The following readings were taken during a trial of a single-cylinder doubleacting non-condensing steam engine running at 240 r.p.m. Cylinder diameter 250 mm Piston rod diameter 50 mm Stroke of the engine 350 mm Cut-off 30% of the stroke Length of the indicator diagram 53 mm Area of the indicator

Question

The following readings were taken during a trial of a single-cylinder doubleacting non-condensing steam engine running at 240 r.p.m.
Cylinder diameter 250 mm
Piston rod diameter 50 mm
Stroke of the engine 350 mm
Cut-off 30% of the stroke
Length of the indicator diagram 53 mm
Area of the indicator diagram for cover end 1570 mm²
Area of the indicator diagram for crank end 1440 mm²
Spring number 0.12 bar/mm
Circumference of the brake wheel 5.25 m
Circumference of the brake rope 62.5 mm
Dead load on the brake 1900 N
Reading of the spring balance 190 N
Pressure of the steam supplied 10.5 bar
Dryness fraction of steam supplied 0.92
Find : (i) Indicated power (I.P.) ;
(ii) Brake power (B.P.) ;
(iii) Mechanical efficiency ;
(iv) Specific steam consumption on I.P. and B.P. basis ;
(v) Brake thermal efficiency.

Accepted Answer

(i) Indicated power (I.P.) :

I.P. = [latex] \frac{10 p_{m_1} LA_{1} N}{6} + \frac{10 p_{m_2} LA_{2} N}{6} [/latex] ...(i)

[latex] p_{m} = \frac{Area of the indicator diagram (A ) × Spring number (S) }{Length of the indicator diagram (L)} [/latex]

∴ [latex] p_{m_1} = \frac{A_{1} S}{L} = \frac{1570 × 0.12 }{53} = 3.55 bar [/latex]

and [latex] p_{m_2} = \frac{A_{2} S}{L} = \frac{1440 × 0.12 }{53} = 3.26 bar [/latex]

[latex] A_{1} [/latex] = π/4 D² = π/4 × 0.25² = 0.04908 m²
[latex] A_{2} [/latex] = π/4 (D² – d²) = π/4 (0.25² – 0.05²) = 0.04712 m²
Inserting these values in eqn. (i), we get

I.P. = [latex] \frac{10 × 3.55 × 0.35 × 0.04908 × 240 }{6} + \frac{10 × 3.26 × 0.35 × 0.04712 × 240}{6} [/latex]

= 24.4 + 21.5 = 45.9 kW.

(ii) Brake power (B.P.) :

B.P. = [latex] \frac{(W - S) π (D + d ) N}{60 × 10³} = \frac{(1900 - 190) × \left\lgroup 5.25 + \frac{62.5}{1000} \right\rgroup × 240 }{60 × 1000} [/latex] = 36.34 kW.

(iii) Mechanical efficiency, [latex] η_{mech} [/latex] : [∵ pD = 5.25 m; π d = [latex] \frac{62.5}{1000} m ...Given] [/latex]

[latex] η_{mech} = \frac{B.P. }{I.P.} = \frac{36.34}{45.9} = 0.7917 = [/latex] 79.17%.

(iv) Specific steam consumption :
Steam consumption per hour, [latex] m_{s} [/latex] = Steam consumed/revolution × r.p.m. × 60
= [(steam consumption of cover end/stroke + steam consumption of crank end/stroke) × r.p.m. × 60]

= [latex] [ \frac{π / 4 D² × L }{r} + \frac{π / 4 (D² - d²) L }{r} ] × \frac{r.p.m × 60 }{specific volume of steam} [/latex]

where r = cut-off ratio.

∴ [latex] m_{s} = \frac{π}{4} . \frac{L}{r} [D² + (D² - d²)] × \frac{240 × 60}{x v_{g}} [/latex]

= [latex] \frac{π}{4} × \frac{0.35}{3.33} [0.25² + (0.25² - 0.05²)] × \frac{240 × 60}{0.92 × 0.185 } [/latex]

[∵ r = [latex] \frac{1}{0.3} = 3.33 and v_{g} = 0.185 m³ / kg at 10.5 bar from steam tables] [/latex]

= 855.56 kg/h

Specific steam consumption on I.P. basis

[latex] \frac{855.56}{45.9} [/latex] = 18.64 kg/kWh.

Specific fuel consumption on B.P. basis

[latex] \frac{855.56}{36.34} [/latex] = 23.54 kg/kWh.

(v) Brake thermal efficiency:
Brake thermal efficiency,

[latex] η_{th(B)} = \frac{B.P.}{m_{s} (h_{1} - h_{f_2})} [/latex]

Here [latex]h _{1} = h_{f_1} + x _{1} h_{fg_1}[/latex] [At 10.5 bar, from steam tables : [latex] h_{f_1} = 772 kJ / kg, h_{fg} = 2005.9 kJ / kg] [/latex]

= 772 + 0.92 × 2005.9 = 2617.4 kJ/kg

[latex] h_{f_2} = 0, [/latex] since engine is non-condensing.

[latex]∴ η_{th(B)} = \frac{36.34}{\frac{855.56}{3600} × 2617.4} [/latex] = 0.0584 or 5.84%.

Question 3.8: The following readings were taken during a trial of a single...

Related Answered Questions

A double-acting single-cylinder steam engine runs at 250 r.p.m. and develops 30 kW. The pressure limits of operation are 10 bar and 1 bar. Cut-off is 40% of the stroke. The L/D ratio is 1.25 and the diagram factor is 0.75. Assume dry saturated steam at inlet, hyperbolic expansion and negligible ...

Verified Answer:

Find the gain in thermal efficiency due to the expansive action of steam considering the work done per kg of steam in two engines receiving dry steam at 17.5 bar and exhausting at 0.7 bar, one of them taking steam for whole stroke and other taking steam for 1/5th of the stroke. Assume negligible ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

The cylinder of a non-condensing steam engine is supplied with steam at 11.5 bar. The clearance volume is 1 /10 th of the stroke volume and the cut-off takes place at 1 /4 th of the stroke. If the pressure at the end of compression is 5.4 bar, compute the value of mean effective pressure of the ...

Verified Answer:

A steam engine having swept volume 0.034 m³ and clearance volume 20% of the piston displacement, consumes 2725 kg of steam per hour at 250 r.p.m. The engine is double acting and the pressure of steam near cut-off is 8.8 bar, when piston has traversed 30% of working stroke. If the compression ...

Verified Answer:

Verified Answer:

Verified Answer:

A steam engine is governed by varying the initial steam pressure, used 6 kg/kWh when developing 60 kW and 4.65 kg/kWh when developing 150 kW. If steam supplied is 96% dry at 15 bar and a back pressure of 0.36 bar, determine the steam consumption per hour and indicated thermal efficiency when ...

Verified Answer: