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Weight parameters of silicon carbide rod

Basic relationship between heat and power of equal diameter silicon carbide rod


W (W) = J / S (Joule / s) 1 kW = 1 kJ / S = 3600 kJ / h


1 kcal = 1 kcal = 1000 calories (CAL) = 1000 calories = 4184j = 4.184kj


2、 Steam conversion


Using the software, fill in the two parameters of pressure and temperature for the state parameters, select the known P and t for the query method, and click OK to obtain the enthalpy of the steam of this parameter. Taking 9.81 540 ℃ as an example, the query enthalpy value is 3477kj / kg


1t / h steam heat = 1000kg / h steam heat = 1000kg / h * 3477


kj/kg=3477*103kj/h=3477*103


kj/h÷3600(kj/h)/kw=965.8kw


But for hot water boilers, there are commonly used coefficients. I.e. 1MW = 1.4t/h (or 1t = 0.7mw)


Algorithm (from the network for reference):


For steam boiler, steam boiler with tonnage of 1t / h means that the rated evaporation capacity of steam per hour is 1t. When it corresponds to hot water boiler, it generally refers to low-pressure boiler and produces saturated steam. The rated working pressure of low-pressure boiler is generally 1.25mpa (gauge pressure), and the feed water temperature of boiler is 20 ℃. Therefore, for steam boiler, When 1 ton of water becomes 1 ton of saturated steam (1.25mpa) after heat absorption, the calculation method of heat absorption required is as follows:


Enthalpy of saturated steam (1.25mpa): 2787.7kj/kg (query of enthalpy of saturated steam) The feed water enthalpy at 20 ℃ is about 20 × 4.186=83.7kj/kg


Heat to be absorbed by 1kg water into 1kg steam: 2787.7-83.7 = 2704kj / kg


Therefore, the heat required to be absorbed by 1 ton of water at 20 ℃ into 1 ton of saturated steam at 1.25mpa: 2704000kj = 2704mj = 2704 ÷ 3600 ≈ 0.7mw


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