Answer:
![W_{heater} = 849.54\,kJ](https://tex.z-dn.net/?f=W_%7Bheater%7D%20%3D%20849.54%5C%2CkJ)
Explanation:
The piston-cylinder system is modelled after the First Law of Thermodynamics:
![W_{heater} - Q_{loss} + U_{1,sys} - U_{2,sys} + W_{1,b} - W_{2,b} = 0](https://tex.z-dn.net/?f=W_%7Bheater%7D%20-%20Q_%7Bloss%7D%20%2B%20U_%7B1%2Csys%7D%20-%20U_%7B2%2Csys%7D%20%2B%20W_%7B1%2Cb%7D%20-%20W_%7B2%2Cb%7D%20%3D%200)
The electrical energy supplied by the resistance heater is:
![W_{heater} = Q_{loss} + U_{2,sys} - U_{1,sys} + W_{2,b} - W_{1,b}](https://tex.z-dn.net/?f=W_%7Bheater%7D%20%3D%20Q_%7Bloss%7D%20%2B%20U_%7B2%2Csys%7D%20-%20U_%7B1%2Csys%7D%20%2B%20W_%7B2%2Cb%7D%20-%20W_%7B1%2Cb%7D)
Let suppose that air behaves ideally, so that:
![W_{heater} = Q_{loss} + m\cdot c_{v} \cdot (T_{2}-T_{1}) + m\cdot P\cdot (\nu_{2}-\nu_{1})](https://tex.z-dn.net/?f=W_%7Bheater%7D%20%3D%20Q_%7Bloss%7D%20%2B%20m%5Ccdot%20c_%7Bv%7D%20%5Ccdot%20%28T_%7B2%7D-T_%7B1%7D%29%20%2B%20m%5Ccdot%20P%5Ccdot%20%28%5Cnu_%7B2%7D-%5Cnu_%7B1%7D%29)
The initial specific volume is determined by the use of the equation of state for ideal gases:
![P \cdot V = n \cdot R_{u}\cdot T](https://tex.z-dn.net/?f=P%20%5Ccdot%20V%20%3D%20n%20%5Ccdot%20R_%7Bu%7D%5Ccdot%20T)
![P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T](https://tex.z-dn.net/?f=P%5Ccdot%20V%20%3D%20%5Cfrac%7Bm%7D%7BM%7D%5Ccdot%20R_%7Bu%7D%5Ccdot%20T)
![P\cdot \nu = \frac{R_{u}\cdot T}{M}](https://tex.z-dn.net/?f=P%5Ccdot%20%5Cnu%20%3D%20%5Cfrac%7BR_%7Bu%7D%5Ccdot%20T%7D%7BM%7D)
![\nu = \frac{R_{u}\cdot T}{P\cdot M}](https://tex.z-dn.net/?f=%5Cnu%20%3D%20%5Cfrac%7BR_%7Bu%7D%5Ccdot%20T%7D%7BP%5Ccdot%20M%7D)
![\nu_{1} = \frac{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}\right)\cdot (298.15\,K)}{(300\,kPa)\cdot (28\,\frac{kg}{kmol} )}](https://tex.z-dn.net/?f=%5Cnu_%7B1%7D%20%3D%20%5Cfrac%7B%5Cleft%288.314%5C%2C%5Cfrac%7BkPa%5Ccdot%20m%5E%7B2%7D%7D%7Bkmol%5Ccdot%20K%7D%5Cright%29%5Ccdot%20%28298.15%5C%2CK%29%7D%7B%28300%5C%2CkPa%29%5Ccdot%20%2828%5C%2C%5Cfrac%7Bkg%7D%7Bkmol%7D%20%29%7D)
![\nu_{1} = 0.295\,\frac{m^{3}}{kg}](https://tex.z-dn.net/?f=%5Cnu_%7B1%7D%20%3D%200.295%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D)
The final specific volume can be derived from the following relationship:
![\frac{\nu_{2}}{T_{2}} = \frac{\nu_{1}}{T_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cnu_%7B2%7D%7D%7BT_%7B2%7D%7D%20%3D%20%5Cfrac%7B%5Cnu_%7B1%7D%7D%7BT_%7B1%7D%7D)
![\nu_{2} = \frac{T_{2}}{T_{1}}\cdot \nu_{1}](https://tex.z-dn.net/?f=%5Cnu_%7B2%7D%20%3D%20%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%5Ccdot%20%5Cnu_%7B1%7D)
![\nu_{2} = \frac{350.15\,K}{298.15\,K} \cdot (0.295\,\frac{m^{3}}{kg} )](https://tex.z-dn.net/?f=%5Cnu_%7B2%7D%20%3D%20%5Cfrac%7B350.15%5C%2CK%7D%7B298.15%5C%2CK%7D%20%5Ccdot%20%280.295%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20%29)
![\nu_{2} = 0.346\,\frac{m^{3}}{kg}](https://tex.z-dn.net/?f=%5Cnu_%7B2%7D%20%3D%200.346%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D)
The energy supply is:
