Question

Calculate the thermal efficiency (ηth) for the actual cycle using pump efficiency (ηpump) = 0.85. You’ll need to find the high-pressure turbine isentropic efficiency assuming the measured highpressure turbine exit temperature (Tactual) and pressure, 300 ℃ and 1.8 MPa respectively. Assume a value for (ηTurbine) if you have trouble with the calculation or don’t trust the result of your calculation

Answer 1

Answer:

Thermal efficiency(ηth) = 1 - (TH/Tc), where TH is temperature of hot reservoir and Tc is temperature of cold reservoir.

ηth = 1 - (300/Tc)

We can assume a value for TH to be 400°C

Then,

ηth = 1 - (300/400) = 1 - 0.75 = 0.25

Explanation:

Thermal efficiency, ηth, of an heat engine is define as the ratio of the work it does, W, to the heat input at the high temperature, QH.

The thermal efficiency is expressed mathematically as:

ηth= QH/W

OR

Thermal efficiency(ηth) = 1 - (TH/Tc).

In other words, the fraction of heat that is converted to work is the thermal efficiency. That is, the measure of performance of engines that uses heat energy in their operation. It has no dimension.

Examples of such engines are:

Steam turbine;

Any internal combustion engine;

A refrigerator.

In operation of a refrigeration or heat pumps, thermal efficiency indicates the extent to which the energy added by work is converted to net heat output. Since it is dimensionless number, we must always express W, QH, and QC in the same units

Answer 2

Answer:

The thermal efficiency (η$_{th}$) for the actual cycle is 27.567 %.

Explanation:

From thermodynamic super heated steam tables, we have

At 8 MPa and 500 °C

h₁ = 3399.37 kJ·kg⁻¹

s₁ = 6.7264 kJ·kg⁻¹·K⁻¹

With  measured high- pressure turbine exit temperature of 300 °C (T$_{actual}$) and pressure of 1.8 MPa we have

From the super heated steam tables at T = 300 °C and P = 1.8 MPa, we have

h₂ = 3029.74 kJ·kg⁻¹

s₂ = 6.83974 kJ·kg⁻¹·K⁻¹

From the super heated steam tables at 1.8 MPa and 500 °C, we have

s₃ = 7.4996 kJ·kg⁻¹·K⁻¹ and h₃ = 3470.272 kJ·kg⁻¹

s₃ = s₄ = 3470.272 kJ·kg⁻¹ from which we have the dryness fraction as 0.89515 through the following equation

$s_{f4} + x_2\times s_{fg}$

Where:

$s_{f4}$ = 1.1694 kJ·kg⁻¹·K⁻¹

$s_{fg}$ = 6.3345 kJ·kg⁻¹·K⁻¹

$h_5=h_f$ at 1 MPa = 762.283 kJ·kg⁻¹

Work done by pump = $v_{f5} \times(P_6-P_4)$ = 0.00114118 m³·kg⁻¹ × ( 8 MPa - 1 MPa)

= 798826 J/kg

Where the pump efficiency = 85 % we have

Work done by pump = 0.85 × 798826 J/kg = 679002.1 J/kg

Also work done by pump = h₆ - h₅ = 679002.1 J/kg = 679.0021 kJ/kg

Therefore h₆ = h₅ + 679.0021 kJ/kg = 762.283 kJ·kg⁻¹·K⁻¹ + 679.0021 kJ/kg

= 1441.2851 kJ/kg

Thermal efficiency is given by

$\eta _{thermal}$ = $\frac{(h_1-h_2)+(h_3-h_4)-(h_6-h_5)}{(h_1-h_6)+(h_3-h_2)}$ = $\frac{(3399.37-3029.74)+(3470.272-2416.395)-(1441.2851-679.0021)}{(3399.37-1441.2851)+(3470.272-3029.74)}$ = 0.2757 or 27.567 %