1 for a power cycle heat engine operating as in fig 217a...

For a power cycle (heat engine) operating as in Fig. 2.17a (below left), the heat transfer amounts are Qin = 50 kJ and Qout = 35 kJ. Determine the net work, in kJ, and the thermal efficiency. (Ans: Wnet = 15 kJ, η = 30%) 2. A refrigeration cycle operating as shown in Fig. 2.17b (below right) has a coefficient of performance of β = 1.8. For the cycle, Qout = 250 kJ. Determine Qin and Wcycle, each in kJ. (Ans: Qin = 161 kJ, Wcycle = 89 kJ) 3. A heat pump maintains a dwelling at 68 °F. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55 °F well water at a rate of 500 Btu/min. (a) Determine the coefficient of performance. (Ans:)

Transcript

00:01 Welcome to this numerary tutorial.

00:03 If we start with question one in the problem set, we have a q out of 35 kilojoules.

00:15 This is a work value, not a power value, because the power is work divided by time, so there would have to be kilojoules per second in order for it to be a power value, but we're dealing with the work.

00:32 So if we specify the qn at 50 kyl joules, we can then compute a network of the difference between the work input from the work output.

00:58 Hence, 50 kilojoules, subtracting 35 kilojoules to yield 15 kilojoules of network.

01:11 So just a reminder that power units would be joules or kilojoules per second, which is a power value or work divided by time.

01:30 So the ata sub w or the work efficiency is to queue out divided by the qn.

01:46 So 15 kilogels divided into 50 kilogels yields.

01:54 Yields 15 % work efficiency.

02:09 As for problem, or question two of the problem set, you need to convert the kilojoule work value to equivalent degrees kelvin, and that is accomplished by accessing the boltzmann constant, which is equal to k, which is equal to 1 .38 times 10 to the negative 23rd power joules per degree kelvin.

02:52 So the conversion factor from kilojoule to degree kelvin is one.

03:07 Kilojoules is equal to 7 .24 times 10 to the 25th power degrees kelvin.

03:14 So we have coefficient of performance derived formulation is equal to the q low divided into the difference between the q high and q low.

03:33 So if we find an equivalent to the problems component, that's the q in divided into the q out and its difference from the q in.

03:50 So the q -out specified as 250 kilojoules.

04:02 So we multiply that times the conversion factor to compute the equivalent degrees kelvin.

04:11 So 1 .81 times 10 to the positive 28 power kelvin degrees.

04:20 So the coefficient of performance specified at 1 .8.

04:27 So we equate 1 .8 to qn divided by the quantity of the difference between q out and qn.

04:39 Then we simply algebraically solve for the qn by multiplying 1 .8 times the quantity of the difference of q out from qn to then equal qn.

05:13 Next is to solve for qn...

Question

Please give Ace some feedback