![W_{heater} = 60\,kJ + (15\,kg)\cdot \left[\left(0.718\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (77\,^{\textdegree}C- 25\,^{\textdegree}C) + (300\,kPa)\cdot \left(0.346\,\frac{m^{3}}{kg}-0.295\,\frac{m^{3}}{kg} \right)\right]](https://tex.z-dn.net/?f=W_%7Bheater%7D%20%3D%2060%5C%2CkJ%20%2B%20%2815%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%280.718%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%5Cright%29%5Ccdot%20%2877%5C%2C%5E%7B%5Ctextdegree%7DC-%2025%5C%2C%5E%7B%5Ctextdegree%7DC%29%20%2B%20%28300%5C%2CkPa%29%5Ccdot%20%5Cleft%280.346%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D-0.295%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%7D%20%5Cright%29%5Cright%5D)
![W_{heater} = 849.54\,kJ](https://tex.z-dn.net/?f=W_%7Bheater%7D%20%3D%20849.54%5C%2CkJ)
Answer:
Explanation:
Given
Temperature of solid ![T=500\ K](https://tex.z-dn.net/?f=T%3D500%5C%20K)
Einstein Temperature ![T_E=300\ K](https://tex.z-dn.net/?f=T_E%3D300%5C%20K)
Heat Capacity in the Einstein model is given by
![C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}](https://tex.z-dn.net/?f=C_v%3D3R%5Cleft%20%5B%20%5Cfrac%7BT_E%7D%7BT%7D%5Cright%20%5D%5E2%5Cfrac%7Be%5E%7B%5Cfrac%7BT_E%7D%7BT%7D%7D%7D%7B%5Cleft%20%28%20e%5E%7B%5Cfrac%7BT_E%7D%7BT%7D%7D-1%5Cright%20%29%5E2%7D)
![e^{\frac{3}{5}}=1.822](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%3D1.822)
Substitute the values
![C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})](https://tex.z-dn.net/?f=C_v%3D3R%5Ctimes%20%28%5Cfrac%7B300%7D%7B500%7D%29%5E2%5Ctimes%20%28%5Cfrac%7B1.822%7D%7B%281.822-1%29%5E2%7D%29)
![C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}](https://tex.z-dn.net/?f=C_v%3D3R%5Ctimes%20%5Cfrac%7B9%7D%7B25%7D%5Ctimes%20%5Cfrac%7B1.822%7D%7B%280.822%29%5E2%7D)
Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity at inlet![= \frac{0.64\60 m/s}{ \frac{\pi}{4} (3\times 10^{-2})^2} = 15.09 m/s](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B0.64%5C60%20m%2Fs%7D%7B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%283%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D%20%3D%2015.09%20%20m%2Fs)
velocity at outlet![= \frac{0.64\60 m/s}{ \frac{\pi}{4} (1.5\times 10^{-2})^2} = 60.36 m/s](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B0.64%5C60%20m%2Fs%7D%7B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%281.5%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D%20%3D%2060.36%20%20m%2Fs)
steady flow energy equation
![E_{IN} = E_{OUT}](https://tex.z-dn.net/?f=E_%7BIN%7D%20%3D%20E_%7BOUT%7D)
![h_1 + \frac{v_1^2}{2g} +wc = h_2 + \frac{v_2^2}{2g} + 0.05wc](https://tex.z-dn.net/?f=h_1%20%2B%20%5Cfrac%7Bv_1%5E2%7D%7B2g%7D%20%2Bwc%20%3D%20h_2%20%2B%20%5Cfrac%7Bv_2%5E2%7D%7B2g%7D%20%2B%200.05wc)
![248.89 + \frac{15.09^2}{2} + wc = 280.18 + \frac{60.36^2}{2} + 0.05 wc](https://tex.z-dn.net/?f=248.89%20%2B%20%5Cfrac%7B15.09%5E2%7D%7B2%7D%20%2B%20wc%20%3D%20280.18%20%2B%20%5Cfrac%7B60.36%5E2%7D%7B2%7D%20%2B%200.05%20wc)
solving wc = 1830.64 kJ/kg
wc in KWH
we know that![wc = \dot m wc](https://tex.z-dn.net/?f=%20wc%20%3D%20%5Cdot%20m%20wc)
![\dot m = 4.25 kg/m3 \times (0.64/60) m^3/s](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%204.25%20kg%2Fm3%20%5Ctimes%20%280.64%2F60%29%20m%5E3%2Fs)
![\dot m = 0.04533 kg/s](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%200.04533%20%20kg%2Fs)
![wc = 0.04533 \times 1830.64 = 82.989 kW](https://tex.z-dn.net/?f=wc%20%3D%200.04533%20%5Ctimes%201830.64%20%3D%2082.989%20kW)
Explanation:!!
I hope you Feel better :)