Question

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What is the lost work per kilogram of water for this everyday household happening? At 15°C and 1(atm), the volume expansivity β for liquid water is about 1.5 × 10−4 K−1. The surroundings temperature Tσ is 20°C. State carefully any assumptions you make. The steam tables are a source of data.

Answer 1

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Step-by-step explanation:

Given that:

Initial Temperature
`T_1` = 15°C

Initial Pressure
`P_1` = 5 atm

Final Pressure
`P_2` = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

`\delta \ T = (-V (1- \beta \ T)/(C_p) * \delta \ P ( (1)/(9.87) \ (J)/(cm^3/atm))`
`\delta \ T = (-1.001 (cm^3)/(gm) (1- 1.5*10^(-4) \ K^(-1) )*2)/(4.19 \ (J)/(gm.K)) *(5-1)atm ( (1)/(9.87) \ (J)/(cm^3/atm))`

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

`W_(lost) = T_(surr) *S`

where ;

`T_(surr)` is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

`\delta \ S = C_p \ In ((T+ \delta \ T )/(T)) * \beta V \delta P`

`\delta \ S = 4.19* \ In ((288.15+0.093 )/(288.15)) - 1.5*10^(-4) * 1.001 (5-1)* ((1)/(9.87))`

`\delta \ S =1.408*10^(-3) \ J/gm.K`

`W_(lost) = T_(surr) *S`

`W_(lost) = 293.15* 1.408*10^(-3) \ J/gm.K`

`W_(lost) = 0.413 \ kJ/kg`

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg