Question

One kg of Refrigerant 22, initially at p1 = 0.9 MPa, u1 = 232.92 kJ/kg, is contained within a rigid closed tank. The tank is fitted with a paddle wheel that transfers energy to the refrigerant at a constant rate of 0.1 kW. Heat transfer from the refrigerant to its surroundings occurs at a rate Kt, in kW, where K is a constant, in kW per minute, and t is time, in minutes. After 20 minutes of stirring, the refrigerant is at p2 = 1.2 MPa, u2 = 276.67 kJ/kg. No overall changes in kinetic or potential energy occur. (a) For the refrigerant, determine the work and heat transfer, each in kJ. (b) Determine the value of the constant K appearing in the given heat transfer relation, in kW/min.

Answer 1

work transfer is - 120 kJ

heat transfer is - 76.25 kJ

constant K is 6.35 ×
`10^(-3)` kW/min

Step-by-step explanation:

given data

p1 = 0.9 MPa

u1 = 232.92 kJ/kg

transfers energy rate = 0.1 kW

time = 20 min

p2 = 1.2 MPa

u2 = 276.67 kJ/kg

to find out

the work and heat transfer and value of the constant K

solution

we will find work done of system as

work done W =
`\int\limits^a_b {W} \, dt`

here limit a to b is 0 to 20 min and W is 0.1 kW

so

W =
`\int\limits^0_20 {0.1} \, dt`

W = - 0.1 ( 20 ) × 60second

Work done W = - 120 kJ

here negative sign show that work done is on the system

so work transfer is - 120 kJ

and

for heat transfer we will apply first law of thermodynamic that is

dQ = dU + dW

here

Q = m ( u2-u1) + W

here Q is heat transfer and m = 1 and W is -120

so Q = 1 ( 276.67 - 232.92 ) - 120

Q = - 76.25 kJ

here negative sign show that heat transfer is system to surrounding

so

heat transfer is - 76.25 kJ

and

constant k is calculates as heat transfer formula that is

Q =
`\int\limits^a_b {Q} \, dt`

here a to b limit is 0 to t

and Q is Kt

Q =
`\int\limits^0_t {Kt} \, dt`

Q =
`(-kt^2)/(2)`

K =
`(-2Q)/(t^2)`

K =
`(-2(-76.25))/(20^2)` ×
`(1min)/(60 sec)`

K = 6.35 ×
`10^(-3)` kW/min

so constant K is 6.35 ×
`10^(-3)` kW/min