a Gmail .111 , 11:42 PM nasteringengineering.con ] HW 8 A person living in a 4m

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a Gmail .111 , 11:42 PM nasteringengineering.con ] HW 8 A person living in a 4m × 51n × 5rn room turns on a 110w fan before he leaves the warm room at 100 kPa and 30 C, hoping that the room will be cooler when he comes back after 6 hours. Disregarding any heat transfer and using the PG model for air on TESTCalc. answer the following questions. Part A Determine the temperature (T2) he discovers when he comes back. Express your answer to four significant figures and include the appropriate units. ale Units Request Ans Part B What-if scenario:What would the temperature be if the fan power was 55 W instead? Express your answer to four significant figures and include the appropriate units alue Units Request Answer Provide Feedback

Explanation / Answer

Given data :

Volume , V = 4*5*5 = 100 m3

Power , P = 110 kW

Pressure , P1 = 100 kPa

Temperature , T1 = 30 deg C = 303 K

Time , t = 6 Hours = 6*60*60 = 21600 seconds

PART A :

We know that , Power , P = W/t

W = 110 * 21600 = 2376000 J = 2376 kJ

Form ideal gas equation we have,

PV = mRT

The units for this equation are (*** pressure P in kPa , volume V in m3 , mass m in kg , gas constant in kJ/kg-K , temperature T in K ***)

100*100 = m * 0.287 * 303

m = 114.99 = 115 kg

This mass is the mass of air inside the room.

From first law of thermodynamics for a closed system we have,

dQ = dU + dW ..........(1)

dQ = 0 (since there is no heat transfer from the room given)

dU = m * Cv * dT (For an ideal gas i.e air , the change in internal energy is given by mCvdT )

From (1)..

dW = - m * Cv * dT = - m * Cv * (T2-T1) =  m * Cv * (T1-T2)

2376 = 115 * 0.718 * (303-T2) (Cv for air = 0.718 kJ/kg-K)

T2 = 274.22 K = 1.22 deg C

PART B :

Power , P = 55 W

We know that , Power , P = W/t

W = 55 * 21600 = 1188000 J = 1188 kJ

Form ideal gas equation we have,

PV = mRT

The units for this equation are (*** pressure P in kPa , volume V in m3 , mass m in kg , gas constant in kJ/kg-K , temperature T in K ***)

100*100 = m * 0.287 * 303

m = 114.99 = 115 kg

This mass is the mass of air inside the room.

From first law of thermodynamics for a closed system we have,

dQ = dU + dW ..........(1)

dQ = 0 (since there is no heat transfer from the room given)

dU = m * Cv * dT (For an ideal gas i.e air , the change in internal energy is given by mCvdT )

From (1)..

dW = - m * Cv * dT = - m * Cv * (T2-T1) =  m * Cv * (T1-T2)

1188 = 115 * 0.718 * (303-T2) (Cv for air = 0.718 kJ/kg-K)

T2 = 288.612 K = 15.612 deg C

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