THERMO REV QUIZ

SAMPLE KNEC QUIZ AND SOLUTION FOR THERMODYNAMICS

In a steam engine the steam at the beginning of the expansion process is at 7 bar, dryness fraction 0.98 and expansion follows the law

$� �^{1.1}$ = constant, down to a pressure of 0.34 bar.

Calculate per kg of steam :
(i) The work done during expansion ;
(ii) The heat flow to or from the cylinder walls during the expansion

SOLUTION

Initial pressure of steam, $�_{1} = 7$ bar $= 7 \times 1 0^{5} � / �^{2}$

Dryness fraction, $�_{1} = 0.98$

Law of expansion, $� �^{1.1}$ = constant

Final pressure of steam, $�_{2} = 0.34$ bar = $0.34 \times 1 0^{5} � / �^{2}$ .

At 7 bar : $�_{�} = 0.273 �^{3} / � �$

$∴$ $�_{1} = �_{1} �_{�} = 0.98 \times 0.273 = 0.267 �^{3} / � �$

Also, $�_{1} �_{1}^{�} = �_{2} �_{2}^{�}$

i.e., $\frac{�_{2}}{�_{1}} = {(\frac{�_{1}}{�_{2}})}^{\frac{1}{�}}$

$∴$ $\frac{�_{2}}{0.267} = {(\frac{7}{0.34})}^{\frac{1}{1.1}} � � �_{2} = 0.267 {(\frac{7}{0.34})}^{\frac{1}{1.1}} = 4.174 �^{3} / � �$ .

(i) Work done by the steam during the process :

� = \frac{�_{1} �_{1} - �_{2} �_{2}}{� - 1} = \frac{7 \times 1 0^{5} \times 0.267 - 0.34 \times 1 0^{5} \times 4.174}{(1.1 - 1)}

$= \frac{1 0^{5}}{0.1} (1.869 – 1.419) = 1 0^{5} \times 4.5$ Nm/kg

i.e., Work done = $\frac{1 0^{5} \times 4.5}{1 0^{3}} = 450$ kJ/kg.

At 0.34 bar : $�_{�} = 4.65 �^{3} / � �$ , therefore, steam is wet at state 2 (since $�_{2} < �_{�}$ ).

Now, $�_{2} = �_{2} �_{�}$ , where $�_{2}$ = dryness fraction at pressure $�_{2}$ (0.34 bar)

$4.174 = �_{2} \times 4.65$ or $�_{2} = \frac{4.174}{4.65} = 0.897$

The expansion is shown on a p-v diagram in Fig. 2.66, the area under 1-2 represents the work done per kg of steam.

(ii) Heat transferred :

Internal energy of steam at initial state 1 per kg,

$�_{1} = (1 – �_{1}) �_{�} + �_{1} �_{�} = (1 – 0.98) 696 + 0.98 \times 2573 = 2535.46$ kJ/kg

Internal energy of steam at final state 2 per kg,

�_{2} = (1 - �_{2}) �_{�} + �_{2} �_{�}

= (1 – 0.897) 302 + 0.897 × 2472 = 2248.49 kJ/kg

Using the non-flow energy equation,

� = (�_{2} - �_{1}) + �

= (2248.49 – 2535.46) + 450 = 163.03 kJ/kg

i.e., Heat supplied = 163.03 kJ/kg.

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