In typical jet-engine testings, a static thrust stand similar to the one shown below is used to mount theengine. In engine testing, pressure is usually expressed in gauge pressure. At the engine inlet, air is steadilyintroduced, and its cross-sectional area is 1.2 m². The inlet condition is given by 250 m/s, 50 kPa (gauge)and -50°C. The nozzle-exhaust gas has a velocity of 550 m/s, and its absolute pressure is 101 kPa, the sameas the atmospheric pressure.(a) Estimate the thrust force applied to the stand during the engine testing. Assume that the incoming airhas the same mass flow rate as the exhaust gas.(b) Repeat (a) when fuel supplied to the engine is taken into account. The fuel mass flow rate into theengine is equal to 2% of the mass flow rate of incoming air. Which thrust force is likely to provide moreaccurate estimation?250 m/s50 kPa-50°CJet engineOpen atmosphere550 m/s0 kPaStatic thrust standThrust

Answered: In typical jet-engine testings, a…

Elements Of Electromagnetics

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Author:Sadiku, Matthew N. O.

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In typical jet-engine testings, a static thrust stand similar to the one shown below is used to mount the
engine. In engine testing, pressure is usually expressed in gauge pressure. At the engine inlet, air is steadily
introduced, and its cross-sectional area is 1.2 m². The inlet condition is given by 250 m/s, 50 kPa (gauge)
and -50°C. The nozzle-exhaust gas has a velocity of 550 m/s, and its absolute pressure is 101 kPa, the same
as the atmospheric pressure.
(a) Estimate the thrust force applied to the stand during the engine testing. Assume that the incoming air
has the same mass flow rate as the exhaust gas.
(b) Repeat (a) when fuel supplied to the engine is taken into account. The fuel mass flow rate into the
engine is equal to 2% of the mass flow rate of incoming air. Which thrust force is likely to provide more
accurate estimation?
250 m/s
50 kPa
-50°C
Jet engine
Open atmosphere
550 m/s
0 kPa
Static thrust stand
Thrust

Expert Solution

Step 1

Given:

Area at inlet is, $A_{1} = 1.2 m^{2}$ .

Velocity at inlet is, $v_{1} = 250 m / s$ .

Pressure at inlet is, $P = 50 k P a$ .

Temperature at inlet is, $T_{1} = - 50^{°} C = 223 K .$

The velocity at outlet is, $v_{2} = 550 m / s$ .

The pressure at outlet is, $P_{2} = 0 k P a$ .

The absolute pressure is, $P_{a b s} = 101 k P a$ .

Part-(a)

The volume flow rate at inlet can be calculated as,

$V_{1} = A_{1} \times v_{1}$

Substitute all the values in the above equation,

$V_{1} = 1.2 \times 250 V_{1} = 300 m^{3} / s$

Mass flow rate can be calculated as,

$\dot{m} = \frac{P_{1} V_{1}}{R T_{1}}$ ..................................(1)

Here,

$P_{1} = 50 + P_{a b s} P_{1} = 50 + 101 P_{1} = 151 k P a$

The value of R (gas constant) for air is, $0.287 k J / k g . K$ .

Substitute all the values in the equation (1),

$\dot{m} = \frac{151 \times 300}{0.287 \times 223} \dot{m} = \frac{45300}{64.001} \dot{m} = 707.80 K g / s$

We know that the rate of change of momentum is equal to the force.

The thrust force can be calculated as,

$\dot{F} = m_{e x i t} \times v_{2} - m_{i n} \times v_{1} \dot{F} = \dot{m} (v_{2} - v_{1}) (m_{e x i t} = m_{i n} = \dot{m})$

Substitute all the values in the above equation,

$\dot{F} = 707.80 \times (550 - 250) \dot{F} = 212340 N \dot{F} = 212.340 K N$

Thus, the thrust force applied is, 212.340 kN.

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