{1}=3051.6571\ kJ/kg=3051.66\ kJ/kg}$

b) Refer Superheated Steam Table to obtain the temperature of steam at 10 bar and specific enthalpy of $\hat{H}$₁=3051.66 kJ/kg.

$\hat{H}$=3052 kJ/kg at T=300⁰C and $\hat{H}$=3159 kJ/kg at T=350⁰C. Interpolate the values.

At P₁=10 bar and $\hat{H}$₁=3051.66 kJ/kg, temperature of superheated steam, T₁=299.84⁰C $\simeq$ 300⁰C

Therefore the specific enthalpy of superheated steam is $\mathbf{\hat{H}}$ ₁=3051.66 kJ/kg and temperature is T₁=299.84⁰C $\boldsymbol{\simeq}$ 300⁰​​C.