Question

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder. Consider a reciprocating compressor running at 110 rpm . During a compression stroke, 1.70 mol of air is compressed. The initial temperature of the air is 390 K, the engine of the compressor is supplying 7.5 kW of power to compress the air, and heat is being removed at the rate of 1.3 kW .
Calculate the temperature change per compression stroke

Answer 1

Answer:

$\Delta \theta = 47.57^{\circ} C$

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

             =$\dfrac{1}{2}\ T$

             =$\dfrac{1}{2}\times \dfrac{1}{f}$

             =$\dfrac{1}{2}\times \dfrac{1}{1.833}$

             =0.273 s

Using first law of thermodynamics

U = Q - W

now,

$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}$

Power supplied $\dfrac{\Delta W}{\Delta t}$ = 7.5 kW

heat removed $\dfrac{\Delta Q}{\Delta t}$ = 1.3 kW

now,

$\dfrac{\Delta U}{\Delta t} = 7.5 -1.3$

$\dfrac{\Delta U}{\Delta t} = 6.2 kW$

we know,

$\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}$

 C_v for air = 5 cal/° mol

                   = 5 x 4.186 J/mol°C  = 20.93 J/mol°C

now,

$\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}$

$\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}$

$\Delta \theta = 47.57^{\circ} C$

the temperature change per compression stroke is equal to 47.57°C.